https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Dbose1978&feedformat=atom AoPS Wiki - User contributions [en] 2022-01-28T18:43:03Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_25&diff=136027 2020 AMC 10A Problems/Problem 25 2020-10-29T03:51:22Z <p>Dbose1978: /* Solution 2 */</p> <hr /> <div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #23]] and [[2020 AMC 10A Problems|2020 AMC 10A #25]]}}<br /> <br /> ==Problem==<br /> Jason rolls three fair standard six-sided dice. Then he looks at the rolls and chooses a subset of the dice (possibly empty, possibly all three dice) to reroll. After rerolling, he wins if and only if the sum of the numbers face up on the three dice is exactly &lt;math&gt;7.&lt;/math&gt; Jason always plays to optimize his chances of winning. What is the probability that he chooses to reroll exactly two of the dice?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{7}{36} \qquad\textbf{(B) } \frac{5}{24} \qquad\textbf{(C) } \frac{2}{9} \qquad\textbf{(D) } \frac{17}{72} \qquad\textbf{(E) } \frac{1}{4}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Consider the probability that rolling two dice gives a sum of &lt;math&gt;s&lt;/math&gt;, where &lt;math&gt;s \leq 7&lt;/math&gt;. There are &lt;math&gt;s - 1&lt;/math&gt; pairs that satisfy this, namely &lt;math&gt;(1, s - 1), (2, s - 2), ..., (s - 1, 1)&lt;/math&gt;, out of &lt;math&gt;6^2 = 36&lt;/math&gt; possible pairs. The probability is &lt;math&gt;\frac{s - 1}{36}&lt;/math&gt;.<br /> <br /> Therefore, if one die has a value of &lt;math&gt;a&lt;/math&gt; and Jason rerolls the other two dice, then the probability of winning is &lt;math&gt;\frac{7 - a - 1}{36} = \frac{6 - a}{36}&lt;/math&gt;.<br /> <br /> In order to maximize the probability of winning, &lt;math&gt;a&lt;/math&gt; must be minimized. This means that if Jason rerolls two dice, he must choose the two dice with the maximum values.<br /> <br /> Thus, we can let &lt;math&gt;a \leq b \leq c&lt;/math&gt; be the values of the three dice, which we will call &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; respectively. Consider the case when &lt;math&gt;a + b &lt; 7&lt;/math&gt;. If &lt;math&gt;a + b + c = 7&lt;/math&gt;, then we do not need to reroll any dice. Otherwise,<br /> if we reroll one die, we can roll dice &lt;math&gt;C&lt;/math&gt; in the hope that we get the value that makes the sum of the three dice &lt;math&gt;7&lt;/math&gt;. This happens with probability &lt;math&gt;\frac16&lt;/math&gt;. If we reroll two dice, we will roll &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, and the probability of winning is &lt;math&gt;\frac{6 - a}{36}&lt;/math&gt;, as stated above.<br /> <br /> However, &lt;math&gt;\frac16 &gt; \frac{6 - a}{36}&lt;/math&gt;, so rolling one die is always better than rolling two dice if &lt;math&gt;a + b &lt; 7&lt;/math&gt;.