How many coefficients are in the expansion $(x + y + z)^{10}$

Question

I need to find the number of coefficients in the expansion $(x + y + z)^{10}$.

I had this exercise on a recent assignment.

The answer I gave is: $3^{10} = \binom {3 + 10 - 1}{10} = \binom{12}{10} = 132$, but my tutor says the answer is 66.

I don't know where I went wrong and, therefore, I don't know how to get 66. Can someone show me where I went wrong with this exercise?

Thanks

Answer 1

In general, the number of coefficients of $(x_1+x_2+\ldots+x_m)^n$ is ${n+m-1\choose m-1}={n+m-1\choose n}$.

This can be shown with the Stars and Bars methods. Basically you need to count the number of ways that it is possible to add $m$ nonnegative integers to get $n$.

Answer 2

You can write \begin{align} (x+y+z)^n &= \sum_{k=0}^n \binom{n}{k}x^k(y+z)^{n-k} = \sum_{k=0}^n\sum_{j=0}^{n-k} \binom{n}{k}\binom{n-k}{j}x^k y ^j z^{n-k-j} \\ &= \sum_{k=0}^n\sum_{j=0}^{n-k} \frac{n!}{k!j!(n-k-j)!}x^k y ^j z^{n-k-j} \end{align} using Newton's Binomial formula twice (or the multinomial formula once, equivalently). This shows you have $$ \sum_{k=0}^n\sum_{j=0}^{n-k} 1 = \sum_{k=0}^n (n+1-k) = \frac{(n+1)(n+2)}{2} = \binom{n+2}{n} $$ coefficients in the expansion.