We have $10$ white, $9$ green and $7$ black balls. All balls are identical except for colour. While the solution for selecting number of ways in which one or more balls can be selected from these already exists which is $(1+10)(1+9)(1+7) - 1$, I want to know why we cannot use this method:
$ x_1$ = number of white balls chosen
$ x_2$ = number of green balls chosen
$ x_3$ = number of black balls chosen
$$ x_1 + x_2 + x_3 \leq 26 $$
$$x_1,x_2,x_3 \geq 0\ \mathrm{and}\ x_1 + x_2 + x_3 + k = 26, k \geq 0 $$
Answer = Coefficient of $x^{26}$ in $$(x^0 + x^1+\cdots+x^{10})(x^0+\cdots+x^9)(x^0+\cdots+x^7)(x^0 + \cdots+x^{25}) $$ This is giving me the wrong answer and I don’t know why.
What I understood from your question is that you wanna find out the no. of ways of selecting one or more balls. This answer, $$(1+10)(1+9)(1+7)−1=879 $$ is perfectly correct for this.
But, note that, when you find the coefficient of $x^{26}$ in
$$(x^0+x^1.......x^{10})(x^0+..........x^9)(x^0+.........+x^7)$$
what you find is the no. of ways of selecting ALL the balls and not at least one!
To find out the no. of ways of selecting at least one ball, you'll have to find the sum of the coefficients of $x^1,x^2,...x^{26}$, which in this case becomes (taking $x=1$)
$$(1+10)(1+9)(1+7)−1$$
Note that I've put $-1$ since $x^0$'s coefficient has to be excluded from the sum.