<br /> <br /> Now consider the case where &lt;math&gt;a + b \geq 7&lt;/math&gt;. Rerolling one die will not help us win since the sum of the three dice will always be greater than &lt;math&gt;7&lt;/math&gt;. If we reroll two dice, the probability of winning is, once again, &lt;math&gt;\frac{6 - a}{36}&lt;/math&gt;. To find the probability of winning if we reroll all three dice, we can let each dice have &lt;math&gt;1&lt;/math&gt; dot and find the number of ways to distribute the remaining &lt;math&gt;4&lt;/math&gt; dots. By stars and bars, there are &lt;math&gt;{6\choose2} = 15&lt;/math&gt; ways to do this, making the probability of winning &lt;math&gt;\frac{15}{6^3} = \frac5{72}&lt;/math&gt;.<br /> <br /> In order for rolling two dice to be more favorable than rolling three dice, &lt;math&gt;\frac{6 - a}{36} &gt; \frac5{72} \rightarrow a \leq 3&lt;/math&gt;.<br /> <br /> Thus, rerolling two dice is optimal if and only if &lt;math&gt;a \leq 3&lt;/math&gt; and &lt;math&gt;a + b \geq 7&lt;/math&gt;. The possible triplets &lt;math&gt;(a, b, c)&lt;/math&gt; that satisfy these conditions, and the number of ways they can be permuted, are<br /> &lt;math&gt;(3, 4, 4) \rightarrow 3&lt;/math&gt; ways.<br /> &lt;math&gt;(3, 4, 5) \rightarrow 6&lt;/math&gt; ways.<br /> &lt;math&gt;(3, 4, 6) \rightarrow 6&lt;/math&gt; ways.<br /> &lt;math&gt;(3, 5, 5) \rightarrow 3&lt;/math&gt; ways.<br /> &lt;math&gt;(3, 5, 6) \rightarrow 6&lt;/math&gt; ways.<br /> &lt;math&gt;(3, 6, 6) \rightarrow 3&lt;/math&gt; ways.<br /> &lt;math&gt;(2, 5, 5) \rightarrow 3&lt;/math&gt; ways.<br /> &lt;math&gt;(2, 5, 6) \rightarrow 6&lt;/math&gt; ways.<br /> &lt;math&gt;(2, 6, 6) \rightarrow 3&lt;/math&gt; ways.<br /> &lt;math&gt;(1, 6, 6) \rightarrow 3&lt;/math&gt; ways.<br /> <br /> There are &lt;math&gt;3 + 6 + 6 + 3 + 6 + 3 + 3 + 6 + 3 + 3 = 42&lt;/math&gt; ways in which rerolling two dice is optimal, out of &lt;math&gt;6^3 = 216&lt;/math&gt; possibilities, Therefore, the probability that Jason will reroll two dice is &lt;math&gt;\frac{42}{216} = \boxed{\textbf{(A) }\frac7{36}}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> We count the numerator.<br /> Jason will pick up no dice if he already has a 7 as a sum. We need to assume he does not have a 7 to begin with.<br /> If Jason decides to pick up all the dice to re-roll, by the stars and bars rule ways to distribute, &lt;math&gt;{n+k-1 \choose k-1}&lt;/math&gt;, there will be 2 bars and 4 stars(3 of them need to be guaranteed because a roll is at least 1) for a probability of &lt;math&gt;\frac{15}{216}=\frac{2.5}{36}&lt;/math&gt;.<br /> If Jason picks up 2 dice and leaves a die showing &lt;math&gt;k&lt;/math&gt;, he will need the other two to sum to &lt;math&gt;7-k&lt;/math&gt;. This happens with probability &lt;cmath&gt;\frac{6-k}{36}&lt;/cmath&gt; for integers &lt;math&gt;1 \leq k \leq 6&lt;/math&gt;.<br /> If the roll is not 7, Jason will pick up exactly one die to re-roll if there can remain two other dice with sum less than 7, since this will give him a &lt;math&gt;\frac{1}{6}&lt;/math&gt; chance which is a larger probability than all the cases unless he has a 7 to begin with.<br /> We have &lt;cmath&gt;\frac{1}{6} &gt; \underline{\frac{5,4,3}{36}} &gt; \frac{2.5}{36} &gt; \frac{2,1,0}{36}.