Sum of coefficients in an multinomial expression.

Question

Q: The sum of all the coefficients of the terms in the expansion of $(x+y+z+w)^{6}$ which contain $x$ but not $y$ is:

What I tried to do was make pairs of two terms and the expand it as a binomial expression and then again expand the binomial in the resulting series which gave me an expression with lot of unknowns and I got stuck.

Any help would be appreciated. Thanks.

Answer 1

$$(x+y+z+w)^6=\{y+(x+w+z)\}^6=\cdots+(x+w+z)^6$$

$$(x+w+z)^6=\sum_{r=0}^6\binom6rx^{6-r}(w+z)^r$$

We need $r\ne6$

The sum of the reuqired coefficients should be $\sum_{r=0}^5\binom6r(1+1)^r$ (setting $x=w=z=1$)

$$=\sum_{r=0}^6\binom6r2^r-\binom662^6=(1+2)^6-2^6$$

Answer 2

Hint: What happens if you evaluate your expression in $x=1,y=0,z=1,w=1$? And in $x=0,y=0,z=1,w=1$?

Answer 3

Q: The sum of all the coefficients of the terms in the expansion of $(x+y+z+w)^{6}$ which contain $x$ but not $y$ is:

Sum of terms with no y : $3^6$ (y=0 rest all 1)
Sum of terms with no y and no x: $2^6$ (x,y=0 rest all 1)
Sum of terms with no y but x: $3^6-2^6=665$ (subtract the above)