# 1958 AHSME Problems/Problem 49

## 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 byProblem 48 Followed byProblem 50 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 All AHSME Problems and Solutions

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