https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Crazyboulder&feedformat=atom AoPS Wiki - User contributions [en] 2021-04-14T12:08:51Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_13&diff=142846 2017 AMC 10B Problems/Problem 13 2021-01-20T03:45:20Z <p>Crazyboulder: </p> <hr /> <div>==Problem==<br /> There are &lt;math&gt;20&lt;/math&gt; students participating in an after-school program offering classes in yoga, bridge, and painting. Each student must take at least one of these three classes, but may take two or all three. There are &lt;math&gt;10&lt;/math&gt; students taking yoga, &lt;math&gt;13&lt;/math&gt; taking bridge, and &lt;math&gt;9&lt;/math&gt; taking painting. There are &lt;math&gt;9&lt;/math&gt; students taking at least two classes. How many students are taking all three classes?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> By PIE (Property of Inclusion/Exclusion), we have<br /> <br /> &lt;math&gt;|A_1 \cup A_2 \cup A_3| = \sum |A_i| - \sum |A_i \cap A_j| + |A_1 \cap A_2 \cap A_3|.&lt;/math&gt;<br /> Number of people in at least two sets is &lt;math&gt;\sum |A_i \cap A_j| - 2|A_1 \cap A_2 \cap A_3| = 9.&lt;/math&gt;<br /> So, &lt;math&gt;20 = (10 + 13 + 9) - (9 + 2x) + x,&lt;/math&gt; which gives &lt;math&gt;x = \boxed{\textbf{(C) } 3}.&lt;/math&gt;<br /> <br /> ==Solution 2 (Subtraction)==<br /> The total number of classes taken is &lt;math&gt;10 + 13 + 9 = 32&lt;/math&gt;. Each student is taking at least one class so let's subtract the &lt;math&gt;20&lt;/math&gt; classes ( &lt;math&gt;1&lt;/math&gt; per each of the &lt;math&gt;20&lt;/math&gt; students) from &lt;math&gt;32&lt;/math&gt; classes to get &lt;math&gt;12&lt;/math&gt;. &lt;math&gt;12&lt;/math&gt; classes is the total number of extra classes taken by the students who take &lt;math&gt;2&lt;/math&gt; or &lt;math&gt;3&lt;/math&gt; classes. Since we know that there are &lt;math&gt;9&lt;/math&gt; students taking at least &lt;math&gt;2&lt;/math&gt; classes, there must be &lt;math&gt;12 - 9 = \boxed{\textbf{(C) } 3}&lt;/math&gt; students that are taking all &lt;math&gt;3&lt;/math&gt; classes.<br /> <br /> ==Solution 3 (Algebra)==<br /> Assume there are &lt;math&gt;a&lt;/math&gt; students taking one class, &lt;math&gt;b&lt;/math&gt; students taking two classes, ad &lt;math&gt;c&lt;/math&gt; students taking three classes. Because there are &lt;math&gt;20&lt;/math&gt; students total, &lt;math&gt;a+b+c = 20&lt;/math&gt;. Because each student taking two classes is counted twice, and each student taking three classes is counted thrice in the total class count, &lt;math&gt;a+2b+3c = 10+13+9 = 32&lt;/math&gt;. There are &lt;math&gt;9&lt;/math&gt; students taking two or three classes, so &lt;math&gt;b+c = 9&lt;/math&gt;. Solving this system of equations gives us &lt;math&gt;c=\boxed{\textbf{(C) 3}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=B|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Crazyboulder https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_19&diff=141327 2013 AMC 12B Problems/Problem 19 2021-01-02T18:31:46Z <p>Crazyboulder: /* Solution 1 */ Added clarification</p> <hr /> <div>{{duplicate|[[2013 AMC 12B Problems|2013 AMC 12B #19]] and [[2013 AMC 10B Problems|2013 AMC 10B #23]]}}<br /> <br /> ==Problem==<br /> <br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;AB=13&lt;/math&gt;, &lt;math&gt;BC=14&lt;/math&gt;, and &lt;math&gt;CA=15&lt;/math&gt;. Distinct points &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt; lie on segments &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{CA}&lt;/math&gt;, and &lt;math&gt;\overline{DE}&lt;/math&gt;, respectively, such that &lt;math&gt;\overline{AD}\perp\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{DE}\perp\overline{AC}&lt;/math&gt;, and &lt;math&gt;\overline{AF}\perp\overline{BF}&lt;/math&gt;. The length of segment &lt;math&gt;\overline{DF}&lt;/math&gt; can be written as &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;m+n&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 30&lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution 1==<br /> Since &lt;math&gt;\angle{AFB}=\angle{ADB}=90^{\circ}&lt;/math&gt;, quadrilateral &lt;math&gt;ABDF&lt;/math&gt; is cyclic. It follows that &lt;math&gt;\angle{ADE}=\angle{ABF}&lt;/math&gt;. In addition, triangles &lt;math&gt;ABF&lt;/math&gt; and &lt;math&gt;ADE&lt;/math&gt; are similar, and triangles &lt;math&gt;ADE&lt;/math&gt; and &lt;math&gt;ADC&lt;/math&gt; are similar. We can easily find &lt;math&gt;AD=12&lt;/math&gt;, &lt;math&gt;BD = 5&lt;/math&gt;, and &lt;math&gt;DC=9&lt;/math&gt; using pythagorean triples. So, the ratio of the hypotenuse to the longer leg of all three similar triangles is &lt;math&gt;\frac{15}{12} = \frac{4}{5}&lt;/math&gt;, and the ratio of the hypotenuse to the shorter leg is &lt;math&gt;\frac{15}{9} = \frac{3}{5}&lt;/math&gt;. It follows that &lt;math&gt;AF=(13)(\frac{4}{5}), BF=(13)(\frac{3}{5})&lt;/math&gt;. By [[Ptolemy's Theorem|Ptolemy's Theorem]], we have &lt;math&gt;13DF+(5)(13)(\frac{4}{5})=(12)(13)(\frac{3}{5})&lt;/math&gt;. Cancelling &lt;math&gt;13&lt;/math&gt;, we obtain &lt;math&gt;DF=\frac{16}{5}&lt;/math&gt;, so our answer is &lt;math&gt;16+5=\boxed{21\,\textbf{(B)}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Using the similar triangles in triangle &lt;math&gt;ADC&lt;/math&gt; gives &lt;math&gt;AE = \frac{48}{5}&lt;/math&gt; and &lt;math&gt;DE = \frac{36}{5}&lt;/math&gt;. Quadrilateral &lt;math&gt;ABDF&lt;/math&gt; is cyclic, implying that &lt;math&gt;\angle{B} + \angle{DFA}&lt;/math&gt; = 180°. Therefore, &lt;math&gt;\angle{B} = \angle{EFA}&lt;/math&gt;, and triangles &lt;math&gt;AEF&lt;/math&gt; and &lt;math&gt;ADB &lt;/math&gt; are similar. Solving the resulting proportion gives &lt;math&gt;EF = 4&lt;/math&gt;. Therefore, &lt;math&gt;DF = ED - EF = \frac{16}{5}&lt;/math&gt; and our answer is &lt;math&gt;\boxed{\textbf{(B)}}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> If we draw a diagram as given, but then add point &lt;math&gt;G&lt;/math&gt; on &lt;math&gt;\overline{BC}&lt;/math&gt; such that &lt;math&gt;\overline{FG}\perp\overline{BC}&lt;/math&gt; in order to use the Pythagorean theorem, we end up with similar triangles &lt;math&gt;\triangle{DFG}&lt;/math&gt; and &lt;math&gt;\triangle{DCE}&lt;/math&gt;. Thus, &lt;math&gt;FG=\tfrac35x&lt;/math&gt; and &lt;math&gt;DG=\tfrac45x&lt;/math&gt;, where &lt;math&gt;x&lt;/math&gt; is the length of &lt;math&gt;\overline{DF}&lt;/math&gt;. Using the Pythagorean theorem, we now get &lt;cmath&gt;BF = \sqrt{\left(\frac45x+ 5\right)^2 + \left(\frac35x\right)^2}&lt;/cmath&gt; and &lt;math&gt;AF&lt;/math&gt; can be found out noting that &lt;math&gt;AE&lt;/math&gt; is just &lt;math&gt;\tfrac{48}5&lt;/math&gt; through base times height (since &lt;math&gt;12\cdot 9 = 15 \cdot \tfrac{36}5&lt;/math&gt;, similar triangles gives &lt;math&gt;AE = \tfrac{48}5&lt;/math&gt;), and that &lt;math&gt;EF&lt;/math&gt; is just &lt;math&gt;\tfrac{36}5 - x&lt;/math&gt;. From there, &lt;cmath&gt;AF = \sqrt{\left(\frac{36}5 - x\right)^2 + \left(\frac{48}5\right)^2}.