Each of 3 committees has 1 vacancy which is to be filled from a group of 6 people. Find the number of ways the 3 vacancies can be filled if ,
(i) Each person can serve on atmost 1 committee.
(ii) There is no restriction on the number of com- mittees on which a person can serve.
(iii) Each person can serve on atmost 2 commit- tees.

Answer

To solve the problem, we will break it down into three parts based on the conditions provided.

### Part (i): Each person can serve on at most 1 committee.

1. **Select 3 people from 6**: Since each committee has 1 vacancy and each person can serve on at most 1 committee, we first need to select 3 people from the group of 6. The number of ways to choose 3 people from 6 is given by the combination formula \( \binom{n}{r} \):

\[ \text{Ways to choose 3 people} = \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]

2. **Assign the selected people to the committees**: After selecting 3 people, we can assign them to the 3 committees. The number of ways to assign 3 people to 3 committees is given by the permutation of 3, which is \( 3! \):

\[ \text{Ways to assign} = 3! = 6 \]

3. **Total ways for part (i)**: The total number of ways to fill the vacancies under this condition is the product of the two calculations:

\[ \text{Total ways} = \binom{6}{3} \times 3! = 20 \times 6 = 120 \]

### Part (ii): There is no restriction on the number of committees on which a person can serve.

1. **Select 1 person to fill all 3 vacancies**: We can choose 1 person from the 6 to fill all 3 vacancies. The number of ways to choose 1 person is:

\[ \text{Ways to choose 1 person} = \binom{6}{1} = 6 \]

2. **Select 2 people**: If we select 2 people, one of them will fill 2 vacancies and the other will fill 1. The number of ways to choose 2 people from 6 is:

\[ \text{Ways to choose 2 people} = \binom{6}{2} = 15 \]

Now, we need to assign these 2 people to the committees. We can choose which of the 3 committees will have 2 members (the other will have 1 member). The number of ways to choose which committee gets 2 members is:

\[ \text{Ways to assign} = \binom{3}{1} = 3 \]

Thus, the total ways for this case is:

\[ \text{Total for 2 people} = 15 \times 3 = 45 \]

3. **Select 3 people**: If we select 3 people, we can assign them to the committees in \( 3! \) ways:

\[ \text{Ways to choose 3 people} = \binom{6}{3} = 20 \]
\[ \text{Ways to assign} = 3! = 6 \]

Thus, the total ways for this case is:

\[ \text{Total for 3 people} = 20 \times 6 = 120 \]

4. **Total ways for part (ii)**: Adding all cases together gives us:

\[ \text{Total ways} = 6 + 45 + 120 = 171 \]

### Part (iii): Each person can serve on at most 2 committees.

1. **Select 2 people**: Similar to part (ii), we can select 2 people from 6. The number of ways to choose 2 people is:

\[ \text{Ways to choose 2 people} = \binom{6}{2} = 15 \]

We can assign them to the committees in the same way as before, which gives us:

\[ \text{Total for 2 people} = 15 \times 3 = 45 \]

2. **Select 3 people**: If we select 3 people, we can assign them to the committees in \( 3! \) ways:

\[ \text{Ways to choose 3 people} = \binom{6}{3} = 20 \]
\[ \text{Ways to assign} = 3! = 6 \]

Thus, the total ways for this case is:

\[ \text{Total for 3 people} = 20 \times 6 = 120 \]

3. **Total ways for part (iii)**: Adding all cases together gives us:

\[ \text{Total ways} = 45 + 120 = 165 \]

### Summary of Results:
- Part (i): 120 ways
- Part (ii): 171 ways
- Part (iii): 165 ways