Combination question on forming a committee

Question

A committee of 3 experts is to be selected out of a panel of 7 persons. 3 of them are engineers, 3 are managers, and 1 is both an engineer and manager. In how many ways can the committee be selected if it must have at least one engineer and one manager?

Answer 1

The total number of ways is:

$\binom {3} {1} \binom {3} {1} \binom {1} {1} + 2 \times \binom {3} {2} \binom {1} {1} + 2 \times\binom {3} {2} \binom {3} {1} =9 + (2 \times 3) + (2 \times 9) = 9 + 6 + 18 = 33.$

Answer 2

$$\text{Total ways}={1\choose 1}{3\choose 1}{3\choose 1} + 2\times{1\choose 1}{3\choose 2} + 2\times{3\choose 1}{3\choose 2} = 33$$

Answer 3

There are ${7\choose 3}=35$ ways to form a committee in total.

There are $2$ ways to form unfavorable committes: all three unique engineers or managers.

Hence: $35-2=33$.