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Difference between revisions of "1958 AHSME Problems/Problem 49"
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==Solution 2 (Stars and Bars)==
 
==Solution 2 (Stars and Bars)==
  
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Each term in the expansion of <math>(a+b+c)^10</math> will have the form <math>a^i\timesb^jtimesc^k</math>, where $0\lei, j, k/le10
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Each term in the expansion of <math>(a+b+c)^{10}</math> will have the form <math>a^i \times b^j \times c^k</math>, where <math>0\le i, j, k\le 10</math> and <math>a+b+c=10</math>. So, we need to find the number of triplets of nonnegative integers <math>(a, b, c)</math> such that <math>a+b+c=10</math>. Using Stars and Bars, this value is <math>\binom{12}{2}=66</math>.
  
 
==See also==
 
==See also==

Latest revision as of 23:59, 31 December 2023

Problem

In the expansion of $(a + b)^n$ there are $n + 1$ dissimilar terms. The number of dissimilar terms in the expansion of $(a + b + c)^{10}$ is:

$\textbf{(A)}\ 11\qquad \textbf{(B)}\ 33\qquad \textbf{(C)}\ 55\qquad \textbf{(D)}\ 66\qquad \textbf{(E)}\ 132$

Solution

Expand the binomial $((a+b)+c)^n$ with the binomial theorem. We have:

\[\sum\limits_{k=0}^{10} \binom{10}{k} (a+b)^k c^{10-k}\]

So for each iteration of the summation operator, we add k+1 dissimilar terms. Therefore our answer is:

\[\sum\limits_{k=0}^{10} k+1 = \frac{11(1+11)}{2} = 66 \to \boxed{\textbf{D}}\]

Solution 2 (Stars and Bars)

Each term in the expansion of $(a+b+c)^{10}$ will have the form $a^i \times b^j \times c^k$, where $0\le i, j, k\le 10$ and $a+b+c=10$. So, we need to find the number of triplets of nonnegative integers $(a, b, c)$ such that $a+b+c=10$. Using Stars and Bars, this value is $\binom{12}{2}=66$.

See also

1958 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 48 		Followed by
Problem 50
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All AHSME Problems and Solutions

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