Krdytkyu

A security company consists of $7$ men and $6$ women. Determine the number of ways to form the night shift team of $3$ men and $3$ women if it can include at most one of the oldest man or oldest woman.

I know the first step is to pick $3$ from each group which is

$_6C_3 cdot _7C_3 $

but I am not sure how to account for the "include at most one of the oldest man or oldest woman"

The answer is $_6C_3 cdot _7C_3 + _6C_3 cdot _5C_3 + _6C_2 cdot _5C_3 $

Also, how do I account for it in my calculations if I want to include the oldest man and oldest woman from the group?

Method 1: There are three admissible ways to select the teams:

1. The oldest man is included, but the oldest woman is not.


2. The oldest man is included, but the oldest man is not.


3. Neither the oldest man nor the oldest woman is included.

The oldest man is included, but the oldest woman is not: We must select the oldest man, two of the other six men, and three of the six women other than the oldest woman. We can do this in
$$binom{1}{1}binom{6}{2}binom{1}{0}binom{5}{3} = binom{6}{2}binom{5}{3}$$
ways.

The oldest woman is included, but the oldest man is not: We must select the oldest man, two of the other five women, and three of the six men other than the oldest man. We can do this in
$$binom{1}{0}binom{6}{3}binom{1}{1}binom{5}{2} = binom{6}{3}binom{5}{2}$$
ways.

Neither the oldest man nor the oldest woman is included: We must select three of the six men other than the oldest man and three of the five women other than the oldest women. We can do this in
$$binom{1}{0}binom{6}{3}binom{1}{0}binom{5}{3} = binom{6}{3}binom{5}{3}$$
ways.

Total: Since these cases are mutually exclusive and exhaustive, the number of ways a security team of three men and three women can be selected from a team of seven men and six women so that it contains at most one of the oldest man or the oldest woman is
$$binom{6}{2}binom{5}{3} + binom{6}{3}binom{5}{2} + binom{6}{3}binom{5}{3}$$

Method 3: We subtract the number of teams that include most the oldest man and the oldest woman from the total number of teams.

If there were no restrictions, we could choose any three of the seven men and any three of the six women on the team in
$$binom{7}{3}binom{6}{3}$$
ways.

If a team were to include both the oldest man and the oldest women, it would also include two of the other six men and two of the other five women. There are
$$binom{1}{1}binom{6}{2}binom{1}{1}binom{5}{2} = binom{6}{2}binom{5}{2}$$
such teams.

Hence, the number of admissible selections is
$$binom{7}{3}binom{6}{3} - binom{6}{2}binom{5}{2}$$