Let's denote the total number of men as M and the total number of women as W. The problem is to select 3 men and 2 women such that one specific man and one specific woman are always selected. Step 1: Since one man is always selected, we need to select the remaining 2 men from the remaining M-1 men.
Number of ways to select 2 men out of (M-1) men = (M−12)\binom{M-1}{2}(2M−1​) Step 2: Similarly, since one woman is always selected, we need to select the remaining 1 woman from the remaining W-1 women.
Number of ways to select 1 woman out of (W-1) women = (W−11)\binom{W-1}{1}(1W−1​) Step 3: Multiply the results from Step 1 and Step 2 to get the total number of ways:
(M−12)×(W−11)\binom{M-1}{2}\times \binom{W-1}{1}(2M−1​)×(1W−1​) From the sources, a similar example states the number of ways to select is (42)×(51)=30\binom{4}{2}\times \binom{5}{1}=30(24​)×(15​)=30 when 3 men and 2 women are selected such that one man and one woman are always selected (assuming 5 men and 6 women total, for instance). So the formula and logic apply as above. If you provide the total number of men and women, I can calculate the exact number for you. Without those numbers, the general formula is: Number of ways = (M−12)×(W−11)\binom{M-1}{2}\times \binom{W-1}{1}(2M−1​)×(1W−1​) For example, if total men = 5 and total women = 6,
Number of ways = (42)×(51)=6×5=30\binom{4}{2}\times \binom{5}{1}=6\times 5=30(24​)×(15​)=6×5=30 ways. This matches the standard answer for this type of problem. Thus, the number of ways to select 3 men and 2 women such that a particular man and a particular woman are always selected is (M−12)×(W−11)\binom{M-1}{2}\times \binom{W-1}{1}(2M−1​)×(1W−1​). If M=5 and W=6, this equals 30 ways.