Answer

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**Hint:**In the above question you have to find the coefficient of ${a^8}{b^6}{c^4}$ in the expansion of ${(a + b + c)^{18}}$. The general term for expansion which you can use is ${(x + y + z)^z} = \dfrac{{n!}}{{p!q!r!}}{x^p}{y^q}{z^r}$, where p, q and r are natural numbers. So let us see how we can solve this problem.

**Step by step solution:**

Given that we have to find the coefficient of ${a^8}{b^6}{c^4}$ in the expansion of ${(a + b + c)^{18}}$. The general term in expansion of ${(x + y + z)^z} = \dfrac{{n!}}{{p!q!r!}}{x^p}{y^q}{z^r}$ , where p + q + r = n and p, q, r $\in$ {1, 2, 3…, n}

We have to expand ${(a + b + c)^{18}}$

$= \dfrac{{18!}}{{p!q!r!}}{a^p}{b^q}{c^r}$

The term contains ${a^8}{b^6}{c^4}$ according to the question

$= \dfrac{{18!}}{{8!6!4!}}{a^8}{b^6}{c^4}$

**Therefore, the coefficient of ${a^8}{b^6}{c^4}$ in the expansion of ${(a + b + c)^{18}}$ is $\dfrac{{18!}}{{8!6!4!}}$.**

**Note:**

There is an alternative method of solving the above question, let us see that also.

Given the term for expansion is ${(a + (b + c))^{18}}$

$= {}^{18}{C_r}{a^{18 - r}}{(b + c)^r}$

$= {}^{18}{C_r}{}^r{C_k}{a^{18 - r}}{b^{r - k}}{c^k}$

From ${a^8}{b^6}{c^4}$ , we know that the exponent of a, b and c are 8, 6 and 4 respectively. So

$\Rightarrow 18{\text{ }} - {\text{ }}r = {\text{ }}8$

Subtracting both sides with 18, we get

$\Rightarrow {\text{ }} - {\text{ }}r = {\text{ }} - 10$

$\Rightarrow r = 10$

Also, r – k = 6 since the coefficient of b is 6 and k = 10 – 6 = 4

Therefore, the coefficient becomes,

$^{18}{C_{10}}^{10}{C_4}$

${ = ^{18}}{C_{10}}^{10}{C_6}$

Therefore, we get $= \dfrac{{18!}}{{8!6!4!}}$ as the coefficient of ${a^8}{b^6}{c^4}$.

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