How to find the coefficient of $x^2y^3z^6$ in the expression $(x^2+y+z)^{10}$?

Question

How can find the coefficient of $x^2y^3z^6$ in the expression $(x^2+y+z)^{10}$?

I know this involves the binomial theorem, but I can't figure out how to solve it.

Answer 1

Hint:

The formula to find the coefficient of $x_1^{r_1}x_2^{r^2}\dots x_k^{r_k}$ in $(x_1+x_2+\dots+x_k)^n$ is:

$$\frac{n!}{r_1!r_2!\dots r_k!}$$

Here you take $x^2 =p$ say, then the problem reduced to finding coeff of $py^3z^6$ in the expression $(p+y+z)^{10}$.

Ans: $$\frac{10!}{1!3! 6!}$$

Answer 2

Think of it this way may be intuitively helpful. How many different $10$-tuples can you make out of $x^{2}, y, y, y, z, z, z, z, z, z$? The number of each "character" corresponds to each of the exponent given; using the notion of tuple is due to the commutative multiplication law. Now treat the same letters as if they were different (something like $y_{1},y_{2},y_{3},...$, which you consider mentally.); then you have got $10!$ tuples certainly. But this is way too many! We have to tailor it by observing that this number is actually $3! \times 6!$ (due to the duplicates of $y$ and $z$, respectively.) times the required number! So the required number is $$ \frac{10!}{3!6!}. $$

Answer 3

We can apply the binomial theorem twice in order to find the coeffcient. It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ of an expression.

We obtain \begin{align*} [x^2y^3z^6](x^2+y+z)^{10}&=[x^2y^3z^6]\sum_{k=0}^{10}\binom{10}{k}x^{2k}(y+z)^{10-k}\tag{1}\\ &=[y^3z^6]\binom{10}{1}(y+z)^9\tag{2}\\ &=10[y^3z^6]\sum_{k=0}^9\binom{9}{k}y^kz^{9-k}\tag{3}\\ &=10\binom{9}{3}\tag{4}\\ &=840 \end{align*}

Comment:

• In (1) we apply the binomial theorem once to $(x^2+(y+z))^{10}$.

• In (2) we select the coefficient of $x^2$ which is given when the index $k=1$.

• In (3) apply the binomial theorem again.

• In (4) we select the coefficient of $y^3z^6$ corresponding to the index $k=3$.