In how many ways we can place $9$ different balls in $3$ different boxes such that in every box at least $2$ balls are placed?
Approach 1:
Let $x_1$, $x_2$, $x_3$ denote the number of balls in boxes $1$,$2$, and $3$ respectively. Now,solving $x_1+x_2+x_3 =9$ for $2≤x_1≤x_2≤x_3≤5$ ($5= 9-$(min of $x_1$) $-$ (min of $x_2$)), we get $(x_1,x_2,x_3) ≡ (2,2,5),(2,3,4),(3,3,3)$
Now, we have to divide $9$ objects in these groups and then distribute into boxes. So the required answer is $$\frac{9!}{2!2!5!}×\frac{1}{2!}×3! +\frac{9!}{2!3!4!}×3! + \frac{9!}{3!3!3!}×\frac{1}{3!}×3!$$ which simplifies to be $\boxed{11508}$. This is the correct answer.
Approach 2 (PIE):
At least $2$ $=$ all $-$ at most $1$. Now, considering $1$ box contains no ($0$) elements, the number of cases corresponding to that is ${3 \choose 1}×2^9$ and the number of cases when $2$ boxes contain no elements will be ${3 \choose 2}×1^9$. Now, number of cases when $1$ box contains only $1$ element is ${3 \choose 1}×{9 \choose 1}×2^8$ (I.e., number of ways to choose 1 box × number of ways to choose 1 ball × remaining distribution) and the number of cases when 2 boxes contain only 1 element is ${3 \choose 2}×{9 \choose 2}×1^7$ . So, using PIE, the required answer should be $$3^9 -({3 \choose 1}2^9 – {3 \choose 2}1^9) – ({3 \choose 1}{9 \choose 1}2^8 – {3 \choose 2}{9 \choose 2}1^7)$$ that is $11346$. This answer is wrong but the "closeness" of the answer to the correct one indicates that I have deducted some cases. Please point out my mistake in approach 2.
Best Answer
There are two small mistakes otherwise your work using the second approach is also correct. It should be -
$ \displaystyle \small 3^9 - \underbrace{\left[{3 \choose 1}2^9 - {3 \choose 2}1^9 \right]}_{\Large {(a)}} - \underbrace{\left[{3 \choose 1}{9 \choose 1} (2^8 - \color { blue } {2}) - \color { blue } {2} \cdot {3 \choose 2}{9 \choose 2}1^7 \right]}_{\Large (b)}$
Explanation on what I added in $ \color {blue} { \text {blue}}$ in $(b)$:
$1)$ Please note $2^8$ also includes two arrangements where one of the two boxes is empty but those are already counted in $(a)$. So we need to subtract those two arrangements.
$2)$ Please note when we choose two balls for two boxes, there are $2$ ways to place them in two boxes such that each box has one ball.