Question

Coefficient of x^2y^3z^4 in the expansion of (x+y+z) ^9 is equal to

Answer 1

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Step-by-step explanation:

Answer 2

The coefficient of the term $x^2y^3z^4$ in the expansion of $(x+y+z) ^9$ is 1260.

Given,

An expression: $(x+y+z) ^9$ .

To Find,

The coefficient of the term $x^2y^3z^4$.

Solution,

The method of finding the coefficient of the term $x^2y^3z^4$ in the expansion of $(x+y+z) ^9$  is as follows -

We know by the multi-binominal theorem that the coefficient of the term $x^py^mz^l$ in the expansion of $(ax+by+cz)^{n}$ is $\frac{n!*a^pb^mc^l}{p! \ m! \ l!}$.

Here in this problem, p = 2, m = 3, l = 4, n = 9, a = b = c = 1.

So putting all the values in the formula, the coefficient will be $\frac{9!*1^21^31^4}{2! \ 3! \ 4!} = \frac{1*2*3*4*5*6*7*8*9}{2*6*24}$

$=1260$

Hence, the coefficient of the term $x^2y^3z^4$ in the expansion of $(x+y+z) ^9$ is 1260.

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