Find the number of combinations and permutations of 4 letters taken from the word EXAMINATION.

Explanation

There are 11 letters
A,A,N,N,X,M,T,O.
then, number of combinations.
=coefficient of `x^(4)` in `(1+x+x^(2))^(3)(1+x)^(5)`
`[because 2A's,2I's,2N's,1E,1X,1M,1T and 1O]`
=Coefficient of `x^(4)` in `{(1+x)^(3)+x^(6)+3(1+x)^(2)x^(2)+3(1+x)x^(4)}(1+x)^(5)`
=Coeffient of `x^(4)` in
`{(1+x)^(8)+x^(6)(1+x)^(5)+3x^(2)(1+x)^(7)+3x^(4)(1+x)^(6)}`
`=.^(8)C_(4)+0+3.^(7)C_(2)+3=(8*7*6*5)/(1*2*3*4)+3*(7*6)/(1*2)+3=70+63+3`
`=136`
annd number of permutations.
=Coefficient of `x^(4)` in `4!(1+(x)/(1!)+(x^(2))/(2!))^(3)(1+(x)/(1!))^(5)`
=Coefficient of `x^(4)` in `4!(1+x+(x^(2))/(2))^(3)(1+x)^(5)`
=Coefficient of `x^(4)` in
`4!{(1+x)^(3)+(x^(6))/(8)+(3)/(2)(1+x)^(2)x^(2)+(3)/(4)x^(4)(1+x)}(1+x)^(5)`
=Coefficient of `x^(4)` in
`4!{(1+x)^(8)+(x^(6))/(8)(1+x)^(5)+(3)/(2)x^(2)(1+x)^(7)+(3)/(4)x^(4)(1+x)^(6)}`
`=4!{.^(8)C_(4)+0+(3)/(2)*.^(7)C_(2)+(3)/(4)}=24{(8*7*6*5)/(1*2*3*4)+(3)/(2)*(7*6)/(1*2)+(3)/(4)}`
`=8*7*6*5+(6(3*7*6)+6*3=1680+756+18=2454`