Question Number 48443 by rahul 19 last updated on 24/Nov/18

Commented by rahul 19 last updated on 24/Nov/18

Commented by Meritguide1234 last updated on 24/Nov/18

Commented by rahul 19 last updated on 24/Nov/18

Answered by tanmay.chaudhury50@gmail.com last updated on 24/Nov/18
![((6!)/(a!b!c!))(2)^a (−x)^b (3x^2 )^c =((6!)/(a!b!c!))(2)^a (3)^c (−1)^b (x)^(b+2c) a+b+c=6 b+2c=4 b=0 c=2 so a=4 b=2 c=1 a=3 b=4 c=0 a=2 so required coefficient is [((6!)/(4!0!2!))×(2)^4 (3)^2 (−1)^0 +((6!)/(3!2!1!))×(2)^3 (3)^1 (−1)^2 + ((6!)/(2!4!0!))×(2)^2 (3)^0 (−1)^4 ] pls check and calculate... [15×16×9+60×8×3+15×4 = 2160 +1440+60 =3660](https://www.tinkutara.com/question/Q48445.png)
Commented by rahul 19 last updated on 24/Nov/18
=3660. (2160+1440+60)...I forgot the last case!
thanks sir ����
Commented by tanmay.chaudhury50@gmail.com last updated on 24/Nov/18
