<s>[INST] Combinations company A company has $4$ workers who can build a wall, $5$ workers who can paint a wall and 2 workers who can do both. A job requires $ 3$ builders and $3$ painters. In how many ways can the company choose the workers for the job?
I found this resolution, is it correct?
We will have three disjoint cases:
Case 1:
When $0$ worker is present who can do both and this can be done in ${2\choose 0}$, then we must select more $3$ builders and painters and this can be done in ${4\choose 3}$ and ${5\choose 3}$ ways respectively. Hence, total number of ways is ${2\choose 0}\cdot {4\choose 3}\cdot {5\choose 3}=40$
Case 2:
When $1$ worker is present who can do both and this can be done in ${2\choose 1}$, then we must select more $2$ builders and painters and this can be done in ${4\choose 2}$ and ${5\choose 2}$ ways respectively. Hence, total number of ways is ${2\choose 1}\cdot {4\choose 2}\cdot {5\choose 2}=120$.
Case 3:
When $2$ worker is present who can do both and this can be done in ${2\choose 2}$, then we must select more $1$ builders and painters and this can be done in ${4\choose 1}$ and ${5\choose 1}$ ways respectively. Hence, total number of ways is ${2\choose 2}\cdot {4\choose 1}\cdot {5\choose 1}=20$.
Hence, total number of ways of selecting the workers are $40+120+20=\boxed{180}.$
 [/INST] Your solution is correct. I have the same result by making another approach. I´ve made a table which shows the number of builder (B), painter (P) and mixed workers (M)
      B P M

1.    3 3 0
2.    2 3 1
3.    3 2 1
4.    2 2 2
5.    1 3 2
6.    3 1 2

For 1. we calculate the number of ways to select $3$ builder out of $4$ builder and $3$ painter out of 4 painter: $\binom{4}{3}\cdot \binom{4}{3}=16$. Similar calculations for 2.-6.
$2. \ \binom{4}{2}\cdot \binom{4}{3}\cdot \binom{2}{1}=48 \qquad 3. \ \binom{4}{3}\cdot \binom{4}{2}\cdot \binom{2}{1}=48 \qquad 4. \ \binom{4}{2}\cdot \binom{4}{2}\cdot \binom{2}{2}=36$
$5. \ \binom{4}{1}\cdot \binom{4}{3}\cdot \binom{2}{2}=16 \qquad 6. \ \binom{4}{3}\cdot \binom{4}{1}\cdot \binom{2}{2}=16 $
The sum is $16+48+48+36+16+16=\boxed{180}$
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