Find the coefficient of `a^4` in the product `(1+a)^4(2-a)^5` using binomial theorem.

Explanation

We know that

`(a+b)^n=^nC_0a^n+^nC_1a^(n-1)b^1+^nC_2a^(n-2)b^2+.......+^nC_(n-1)a^1b^(n-1)+^nC_nb^n`

Hence,
`(a+b)^4=^4C_0a^4+^4C_1a^3b^1+^4C_2a^2b^2+^4C_3a^1b^3+^4C_4b^4`

`(a+b)^4=(4!)/(0!(4-0)!)a^4+(4!)/(1!(4-1)!)a^3b^1+(4!)/(2!(4-2)!)a^2b^2+(4!)/(3!(4-3)!)a^1b^3+(4!)/(4!(4-4)!)b^4`

`(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4`

Putting `a=1` & `b=2a`

`(1+2a)^4=1^4+4(1)^3(2a)+6(1)^2(2a)^2+4(1)(2a)^3+(2a)^4`

`(1+2a)^4=1+8a+24a^2+32a^3+16a^4`

Also,
`(a+b)^n=^nC_0a^n+^nC_1a^(n-1)b^1+^nC_2a^(n-2)b^2+.......+^nC_(n-1)a^1b^(n-1)+^nC_nb^n`

`(a+b)^5=^5C_0a^5+^5C_1a^4b^1+^5C_2a^3b^2+^5C_3a^2b^3+^5C_4a^1b^4+^5C_5b^5`

`(a+b)^5=(5!)/(0!(5-0)!)a^5+(5!)/(1!(5-1)!)a^4b^1+(5!)/(2!(5-2)!)a^3b^2+(5!)/(3!(5-3)!)a^2b^3+(5!)/(4!(5-4)!)ab^4+(5!)/(5!(5-5)!)b^5`

`(a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5`

Putting `a=2` and `b=-a`

`(2-a)^5=(2)^5+5(2)^4(-a)+10(2)^3(-a)^2+10(2)^2(-a)^3+5(2)(-a)^4+(-a)^5`

`(2-a)^5=32-80a+80a^2-40a^3+10a^4-a^5`

`(1+2a)^4(2-a)^5=(1+8a+24a^2+32a^3+16a^4)(32-80a+80a^2-40a^3+10a^4-a^5)`

Coefficient of `a^4` is possible as follows



Therefore, coefficient of `a^4` is `-438`