https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems/Problem_5&feed=atom&action=history2005 AIME I Problems/Problem 5 - Revision history2024-03-29T09:54:38ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems/Problem_5&diff=190403&oldid=prevNiudashu: /* Intuition */2023-03-01T05:51:07Z<p><span dir="auto"><span class="autocomment">Intuition</span></span></p>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>~th1nq3r</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>~th1nq3r</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;"></del></div></td><td colspan="2"> </td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">~Arcticturn (minor formatting issues)</del></div></td><td colspan="2"> </td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td></tr>
</table>Niudashuhttps://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems/Problem_5&diff=173340&oldid=prevSiddhartha.basu: /* Solution 3 */2022-04-13T00:21:05Z<p><span dir="auto"><span class="autocomment">Solution 3</span></span></p>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Solution 3 ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Solution 3 ==</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>First, we can break this problem up into two parts: the amount of ways to order the coins based on color, and which side is facing up for each coin (assuming they are indistinguishable). <del class="diffchange diffchange-inline">I’m </del>the end, we multiply these values together.  </div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>First, we can break this problem up into two parts: the amount of ways to order the coins based on color, and which side is facing up for each coin (assuming they are indistinguishable). <ins class="diffchange diffchange-inline">In </ins>the end, we multiply these values together.  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>First, there are obviously <math>\binom{8}{4}</math> ways to order the coins based on color.  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>First, there are obviously <math>\binom{8}{4}</math> ways to order the coins based on color.  </div></td></tr>
</table>Siddhartha.basuhttps://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems/Problem_5&diff=170829&oldid=prevAwesomelife math: /* Solution 3 */2022-02-01T16:29:27Z<p><span dir="auto"><span class="autocomment">Solution 3</span></span></p>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>First, we can break this problem up into two parts: the amount of ways to order the coins based on color, and which side is facing up for each coin (assuming they are indistinguishable). I’m the end, we multiply these values together.  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>First, we can break this problem up into two parts: the amount of ways to order the coins based on color, and which side is facing up for each coin (assuming they are indistinguishable). I’m the end, we multiply these values together.  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>First, there are obviously <math>\binom{8}{<del class="diffchange diffchange-inline">5</del>}</math> ways to order the coins based on color.  </div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>First, there are obviously <math>\binom{8}{<ins class="diffchange diffchange-inline">4</ins>}</math> ways to order the coins based on color.  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Next, we set up a recurrence. Let <math>a_n</math> be the number of ways <math>n</math> indistinguishable coins to be stacked such that no heads are facing each other.  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Next, we set up a recurrence. Let <math>a_n</math> be the number of ways <math>n</math> indistinguishable coins to be stacked such that no heads are facing each other.  </div></td></tr>
</table>Awesomelife mathhttps://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems/Problem_5&diff=169153&oldid=prevFuzimiao2013: /* Note: */2022-01-02T23:36:15Z<p><span dir="auto"><span class="autocomment">Note:</span></span></p>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>~superagh</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>~superagh</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>== <del class="diffchange diffchange-inline">Note: </del>==</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>== <ins class="diffchange diffchange-inline">Intuition </ins>==</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Really try not to overthink especially in combinatorics. In this case, we just have to notice that after we place a coin facing up ANYWHERE, the rest of the coins to the right of the row of coins MUST be facing up. (Or else we have a coin facing up followed by a coin facing down, and that is not cool).  