# 1958 AHSME Problems/Problem 49

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## 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 $\binom{12}{2}=66$.