How many variables in the multinomial expansion have an exponent that's different from 2? Multinominal: $(a + b + c + d)^{10}$
Question: How many variables in the multinomial expansion have an exponent that's different from 2?
I want to solve it using generating functions.
So we can write: $t_1 + t_2 + t_3 + t_4 = 10$, When $t_i$ cannot use $x^2$.
That equals to the generating function:
$$
(1+x+x^3+x^4...)^4 = (1+x+\frac{x^3}{1-x})^4
$$
But now what? How can I keep expending this till I find the coefficient of $x^{10}$?
 A: $(1 + x + x^3 + x^4 + ...)$ can be simplified as $[(1 - x)^{-1} - x^2]$.
Now, we had to find coefficient of $x^{10}$ in ${[(1 - x)^{-1} - x^2]}^4$
${[(1 - x)^{-1} - x^2]}^4$ = $\binom{4}{0}$ $[({1 - x}) ^ {-4}]$ - $\binom{4}{1}$ $[({1 - x}) ^ {-3}]$ $x^2$ + $\binom{4}{2}$ $[({1 - x}) ^ {-2}]$ $x^4$ - $\binom{4}{3}$ $[({1 - x}) ^ {-1}]$ $x^6$ + $\binom{4}{4}$ $x^8$
Hence the answer will be $\binom{13}{3}$ - 4 $\binom{10}{2}$ + 6 $\binom{7}{1}$ - 4 $\binom{4}{0}$ i.e. 144
A: You can rewrite $$(1+x+x^3+x^4...)^4=(1+x)^4(1+x^3+x^5+x^7+...)^4$$
$$=(x^4+4x^3+6x^2+4x+1)(1+x^3+x^5+x^7+x^9+...)^4$$
Now first look at the $x^4$ term: the only way to add exponent up to $10$ is $4+3+3=10$, therefore we have ${4\choose 2}=6$ ways.
Next look at $4x^3$ term, the only way is $3+7=10$, therefore we have $4{4\choose 1}=16$ ways.
Next look at $6x^2$ term, the only way is $2+3+5=10$, therefore $6{4\choose 1}{3\choose 1}=72$ ways.
Next look at $4x$ term, we have $1+9=1+3+3+3=10$, therefore $4({4\choose 1}+{4\choose 3})=32$ ways.
Next look at $1$ term, we have $0+3+7=0+5+5=10$, therefore ${4\choose1}{3\choose1}+{4\choose 2}=18$ ways.
Finally add them all up, $6+16+72+32+18=144$ ways.