&lt;/cmath&gt; Now, &lt;math&gt;BF^2 + AF^2 = 169&lt;/math&gt;, and squaring and adding both sides and subtracting a 169 from both sides gives &lt;math&gt;2x^2 - \tfrac{32}5x = 0&lt;/math&gt;, so &lt;math&gt;x = \tfrac{16}5&lt;/math&gt;. Thus, the answer is &lt;math&gt;\boxed{\textbf{(B)}}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=B|num-b=18|num-a=20}}<br /> <br /> {{AMC10 box|year=2013|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Crazyboulder https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_16&diff=141294 2013 AMC 10B Problems/Problem 16 2021-01-02T03:38:09Z <p>Crazyboulder: Added solution</p> <hr /> <div>==Problem==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, medians &lt;math&gt;AD&lt;/math&gt; and &lt;math&gt;CE&lt;/math&gt; intersect at &lt;math&gt;P&lt;/math&gt;, &lt;math&gt;PE=1.5&lt;/math&gt;, &lt;math&gt;PD=2&lt;/math&gt;, and &lt;math&gt;DE=2.5&lt;/math&gt;. What is the area of &lt;math&gt;AEDC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\qquad\textbf{(A) }13\qquad\textbf{(B) }13.5\qquad\textbf{(C) }14\qquad\textbf{(D) }14.5\qquad\textbf{(E) }15&lt;/math&gt;<br /> &lt;asy&gt;<br /> pair A,B,C,D,E,P;<br /> A=(0,0);<br /> B=(80,0);<br /> C=(20,40);<br /> D=(50,20);<br /> E=(40,0);<br /> P=(33.3,13.3);<br /> draw(A--B);<br /> draw(B--C);<br /> draw(A--C);<br /> draw(C--E);<br /> draw(A--D);<br /> draw(D--E);<br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(D);<br /> dot(E);<br /> dot(P);<br /> label(&quot;A&quot;,A,NNW);<br /> label(&quot;B&quot;,B,NNE);<br /> label(&quot;C&quot;,C,ENE);<br /> label(&quot;D&quot;,D,ESE);<br /> label(&quot;E&quot;,E,SSE);<br /> label(&quot;P&quot;,P,SSE);<br /> &lt;/asy&gt;<br /> <br /> <br /> ==Solution 1==<br /> Let us use mass points:<br /> Assign &lt;math&gt;B&lt;/math&gt; mass &lt;math&gt;1&lt;/math&gt;. Thus, because &lt;math&gt;E&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt; also has a mass of &lt;math&gt;1&lt;/math&gt;. Similarly, &lt;math&gt;C&lt;/math&gt; has a mass of &lt;math&gt;1&lt;/math&gt;. &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; each have a mass of &lt;math&gt;2&lt;/math&gt; because they are between &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; respectively. Note that the mass of &lt;math&gt;D&lt;/math&gt; is twice the mass of &lt;math&gt;A&lt;/math&gt;, so AP must be twice as long as &lt;math&gt;PD&lt;/math&gt;. PD has length &lt;math&gt;2&lt;/math&gt;, so &lt;math&gt;AP&lt;/math&gt; has length &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt; has length &lt;math&gt;6&lt;/math&gt;. Similarly, &lt;math&gt;CP&lt;/math&gt; is twice &lt;math&gt;PE&lt;/math&gt; and &lt;math&gt;PE=1.5&lt;/math&gt;, so &lt;math&gt;CP=3&lt;/math&gt; and &lt;math&gt;CE=4.5&lt;/math&gt;. Now note that triangle &lt;math&gt;PED&lt;/math&gt; is a &lt;math&gt;3-4-5&lt;/math&gt; right triangle with the right angle &lt;math&gt;DPE&lt;/math&gt;. Since the diagonals of quadrilaterals &lt;math&gt;AEDC&lt;/math&gt;, &lt;math&gt;AD&lt;/math&gt; and &lt;math&gt;CE&lt;/math&gt;, are perpendicular, the area of &lt;math&gt;AEDC&lt;/math&gt; is &lt;math&gt;\frac{6 \times 4.5}{2}=\boxed{\textbf{(B)} 13.5}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Note that triangle &lt;math&gt;DPE&lt;/math&gt; is a right triangle, and that the four angles (angles &lt;math&gt;APC, CPD, DPE,&lt;/math&gt; and &lt;math&gt;EPA&lt;/math&gt;) that have point &lt;math&gt;P&lt;/math&gt; are all right angles. Using the fact that the centroid (&lt;math&gt;P&lt;/math&gt;) divides each median in a &lt;math&gt;2:1&lt;/math&gt; ratio, &lt;math&gt;AP=4&lt;/math&gt; and &lt;math&gt;CP=3&lt;/math&gt;. Quadrilateral &lt;math&gt;AEDC&lt;/math&gt; is now just four right triangles. The area is &lt;math&gt;\frac{4\cdot 1.5+4\cdot 3+3\cdot 2+2\cdot 1.5}{2}=\boxed{\textbf{(B)} 13.5}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> From the solution above, we can find that the lengths of the diagonals are &lt;math&gt;6&lt;/math&gt; and &lt;math&gt;4.5&lt;/math&gt;. Now, since the diagonals of AEDC are perpendicular, we use the area formula to find that the total area is &lt;math&gt;\frac{6\times4.5}{2} = \frac{27}{2} = 13.5 = \boxed{\textbf{(B)} 13.5}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> <br /> From the solutions above, we know that the sides CP and AP are 3 and 4 respectively because of the properties of medians that divide cevians into 1:2 ratios. We can then proceed to use the heron's formula on the middle triangle EPD and get the area of EPD as 3/2, (its simple computation really, nothing large). Then we can find the areas of the remaining triangles based on side and ratio length of the bases.<br /> <br /> ==Solution 5==<br /> We know that &lt;math&gt;[AEDC]=\frac{3}{4}[ABC]&lt;/math&gt;, and &lt;math&gt;[ABC]=3[PAC]&lt;/math&gt; using median properties. So Now we try to find &lt;math&gt;[PAC]&lt;/math&gt;. Since &lt;math&gt;\triangle PAC\sim \triangle PDE&lt;/math&gt;, then the side lengths of &lt;math&gt;\triangle PAC&lt;/math&gt; are twice as long as &lt;math&gt;\triangle PDE&lt;/math&gt; since &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; are midpoints. Therefore, &lt;math&gt;\frac{[PAC]}{[PDE]}=2^2=4&lt;/math&gt;. It suffices to compute &lt;math&gt;[PDE]&lt;/math&gt;. Notice that &lt;math&gt;(1.5, 2, 2.5)&lt;/math&gt; is a Pythagorean Triple, so &lt;math&gt;[PDE]=\frac{1.5\times 2}{2}=1.5&lt;/math&gt;. This implies &lt;math&gt;[PAC]=1.5\cdot 4=6&lt;/math&gt;, and then &lt;math&gt;[ABC]=3\cdot 6=18&lt;/math&gt;. Finally, &lt;math&gt;[AEDC]=\frac{3}{4}\times 18=\boxed{13.5}&lt;/math&gt;.<br /> <br /> ~CoolJupiter<br /> <br /> ==Solution 6==<br /> As from Solution 4, we find the area of &lt;math&gt;\triangle DPE&lt;/math&gt; to be &lt;math&gt;\frac{3}{2}&lt;/math&gt;. Because &lt;math&gt;DE = \frac{5}{2}&lt;/math&gt;, the altitude perpendicular to &lt;math&gt;DE = \frac{6}{5}&lt;/math&gt;. Also, because &lt;math&gt;DE || AC&lt;/math&gt;, &lt;math&gt;\triangle ABC&lt;/math&gt; is similar to &lt;math&gt;\triangle{DBE}&lt;/math&gt; with side length ratio &lt;math&gt;2:1&lt;/math&gt;, so &lt;math&gt;AC=5&lt;/math&gt; and the altitude perpendicular to &lt;math&gt;AC = \frac{12}{5}&lt;/math&gt;. The altitude of trapezoid &lt;math&gt;ACDE&lt;/math&gt; is then &lt;math&gt;\frac{18}{5}&lt;/math&gt; and the bases are &lt;math&gt;\frac{5}{2}&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt;. So, we use the formula for the area of a trapezoid to find the area of &lt;math&gt;ACDE = \boxed{13.5}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC10 box|year=2013|ab=B|num-b=15|num-a=17}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:Area Problems]]<br /> {{MAA Notice}}</div> Crazyboulder https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_17&diff=140305 2010 AMC 12B Problems/Problem 17 2020-12-22T23:41:30Z <p>Crazyboulder: /* Solution 3 */</p> <hr /> <div>{{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #17]] and [[2010 AMC 10B Problems|2010 AMC 10B #23]]}}<br /> <br /> == Problem ==<br /> The entries in a &lt;math&gt;3 \times 3&lt;/math&gt; array include all the digits from &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;9&lt;/math&gt;, arranged so that the entries in every row and column are in increasing order. How many such arrays are there?