&lt;/cmath&gt;<br /> We count the underlined part's frequency for the numerator without upsetting the probability greater than it.<br /> Let &lt;math&gt;a&lt;/math&gt; be the roll we keep. We know &lt;math&gt;a&lt;/math&gt; is at most 3 since 4 would cause Jason to pick up all the dice.<br /> When &lt;math&gt;a=1&lt;/math&gt;, there are 3 choices for whether it is rolled 1st, 2nd, or 3rd, and in this case the other two rolls have to be at least 6(or he would have only picked up 1). This give &lt;math&gt;3 \cdot 1^{2} =3&lt;/math&gt; ways.<br /> Similarly, &lt;math&gt;a=2&lt;/math&gt; gives &lt;math&gt;3 \cdot 2^{2} =12&lt;/math&gt; because the 2 can be rolled in 3 places and the other two rolls are at least 5.<br /> &lt;math&gt;a=3&lt;/math&gt; gives &lt;math&gt;3 \cdot 3^{2} =27&lt;/math&gt;. <br /> Summing together gives the numerator of 42.<br /> The denominator is &lt;math&gt;6^3=216&lt;/math&gt;, so we have &lt;math&gt;\frac{42}{216}=\boxed{(A) \frac{7}{36}}&lt;/math&gt;<br /> <br /> == Video Solution ==<br /> <br /> https://youtu.be/3W4jOpCiBx8<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2020|ab=A|num-b=24|after=Last Problem}}<br /> {{AMC12 box|year=2020|ab=A|num-b=22|num-a=24}}<br /> <br /> [[Category: Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Dbose1978 https://artofproblemsolving.com/wiki/index.php?title=Binomial_Theorem&diff=120842 Binomial Theorem 2020-04-12T00:30:10Z <p>Dbose1978: Undo revision 120841 by Dbose1978 (talk)</p> <hr /> <div>The '''Binomial Theorem''' states that for [[real]] or [[complex]] &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and [[non-negative]] [[integer]] &lt;math&gt;n&lt;/math&gt;,<br /> &lt;center&gt;&lt;math&gt;(a+b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k&lt;/math&gt;&lt;/center&gt;<br /> <br /> where &lt;math&gt;\binom{n}{k} = \frac{n!}{k!(n-k)!}&lt;/math&gt; is a [[binomial coefficient]]. In other words, the coefficients when &lt;math&gt;(a + b)^n&lt;/math&gt; is expanded and like terms are collected are the same as the entries in the &lt;math&gt;n&lt;/math&gt;th row of [[Pascal's Triangle]].<br /> <br /> For example, &lt;math&gt;(a + b)^5 = a^5 + 5 a^4 b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a b^4 + b^5&lt;/math&gt;, with coefficients &lt;math&gt;1 = \binom{5}{0}&lt;/math&gt;, &lt;math&gt;5 = \binom{5}{1}&lt;/math&gt;, &lt;math&gt;10 = \binom{5}{2}&lt;/math&gt;, etc.<br /> <br /> ==Proof==<br /> There are a number of different ways to prove the Binomial Theorem, for example by a straightforward application of [[mathematical induction]]. The Binomial Theorem also has a nice combinatorial proof: <br /> <br /> We can write &lt;math&gt;(a+b)^n=\underbrace{ (a+b)\cdot(a+b)\cdot(a+b)\cdot\cdots\cdot(a+b) }_{n}&lt;/math&gt;. Repeatedly using the [[distributive property]], we see that for a term &lt;math&gt;a^m b^{n-m}&lt;/math&gt;, we must choose &lt;math&gt;m&lt;/math&gt; of the &lt;math&gt;n&lt;/math&gt; terms to contribute an &lt;math&gt;a&lt;/math&gt; to the term, and then each of the other &lt;math&gt;n-m&lt;/math&gt; terms of the product must contribute a &lt;math&gt;b&lt;/math&gt;. Thus, the