<s>[INST] Number of ways to pick $3$ balls from a box with $8$ balls if $3$ of the balls are identical and the other $5$ are all different 
A box contains 8 balls, of which 3 are identical and the remaining 5 are different from each other. 3 balls are to be picked out of the box; the order in which they are picked out does not matter. Find the number of different possible selections of 3 balls.

My Solution: I grouped them into cases
Case 1: All three balls are identical in which number of ways $= 1$
Case 2: Two of the balls are identical in which number of ways $= 5$
Case 3: One of the ball is from the $3$ identical ones in which number of ways $= 5 \cdot 4$ since there are $5$ ways to choose $1$ ball from the remaining $5$ balls and another $4$ ways to choose again from the remaining $4$ balls.
Case 4: I do not take from any of the $3$ identical balls so that leaves me with $5$ balls to choose $3$ from in which number of ways $= 5 \cdot 4 \cdot 3 = 60$.
Total number of ways $= 1+5+20+60 = 86$.
However, the answer given is $26$.
 [/INST] $\mathbf{\text{Generating Function Approach:}}$
The generating function of identical objects : $(1+x+x^2+x^3)$
The generating function of distinct objects : $(\binom{5}{0}x^0+ \binom{5}{1}x^1+\binom{5}{2}x^2+\binom{5}{3}x^3+\binom{5}{4}x^4 +\binom{5}{5}x^5)$
Then , find the coefficent of $x^3$ in the expansion of $$(1+x+x^2+x^3)\bigg(\binom{5}{0}x^0+ \binom{5}{1}x^1+\binom{5}{2}x^2+\binom{5}{3}x^3+\binom{5}{4}x^4 +\binom{5}{5}x^5\bigg)$$
So , the answer is $26$
$\mathbf{\text{NOTE:}}$ Realize that the coefficents of generating functions determine th number of ways to select desired objects , for example , the coefficients of identical objects is always $1$ , because there is only one way to select identical objects , but when they are distinct , the coeffients are determined by combination.For example $\binom{5}{3}x^3$ means the number of selection of three distinct objects among $5$ distinct objects
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