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 42 \qquad \textbf{(E)}\ 60&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> Observe that all tableaus must have 1s and 9s in the corners, 8s and 2s next to those corner squares, and 4-6 in the middle square. Also note that for each tableau, there exists a valid tableau diagonally symmetrical across the diagonal extending from the top left to the bottom right. <br /> <br /> <br /> *'''Case 1: Center 4'''<br /> &lt;cmath&gt;\begin{tabular}{|c|c|c|} \hline 1&amp;2&amp;\\<br /> \hline 3&amp;4&amp;8\\<br /> \hline &amp;&amp;9\\<br /> \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&amp;2&amp;\\<br /> \hline 3&amp;4&amp;\\<br /> \hline &amp;8&amp;9\\<br /> \hline \end{tabular}&lt;/cmath&gt;<br /> <br /> 3 necessarily must be placed as above. Any number could fill the isolated square, but the other 2 are then invariant. So, there are 3 cases each and 6 overall cases. Given diagonal symmetry, alternate 2 and 8 placements yield symmetrical cases. &lt;math&gt;2*6=12&lt;/math&gt;<br /> <br /> *'''Case 2: Center 5'''<br /> &lt;cmath&gt;\begin{tabular}{|c|c|c|} \hline 1&amp;2&amp;3\\<br /> \hline 4&amp;5&amp;\\<br /> \hline &amp;8&amp;9\\<br /> \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&amp;2&amp;\\<br /> \hline 3&amp;5&amp;\\<br /> \hline &amp;8&amp;9\\<br /> \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&amp;2&amp;\\<br /> \hline 3&amp;5&amp;8\\<br /> \hline &amp;&amp;9\\<br /> \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&amp;2&amp;3\\<br /> \hline 4&amp;5&amp;8\\<br /> \hline &amp;&amp;9\\<br /> \hline \end{tabular}&lt;/cmath&gt;<br /> <br /> Here, no 3s or 7s are assured, but this is only a teensy bit trickier and messier. WLOG, casework with 3 instead of 7 as above. Remembering that &lt;math&gt;4&lt;5&lt;/math&gt;, logically see that the numbers of cases are then 2,3,3,1 respectively. By symmetry, &lt;math&gt;2*9=18&lt;/math&gt;<br /> <br /> *'''Case 3: Center 6'''<br /> By inspection, realize that this is symmetrical to case 1 except that the 7s instead of the 3s are assured.<br /> &lt;math&gt;2*6=12&lt;/math&gt;<br /> <br /> &lt;cmath&gt;12+18+12=\boxed{\textbf{D)}42}&lt;/cmath&gt;<br /> <br /> <br /> ~BJHHar<br /> <br /> == Solution 2==<br /> This solution is trivial by the hook length theorem. The hooks look like this:<br /> <br /> &lt;math&gt; \begin{tabular}{|c|c|c|} \hline 5 &amp; 4 &amp; 3 \\<br /> \hline 4 &amp; 3 &amp; 2\\<br /> \hline 3 &amp; 2 &amp; 1\\<br /> \hline \end{tabular}&lt;/math&gt;<br /> <br /> So, the answer is &lt;math&gt;\frac{9!}{5 \cdot 4 \cdot 3 \cdot 4 \cdot 3 \cdot 2 \cdot 3 \cdot 2 \cdot 1}&lt;/math&gt; = &lt;math&gt;\boxed{\text{(D) }42}&lt;/math&gt;<br /> <br /> P.S. The hook length formula is a formula to calculate the number of standard Young tableaux of a Young diagram. Numberphile has an easy-to-understand video about it here: https://www.youtube.com/watch?v=vgZhrEs4tuk The full proof is quite complicated and is not given in the video, although the video hints at possible proofs.<br /> <br /> ==Video Solution==<br /> https://youtu.be/ZfnxbpdFKjU?t=422<br /> <br /> ~IceMatrix<br /> <br /> == See also ==<br /> {{AMC12 box|year=2010|num-b=16|num-a=18|ab=B}}<br /> {{AMC10 box|year=2010|num-b=22|num-a=24|ab=B}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Crazyboulder