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Really try not to overthink especially in combinatorics. In this case, we just have to notice that after we place a coin facing up ANYWHERE, the rest of the coins to the right of the row of coins MUST be facing up. (Or else we have a coin facing up followed by a coin facing down, and that is not cool).  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>So for example, suppose we started with a coin facing up. Then the next coin must also be facing up. Then so must the next coin. And so on. Now we know what happens if it starts with a coin facing up. Hmmmmmmm. What if it starts with a coin facing down?  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>So for example, suppose we started with a coin facing up. Then the next coin must also be facing up. Then so must the next coin. And so on. Now we know what happens if it starts with a coin facing up. Hmmmmmmm. What if it starts with a coin facing down?  </div></td></tr>
</table>Fuzimiao2013https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems/Problem_5&diff=167007&oldid=prevTh1nq3r: /* Note: */2021-11-29T17:21:00Z<p><span dir="auto"><span class="autocomment">Note:</span></span></p>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>So we can have the following configurations: <math>9U, 1D+8U, 2D+7U, 3D+6U, ..., 9D+0U</math>. These are the total number of configurations, which is <math>9</math>. (As a sidenote, we really count the number of solutions to <math>a+b=8</math>, which is <math>8+1=9</math>). So we must now put the 4 gold coins as a part of this sequence of ups and downs. This can be done in <math>8\choose 4</math> ways, and the rest of the silver coins can be put in <math>8-4</math>choose <math>4</math> ways. (We subtract the <math>4</math> from the <math>8</math> for the silver coin placement, because we already used up <math>4</math> coins, namely the gold ones). So we have the answer as <math>9 \binom {8}{4}=630</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>So we can have the following configurations: <math>9U, 1D+8U, 2D+7U, 3D+6U, ..., 9D+0U</math>. These are the total number of configurations, which is <math>9</math>. (As a sidenote, we really count the number of solutions to <math>a+b=8</math>, which is <math>8+1=9</math>). So we must now put the 4 gold coins as a part of this sequence of ups and downs. This can be done in <math>8\choose 4</math> ways, and the rest of the silver coins can be put in <math>8-4</math>choose <math>4</math> ways. (We subtract the <math>4</math> from the <math>8</math> for the silver coin placement, because we already used up <math>4</math> coins, namely the gold ones). So we have the answer as <math>9 \binom {8}{4}=630</math>.</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">~th1nq3r</ins></div></td></tr>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>~Arcticturn (minor formatting issues)</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>~Arcticturn (minor formatting issues)</div></td></tr>
</table>Th1nq3rhttps://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems/Problem_5&diff=164017&oldid=prevArcticturn: /* Note: */2021-10-24T01:25:56Z<p><span dir="auto"><span class="autocomment">Note:</span></span></p>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Well, start with a coin facing down. Now, does the next coin HAVE to have a coin facing down, or can it have a coin facing up? It could have both! But wait! What happens if the next coin is heads? Well we saw already that now the next coin must be heads, or else we have a up coin followed by a down coin, which once again, sucks. So now everything to the right of the up coin must be facing up.  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Well, start with a coin facing down. Now, does the next coin HAVE to have a coin facing down, or can it have a coin facing up? It could have both! But wait! What happens if the next coin is heads? Well we saw already that now the next coin must be heads, or else we have a up coin followed by a down coin, which once again, sucks. So now everything to the right of the up coin must be facing up.  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>So we can have the following configurations: <math>9U, 1D+8U, 2D+7U, 3D+6U, ..., 9D+0U</math>. These are the total number of configurations, which is <math>9</math>. (As a sidenote, we really count the number of solutions to <math>a+b=8</math>, which is <math>8+1=9</math>). So we must now put the 4 gold coins as a part of this sequence of ups and downs. This can be done in <math>8\choose 4</math> ways, and the rest of the silver coins can be put in <math>8-4</math>choose <math>4</math> ways. (We subtract the <math>4</math> from the <math>8</math> for the silver coin placement, because we already used up <math>4</math> coins, namely the gold ones). So we have the answer as <math>9 <del class="diffchange diffchange-inline">8</del>\<del class="diffchange diffchange-inline">choose 4 </del>8<del class="diffchange diffchange-inline">-4\choose </del>4=630</math>.