coefficient of &lt;math&gt;a^m b^{n-m}&lt;/math&gt; is the number of ways to choose &lt;math&gt;m&lt;/math&gt; objects from a set of size &lt;math&gt;n&lt;/math&gt;, or &lt;math&gt;\binom{n}{m}&lt;/math&gt;. Extending this to all possible values of &lt;math&gt;m&lt;/math&gt; from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;n&lt;/math&gt;, we see that &lt;math&gt;(a+b)^n = \sum_{m=0}^{n}{\binom{n}{m}}\cdot a^m\cdot b^{n-m}&lt;/math&gt;, as claimed.<br /> <br /> Similarly, the coefficients of &lt;math&gt;(x+y)^n&lt;/math&gt; will be the entries of the &lt;math&gt;n^\text{th}&lt;/math&gt; row of [[Pascal's Triangle]]. This is explained further in the Counting and Probability textbook [AoPS].<br /> <br /> ===Proof via Induction===<br /> Given the constants &lt;math&gt;a,b,n&lt;/math&gt; are all natural numbers, it's clear to see that &lt;math&gt;(a+b)^{1} = a+b&lt;/math&gt;. Assuming that &lt;math&gt;a+b)^{n} = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k}&lt;/math&gt;, &lt;cmath&gt;(a+b)^{n+1} = (\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k})(a+b)&lt;/cmath&gt;<br /> &lt;cmath&gt;=(\binom{n}{0}a^{n}b^{0} + \binom{n}{1}a^{n-1}b^{1} + \binom{n}{2}a^{n-2}b^{2}+\cdots+\binom{n}{n}a^{0}b^{n})(a+b)&lt;/cmath&gt;<br /> &lt;cmath&gt;=(\binom{n}{0}a^{n+1}b^{0} + \binom{n}{1}a^{n}b^{1} + \binom{n}{2}a^{n-1}b^{2}+\cdots+\binom{n}{n}a^{1}b^{n})<br /> + (\binom{n}{0}a^{n}b^{1} + \binom{n}{1}a^{n-1}b^{2} + \binom{n}{2}a^{n-2}b^{3}+\cdots+\binom{n}{n}a^{0}b^{n+1})&lt;/cmath&gt;<br /> &lt;cmath&gt;=(\binom{n}{0}a^{n+1}b^{0} + (\binom{n}{0}+\binom{n}{1})(a^{n}b^{1}) + (\binom{n}{1}+\binom{n}{2})(a^{n-1}b^{2})+\cdots+(\binom{n}{n-1}+\binom{n}{n})(a^{1}b^{n})+\binom{n}{n}a^{0}b^{n+1})&lt;/cmath&gt;<br /> &lt;cmath&gt;=\binom{n+1}{0}a^{n+1}b^{0} + \binom{n+1}{1}a^{n}b^{1} + \binom{n+1}{2}a^{n-1}b^{2}+\cdots+\binom{n+1}{n}a^{1}b^{n} + \binom{n+1}{n+1}a^{0}b^{n+1}&lt;/cmath&gt;<br /> &lt;cmath&gt;=\sum_{k=0}^{n+1}\binom{n+1}{k}a^{(n+1)-k}b^{k}&lt;/cmath&gt;<br /> Therefore, if the theorem holds under &lt;math&gt;n+1&lt;/math&gt;, it must be valid.<br /> (Note that &lt;math&gt;\binom{n}{m} + \binom{n}{m+1} = \binom{n+1}{m+1} &lt;/math&gt; for &lt;math&gt;m\leq n&lt;/math&gt;)<br /> <br /> ==Generalizations==<br /> The Binomial Theorem was generalized by [[Isaac Newton]], who used an [[infinite]] [[series]] to allow for complex [[exponent]]s: For any [[real]] or [[complex]] &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;r&lt;/math&gt;,<br /> &lt;center&gt;&lt;math&gt;(a+b)^r = \sum_{k=0}^{\infty}\binom{r}{k}a^{r-k}b^k&lt;/math&gt;.&lt;/center&gt;<br /> <br /> ===Proof===<br /> Consider the function &lt;math&gt;f(b)=(a+b)^r&lt;/math&gt; for constants &lt;math&gt;a,r&lt;/math&gt;. It is easy to see that &lt;math&gt;\frac{d^k}{db^k}f=r(r-1)\cdots(r-k+1)(a+b)^{r-k}&lt;/math&gt;. Then, we have &lt;math&gt;\frac{d^k}{db^k}f(0)=r(r-1)\cdots(r-k+1)a^{r-k}&lt;/math&gt;. So, the [[Taylor series]] for &lt;math&gt;f(b)&lt;/math&gt; centered at &lt;math&gt;0&lt;/math&gt; is <br /> <br /> &lt;cmath&gt;(a+b)^r=\sum_{k=0}^\infty \frac{r(r-1)\cdots(r-k+1)a^{r-k}b^k}{k!