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  • Find the coefficient of a7b8c5 in (ab+bc+ac)10?Option: 1 <img alt="\frac{10!}{3!5!4!}" src="https://

Find the coefficient of a7b8c5 in (ab+bc+ac)10?

Option: 1

\frac{10!}{3!5!4!}


Option: 2

\frac{10!}{7!5!6!}


Option: 3

\frac{10!}{3!5!6!}


Option: 4

None of these


Answers (1)

best_answer

Multinomial Theorem

\mathrm{\left ( x_1+x_2+x_3+.....+x_k \right )^n=\sum \frac{n!}{\alpha_1!\;\alpha_2!\; \alpha_3!\;...\;\alpha_k!}\mathrm{\left (x_1^{\alpha_1}\cdot x_2^{\alpha_2}\cdot x_3^{\alpha_1}\cdot ....\cdot x_k^{\alpha_k} \right )}}

 

Now,

\\(ab+bc+ac)^{10} \\\\= \frac{10!}{p!q!r!}.(ab)^p.(bc)^q.(ac)^r \\\\=\frac{10!}{p!q!r!}.(a)^{p+r}.(b)^{p+q}.(c)^{q+r}

To get coefficient of a7b8c5, 

p + r = 7

p + q = 8

q + r = 5

Adding these three equations

2 ( p + q + r ) = 20

p + q + r = 10, which matches with the condition needed for multinomial ( p + q + r = n , so we will have a term with given powers of a, b and c in the expansion)

So, q = 3, r = 2, p = 5

Coefficient is \frac{10!}{5!3!2!}

 

 

But, the sum of these powers is not 10

So, no such term is possible in this expansion

Hence, option D is correct

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