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>So we can have the following configurations: <math>9U, 1D+8U, 2D+7U, 3D+6U, ..., 9D+0U</math>. These are the total number of configurations, which is <math>9</math>. (As a sidenote, we really count the number of solutions to <math>a+b=8</math>, which is <math>8+1=9</math>). So we must now put the 4 gold coins as a part of this sequence of ups and downs. This can be done in <math>8\choose 4</math> ways, and the rest of the silver coins can be put in <math>8-4</math>choose <math>4</math> ways. (We subtract the <math>4</math> from the <math>8</math> for the silver coin placement, because we already used up <math>4</math> coins, namely the gold ones). So we have the answer as <math>9 \<ins class="diffchange diffchange-inline">binom {</ins>8<ins class="diffchange diffchange-inline">}{</ins>4<ins class="diffchange diffchange-inline">}</ins>=630</math>.</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">~Arcticturn (minor formatting issues)</ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td></tr>
</table>Arcticturnhttps://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems/Problem_5&diff=164016&oldid=prevArcticturn: /* Note: */2021-10-24T01:25:11Z<p><span dir="auto"><span class="autocomment">Note:</span></span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
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<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 01:25, 24 October 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l31" >Line 31:</td>
<td colspan="2" class="diff-lineno">Line 31:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Well, start with a coin facing down. Now, does the next coin HAVE to have a coin facing down, or can it have a coin facing up? It could have both! But wait! What happens if the next coin is heads? Well we saw already that now the next coin must be heads, or else we have a up coin followed by a down coin, which once again, sucks. So now everything to the right of the up coin must be facing up.  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Well, start with a coin facing down. Now, does the next coin HAVE to have a coin facing down, or can it have a coin facing up? It could have both! But wait! What happens if the next coin is heads? Well we saw already that now the next coin must be heads, or else we have a up coin followed by a down coin, which once again, sucks. So now everything to the right of the up coin must be facing up.  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>So we can have the following configurations: <math>9U, 1D+8U, 2D+7U, 3D+6U, ..., 9D+0U</math>. These are the total number of configurations, which is <math>9</math>. (As a sidenote, we really count the number of solutions to <math>a+b=8</math>, which is <math>8+1=9</math>). So we must now put the 4 gold coins as a part of this sequence of ups and downs. This can be done in <math>8\choose 4</math> ways, and the rest of the silver coins can be put in <math><del class="diffchange diffchange-inline">\</del>8-4 choose 4</math> ways. (We subtract the <math>4</math> from the <math>8</math> for the silver coin placement, because we already used up <math>4</math> coins, namely the gold ones). So we have the answer as <math>9 8\choose 4 8-4\choose 4=630</math>.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>So we can have the following configurations: <math>9U, 1D+8U, 2D+7U, 3D+6U, ..., 9D+0U</math>. These are the total number of configurations, which is <math>9</math>. (As a sidenote, we really count the number of solutions to <math>a+b=8</math>, which is <math>8+1=9</math>). So we must now put the 4 gold coins as a part of this sequence of ups and downs. This can be done in <math>8\choose 4</math> ways, and the rest of the silver coins can be put in <math>8-4<ins class="diffchange diffchange-inline"></math></ins>choose <ins class="diffchange diffchange-inline"><math></ins>4</math> ways. (We subtract the <math>4</math> from the <math>8</math> for the silver coin placement, because we already used up <math>4</math> coins, namely the gold ones). So we have the answer as <math>9 8\choose 4 8-4\choose 4=630</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td></tr>
</table>Arcticturnhttps://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems/Problem_5&diff=157504&oldid=prevTh1nq3r: /* Note: */2021-07-07T20:40:02Z<p><span dir="auto"><span class="autocomment">Note:</span></span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