}=\sum_{k=0}^\infty \binom{r}{k}a^{r-k}b^k.&lt;/cmath&gt;<br /> <br /> ==Usage==<br /> Many [[factoring | factorizations]] involve complicated [[polynomial]]s with [[binomial coefficient]]s. For example, if a contest problem involved the polynomial &lt;math&gt;x^5+4x^4+6x^3+4x^2+x&lt;/math&gt;, one could factor it as such: &lt;math&gt; x(x^4+4x^3+6x^2+4x+1)=x(x+1)^{4}&lt;/math&gt;. It is a good idea to be familiar with binomial expansions, including knowing the first few binomial coefficients.<br /> <br /> ==See also==<br /> *[[Combinatorics]]<br /> *[[Multinomial Theorem]]<br /> <br /> [[Category:Theorems]]<br /> [[Category:Combinatorics]]<br /> [[Category:Algebra]]</div> Dbose1978 https://artofproblemsolving.com/wiki/index.php?title=Binomial_Theorem&diff=120841 Binomial Theorem 2020-04-12T00:29:31Z <p>Dbose1978: </p> <hr /> <div>The '''Binomial Theorem''' states that for [[real]] or [[complex]] &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and [[non-negative]] [[integer]] &lt;math&gt;n&lt;/math&gt;,<br /> &lt;center&gt;&lt;math&gt;(a+b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k&lt;/math&gt;&lt;/center&gt;<br /> <br /> where &lt;math&gt;\binom{n}{k} = \frac{n!}{k!(n-k)!}&lt;/math&gt; is a [[binomial coefficient]]. In other words, the coefficients when &lt;math&gt;(a + b)^n&lt;/math&gt; is expanded and like terms are collected are the same as the entries in the &lt;math&gt;n&lt;/math&gt;th row of [[Pascal's Triangle]].<br /> <br /> For example, &lt;math&gt;(a + b)^5 = a^5 + 5 a^4 b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a b^4 + b^5&lt;/math&gt;, with coefficients &lt;math&gt;1 = \binom{5}{0}&lt;/math&gt;, &lt;math&gt;5 = \binom{5}{1}&lt;/math&gt;, &lt;math&gt;10 = \binom{5}{2}&lt;/math&gt;, etc.<br /> <br /> ==Proof==<br /> There are a number of different ways to prove the Binomial Theorem, for example by a straightforward application of [[mathematical induction]]. The Binomial Theorem also has a nice combinatorial proof: <br /> <br /> We can write &lt;math&gt;(a+b)^n=\underbrace{ (a+b)\cdot(a+b)\cdot(a+b)\cdot\cdots\cdot(a+b) }_{n}&lt;/math&gt;. Repeatedly using the [[distributive property]], we see that for a term &lt;math&gt;a^m b^{n-m}&lt;/math&gt;, we must choose &lt;math&gt;m&lt;/math&gt; of the &lt;math&gt;n&lt;/math&gt; terms to contribute an &lt;math&gt;a&lt;/math&gt; to the term, and then each of the other &lt;math&gt;n-m&lt;/math&gt; terms of the product must contribute a &lt;math&gt;b&lt;/math&gt;. Thus, the coefficient of &lt;math&gt;a^m b^{n-m}&lt;/math&gt; is the number of ways to choose &lt;math&gt;m&lt;/math&gt; objects from a set of size &lt;math&gt;n&lt;/math&gt;, or &lt;math&gt;\binom{n}{m}&lt;/math&gt;. Extending this to all possible values of &lt;math&gt;m&lt;/math&gt; from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;n&lt;/math&gt;, we see that &lt;math&gt;(a+b)^n = \sum_{m=0}^{n}{\binom{n}{m}}\cdot a^m\cdot b^{n-m}&lt;/math&gt;, as claimed.