# Question d17b5

Mar 11, 2017

The Reqd. both Co-effs. are ""_12C_4=495=_12C_8.

#### Explanation:

The General ${\left(r + 1\right)}^{t h}$ Term, ${T}_{r = 1}$ in the Expansion of

${\left(a + b\right)}^{n}$ is, T_(r+1)=""_nC_ra^(n-r)b^r, r=0,1,2,...,n.

Since, n=12, T_(r+1)=""_12C_ra^(12-r)b^r, r=0,1,2,...,12......(star).

To find the co-eff. of ${a}^{8} {b}^{4}$, we compare this with, $\left(\star\right) .$

We find that,12-r=8,r=4; i.e., r=4.

This means that ${a}^{8} {b}^{4}$ is ${T}_{r + 1} = {T}_{5}$ in ${\left(a + b\right)}^{12.}$

Hence, the co-eff. of ${a}^{8} {b}^{4}$ is ""_12C_4=(12!)/{(4!)(12-4)!}

=(12!)/{(4!)(8!)}={cancel(12)(11)(10)(9)cancel(8!)}/{(1)(2)cancel(3)cancel(4)cancel(8!)}=(11)(5)(9)=495.#

Similarly, we can find the co-eff. of ${a}^{4} {b}^{8} ,$ which is,

${\text{_12C_8=}}_{12} {C}_{4} = 495.$

The later follows from the fact that,

$\text{_nC_r""=""_nC_(n-r)} .$

Enjoy Maths.!