<col class="diff-marker" />
<col class="diff-content" />
<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 20:40, 7 July 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l31" >Line 31:</td>
<td colspan="2" class="diff-lineno">Line 31:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Well, start with a coin facing down. Now, does the next coin HAVE to have a coin facing down, or can it have a coin facing up? It could have both! But wait! What happens if the next coin is heads? Well we saw already that now the next coin must be heads, or else we have a up coin followed by a down coin, which once again, sucks. So now everything to the right of the up coin must be facing up.  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Well, start with a coin facing down. Now, does the next coin HAVE to have a coin facing down, or can it have a coin facing up? It could have both! But wait! What happens if the next coin is heads? Well we saw already that now the next coin must be heads, or else we have a up coin followed by a down coin, which once again, sucks. So now everything to the right of the up coin must be facing up.  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>So we can have the following configurations: <math>9U, 1D+8U, 2D+7U, 3D+6U, ..., 9D+0U</math>. These are the total number of configurations, which is <math>9</math>. (As a sidenote, we really count the number of solutions to <math>a+b=8</math>, which is <math>8+1=9</math>). So we must now put the 4 gold coins as a part of this sequence of ups and downs. This can be done in <math>8\choose 4</math> ways, and the rest of the silver coins can be put in <math>\<del class="diffchange diffchange-inline">{</del>8-4<del class="diffchange diffchange-inline">} </del>choose 4</math> ways. (We subtract the <math>4</math> from the <math>8</math> for the silver coin placement, because we already used up <math>4</math> coins, namely the gold ones). So we have the answer as <math>9 8\choose 4 <del class="diffchange diffchange-inline">{</del>8-4<del class="diffchange diffchange-inline">}</del>\choose 4=630</math>.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>So we can have the following configurations: <math>9U, 1D+8U, 2D+7U, 3D+6U, ..., 9D+0U</math>. These are the total number of configurations, which is <math>9</math>. (As a sidenote, we really count the number of solutions to <math>a+b=8</math>, which is <math>8+1=9</math>). So we must now put the 4 gold coins as a part of this sequence of ups and downs. This can be done in <math>8\choose 4</math> ways, and the rest of the silver coins can be put in <math>\8-4 choose 4</math> ways. (We subtract the <math>4</math> from the <math>8</math> for the silver coin placement, because we already used up <math>4</math> coins, namely the gold ones). So we have the answer as <math>9 8\choose 4 8-4\choose 4=630</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td></tr>
</table>Th1nq3rhttps://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems/Problem_5&diff=157503&oldid=prevTh1nq3r: /* Note: */2021-07-07T20:39:29Z<p><span dir="auto"><span class="autocomment">Note:</span></span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
<col class="diff-marker" />
<col class="diff-content" />
<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 20:39, 7 July 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l31" >Line 31:</td>
<td colspan="2" class="diff-lineno">Line 31:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Well, start with a coin facing down. Now, does the next coin HAVE to have a coin facing down, or can it have a coin facing up? It could have both! But wait! What happens if the next coin is heads? Well we saw already that now the next coin must be heads, or else we have a up coin followed by a down coin, which once again, sucks. So now everything to the right of the up coin must be facing up.  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Well, start with a coin facing down. Now, does the next coin HAVE to have a coin facing down, or can it have a coin facing up? It could have both! But wait! What happens if the next coin is heads? Well we saw already that now the next coin must be heads, or else we have a up coin followed by a down coin, which once again, sucks. So now everything to the right of the up coin must be facing up.  