<br /> <br /> Similarly, the coefficients of &lt;math&gt;(x+y)^n&lt;/math&gt; will be the entries of the &lt;math&gt;n^\text{th}&lt;/math&gt; row of [[Pascal's Triangle]]. This is explained further in the Counting and Probability textbook [AoPS].<br /> <br /> ===Proof via Induction===<br /> Given the constants &lt;math&gt;a,b,n&lt;/math&gt; are all natural numbers, it's clear to see that &lt;math&gt;(a+b)^{1} = a+b&lt;/math&gt;. Assuming that &lt;math&gt;a+b)^{n} = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k}&lt;/math&gt;, &lt;cmath&gt;(a+b)^{n+1} = (\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k})(a+b)&lt;/cmath&gt;<br /> &lt;cmath&gt;=(\binom{n}{0}a^{n}b^{0} + \binom{n}{1}a^{n-1}b^{1} + \binom{n}{2}a^{n-2}b^{2}+\cdots+\binom{n}{n}a^{0}b^{n})(a+b)&lt;/cmath&gt;<br /> &lt;cmath&gt;=(\binom{n}{0}a^{n+1}b^{0} + \binom{n}{1}a^{n}b^{1} + \binom{n}{2}a^{n-1}b^{2}+\cdots+\binom{n}{n}a^{1}b^{n})<br /> + (\binom{n}{0}a^{n}b^{1} + \binom{n}{1}a^{n-1}b^{2} + \binom{n}{2}a^{n-2}b^{3}+\cdots+\binom{n}{n}a^{0}b^{n+1})&lt;/cmath&gt;<br /> &lt;cmath&gt;=(\binom{n}{0}a^{n+1}b^{0} + (\binom{n}{0}+\binom{n}{1})(a^{n}b^{1}) + (\binom{n}{1}+\binom{n}{2})(a^{n-1}b^{2})+\cdots+(\binom{n}{n-1}+\binom{n}{n})(a^{1}b^{n})+\binom{n}{n}a^{0}b^{n+1})&lt;/cmath&gt;<br /> &lt;cmath&gt;=\binom{n+1}{0}a^{n+1}b^{0} + \binom{n+1}{1}a^{n}b^{1} + \binom{n+1}{2}a^{n-1}b^{2}+\cdots+\binom{n+1}{n}a^{1}b^{n} + \binom{n+1}{n+1}a^{0}b^{n+1}&lt;/cmath&gt;<br /> &lt;cmath&gt;=\sum_{k=0}^{n+1}\binom{n+1}{k}a^{(n+1)-k}b^{k}&lt;/cmath&gt;<br /> Therefore, if the theorem holds under &lt;math&gt;n+1&lt;/math&gt;, it must be valid.<br /> (Note that &lt;math&gt;\binom{n}{m} + \binom{n}{m+1} = \binom{n+1}{m+1} &lt;/math&gt; for &lt;math&gt;m\leq n&lt;/math&gt;)<br /> <br /> ==Generalizations==<br /> The Binomial Theorem was generalized by [[Isaac Newton]], who used an [[infinite]] [[series]] to allow for complex [[exponent]]s: For any [[real]] or [[complex]] &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;r&lt;/math&gt;,<br /> &lt;center&gt;&lt;math&gt;(a+b)^r = \sum_{k=0}^{\infty}\binom{r}{k}a^{r-k}b^k&lt;/math&gt;.&lt;/center&gt;<br /> <br /> ===Proof===<br /> Consider the function &lt;math&gt;f(b)=(a+b)^r&lt;/math&gt; for constants &lt;math&gt;a,r&lt;/math&gt;. It is easy to see that &lt;math&gt;\frac{d^k}{db^k}f=r(r-1)\cdots(r-k+1)(a+b)^{r-k}&lt;/math&gt;. Then, we have &lt;math&gt;\frac{d^k}{db^k}f(0)=r(r-1)\cdots(r-k+1)a^{r-k}&lt;/math&gt;. So, the [[Taylor series]] for &lt;math&gt;f(b)&lt;/math&gt; centered at &lt;math&gt;0&lt;/math&gt; is <br /> <br /> &lt;cmath&gt;(a+b)^r=\sum_{k=0}^\infty \frac{r(r-1)\cdots(r-k+1)a^{r-k}b^k}{k!}=\sum_{k=0}^\infty \binom{r}{k}a^{r-k}b^k.&lt;/cmath&gt;<br /> <br /> ==Usage==<br /> Many [[factoring | factorizations]] involve complicated [[polynomial]]s with [[binomial coefficient]]s. For example, if a contest problem involved the polynomial &lt;math&gt;x^5+4x^4+6x^3+4x^2+x&lt;/math&gt;, one could factor it as such: &lt;math&gt; x(x^4+4x^3+6x^2+4x+1)=x(x+1)^{4}&lt;/math&gt;. It is a good idea to be familiar with binomial expansions, including knowing the first few binomial coefficients.</div> Dbose1978