Question

A committee of 5 is to be chosen from a group of 9 people. Number of ways in which it can be formed if two particular persons either serve together or not at all and two other particular persons refuse to serve with each other, is

Answer 1

41

Answer 2

To solve this problem, we need to consider the two given conditions simultaneously. Let the group of 9 people be $P_1, P_2, \dots, P_9$.

Let the two particular persons who either serve together or not at all be A and B.
Let the two other particular persons who refuse to serve with each other be C and D.
The remaining 5 people are denoted as $R_1, R_2, R_3, R_4, R_5$.

We need to form a committee of 5 members.

We break down the problem based on the first condition (A and B serving together or not at all).

Case 1: A and B serve together.

If A and B are in the committee, then 2 positions are filled. We need to choose $5 - 2 = 3$ more members for the committee.
The remaining pool of people from whom we can choose these 3 members consists of the 7 people (C, D, and the 5 remaining people $R_1, \dots, R_5$).
Now, we apply the second condition: C and D refuse to serve with each other. This means C and D cannot both be in the committee.
To choose 3 members from these 7 people such that C and D are not both selected:

• Total ways to choose 3 from 7 people: $\binom{7}{3} = \frac{7!}{3!4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.
• Ways where both C and D are selected: If C and D are selected, we need to choose $3 - 2 = 1$ more person from the remaining 5 people ($R_1, \dots, R_5$). This can be done in $\binom{5}{1} = 5$ ways.
• Ways where C and D are not both selected (for Case 1): $\binom{7}{3} - \binom{5}{1} = 35 - 5 = 30$ ways.

Case 2: A and B do not serve at all.

If A and B are not in the committee, then we need to choose all 5 members for the committee from the remaining 7 people (C, D, and the 5 remaining people $R_1, \dots, R_5$).
Again, we apply the second condition: C and D refuse to serve with each other.
To choose 5 members from these 7 people such that C and D are not both selected:

• Total ways to choose 5 from 7 people: $\binom{7}{5} = \binom{7}{7-5} = \binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21$.
• Ways where both C and D are selected: If C and D are selected, we need to choose $5 - 2 = 3$ more people from the remaining 5 people ($R_1, \dots, R_5$). This can be done in $\binom{5}{3} = \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10$ ways.
• Ways where C and D are not both selected (for Case 2): $\binom{7}{5} - \binom{5}{3} = 21 - 10 = 11$ ways.

Total number of ways:

The total number of ways to form the committee is the sum of the ways from Case 1 and Case 2.
Total ways = (Ways from Case 1) + (Ways from Case 2) = $30 + 11 = 41$.

The final answer is $\boxed{41}$.