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>So we can have the following configurations: <math>9U, 1D+8U, 2D+7U, 3D+6U, ..., 9D+0U</math>. These are the total number of configurations, which is <math>9</math>. (As a sidenote, we really count the number of solutions to <math>a+b=8</math>, which is <math>8+1=9</math>). So we must now put the 4 gold coins as a part of this sequence of ups and downs. This can be done in <math>\choose<del class="diffchange diffchange-inline">{8}{</del>4<del class="diffchange diffchange-inline">}</del></math> ways, and the rest of the silver coins can be put in <math>\<del class="diffchange diffchange-inline">choose</del>{8-4}<del class="diffchange diffchange-inline">{</del>4<del class="diffchange diffchange-inline">}</del></math> ways. (We subtract the <math>4</math> from the <math>8</math> for the silver coin placement, because we already used up <math>4</math> coins, namely the gold ones). So we have the answer as <math>9\choose{8<del class="diffchange diffchange-inline">}{</del>4}\choose<del class="diffchange diffchange-inline">{8-</del>4<del class="diffchange diffchange-inline">}{4}</del>=630</math>.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>So we can have the following configurations: <math>9U, 1D+8U, 2D+7U, 3D+6U, ..., 9D+0U</math>. These are the total number of configurations, which is <math>9</math>. (As a sidenote, we really count the number of solutions to <math>a+b=8</math>, which is <math>8+1=9</math>). So we must now put the 4 gold coins as a part of this sequence of ups and downs. This can be done in <math><ins class="diffchange diffchange-inline">8</ins>\choose 4</math> ways, and the rest of the silver coins can be put in <math>\{8-4} <ins class="diffchange diffchange-inline">choose </ins>4</math> ways. (We subtract the <math>4</math> from the <math>8</math> for the silver coin placement, because we already used up <math>4</math> coins, namely the gold ones). So we have the answer as <math>9 <ins class="diffchange diffchange-inline">8</ins>\choose <ins class="diffchange diffchange-inline">4 </ins>{8<ins class="diffchange diffchange-inline">-</ins>4}\choose 4=630</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td></tr>
</table>Th1nq3rhttps://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems/Problem_5&diff=157502&oldid=prevTh1nq3r: /* Note: */2021-07-07T20:37:27Z<p><span dir="auto"><span class="autocomment">Note:</span></span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
<col class="diff-marker" />
<col class="diff-content" />
<col class="diff-marker" />
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 20:37, 7 July 2021</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Well, start with a coin facing down. Now, does the next coin HAVE to have a coin facing down, or can it have a coin facing up? It could have both! But wait! What happens if the next coin is heads? Well we saw already that now the next coin must be heads, or else we have a up coin followed by a down coin, which once again, sucks. So now everything to the right of the up coin must be facing up.  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Well, start with a coin facing down. Now, does the next coin HAVE to have a coin facing down, or can it have a coin facing up? It could have both! But wait! What happens if the next coin is heads? Well we saw already that now the next coin must be heads, or else we have a up coin followed by a down coin, which once again, sucks. So now everything to the right of the up coin must be facing up.  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>So we can have the following configurations: <math>9U, 1D+8U, 2D+7U, 3D+6U, ..., 9D+0U</math>. These are the total number of configurations, which is <math>9</math>. (As a sidenote, we really count the number of solutions to <math>a+b=8</math>, which is <math>8+1=9</math>). So we must now put the 4 gold coins as a part of this sequence of ups and downs. This can be done in <math>\choose{8}{4}</math> ways, and the rest of the silver coins can be put in <math>\choose{8-4}{4}</math> ways. (We subtract the <math>4</math> from the <math>8</math> for the silver coin placement, because we already used up <math>4</math> coins, namely the gold ones). So we have the answer as <math>9\choose{8}{4} \choose{8-4}{4}=630</math>.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>So we can have the following configurations: <math>9U, 1D+8U, 2D+7U, 3D+6U, ..., 9D+0U</math>. These are the total number of configurations, which is <math>9</math>. (As a sidenote, we really count the number of solutions to <math>a+b=8</math>, which is <math>8+1=9</math>). So we must now put the 4 gold coins as a part of this sequence of ups and downs. This can be done in <math>\choose{8}{4}</math> ways, and the rest of the silver coins can be put in <math>\choose{8-4}{4}</math> ways. (We subtract the <math>4</math> from the <math>8</math> for the silver coin placement, because we already used up <math>4</math> coins, namely the gold ones). So we have the answer as <math>9\choose{8}{4}\choose{8-4}{4}=630</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td></tr>
</table>Th1nq3r