If coefficient of 4^{th} term in expansion of {(a + b)^n} is 56 , then n is

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7.Binomial Theorem

easy

If the coefficient of $4^{th}$ term in the expansion of ${(a + b)^n}$ is $56$, then $n$ is

A

$12$

B

$10$

C

$8$

D

$6$

Solution

(c) ${T_4} = {T_{3 + 1}}$

$ = \,{}^n{C_3}\,\,\,{a^{n – 3}}{b^3}$

==> ${}^n{C_3} = 56$

==> $\frac{{n!}}{{3\,!\,\,\,\left( {n – 3} \right)\,!}} = 56$

==> $n(n – 1)(n – 2) = 56.6$

==> $n(n – 1)(n – 2) = 8\,.\,7\,.\,6$

==> $n = 8$.

Standard 11

Mathematics

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