Given $(x+y+z)^{15}$ find the coefficient of $x^2y^{10}z^{3}$

Question

Given $(x+y+z)^{15}$ find the coefficient of $x^2y^{10}z^3$

Weak in this chapter. Don't know how to proceed. Please help

Sorry for the typo made before

Answer 1

All the terms of $(x+y+z)^{15}$ are of order 15. The given term $x^3y^5z$ is of order 9 and so does not appear in $(x+y+z)^{15}$, that is, its coefficient is zero.

This can be seen more formally by using the binomial theorem twice, as suggested in another answer: \begin{eqnarray*} (x+y+z)^{15} &=& ((x+y)+z)^{15}\\ &=& \sum_{k=0}^{15}{15 \choose k}(x+y)^kz^{15-k}\\ &=& \sum_{k=0}^{15}{15 \choose k}\sum_{m=0}^k {k\choose m}x^my^{k-m}z^{15-k} \end{eqnarray*} One can see that the order of each term is $15$. Another way of seeing the result is to recognise that the only way to get $z^1$ in the sum is when $k=14$ in which case the sum reduces to \begin{eqnarray*} {15 \choose 14}\sum_{m=0}^{14} {14\choose m}x^my^{14-m}z \end{eqnarray*} and we see that the sum of the powers on $x^my^{14-m}$ are always $14$ so we can never get an $x^3y^5$ term multiplying $z^1$.

Answer 2

Hint. All terms of that form can be found from considering $x+y+z$ as the binomial $(x+y)+z.$ Taking the $15$th power of this gives a sum whose $k+1$th term is $${15\choose k}(x+y)^{15-k}z^k.$$ Then set $k=3$ and work out the rest.