A group has 10 members and wish to divide into 3 committees. Everyone serves on at least one committee and no one serves on more than two committees. Each committee must have 4 mem... A group has 10 members and wish to divide into 3 committees. Everyone serves on at least one committee and no one serves on more than two committees. Each committee must have 4 members. How many ways are there for the 10 to be split into 3 committees? Read more

Steps to Solve

1. Determine the number of members serving on two committees

Since each committee has 4 members and there are 3 committees, there is a total of $4 \times 3 = 12$ "committee seats". However, there are only 10 members. Thus $12 - 10 = 2$ members must serve on two committees.

2. Choose the two members that serve on two committees

We must choose 2 members out of 10 to be on two committees. The number of ways to choose these 2 members is $\binom{10}{2}$.

$$ \binom{10}{2} = \frac{10!}{2!(10-2)!} = \frac{10!}{2!8!} = \frac{10 \times 9}{2 \times 1} = 45 $$

3. Assign the two members to the committees

Let the two members who serve on two committees be A and B. Since each of them is on two committees out of the three, they must be on committees 1 & 2, 1 & 3, or 2 & 3. We can choose which two committees A is on in $\binom{3}{2}=3$ ways. Then we can choose which two committees B is on in $\binom{3}{2}=3$ ways. Since A and B can be assigned independently, there are $3 \times 3 = 9$ ways to assign them to the committees. However, as we shall later see, the order of the committees doesn't matter so we will correct for this overcounting later.

4. Determine the number of members on only one committee

Since 2 members serve on two committees, that means $10 - 2 = 8$ members serve on only one committee.

5. Determine the remaining members to choose for each committee

Each committee needs 4 members. Since each of the two members are each on two committees there are already two people on two committees. This means that we have $4-2 = 2$ spots to fill in those two committees. The last committee needs all 4 of its members to be on only 1 committee. This means that we have to put 2 members only on committees 1 and 2, and 4 members only on committee 3. This accounts for all 8 members.

6. Fill committees 1 and 2 (accounting for only members on 1 committee)

We need to choose 2 members for committee 1 from the 8 remaining, and then 2 members for committee 2 from the 6 remaining. This can be done in $\binom{8}{2}\binom{6}{2}$ ways.

$$ \binom{8}{2} = \frac{8!}{2!6!} = \frac{8 \times 7}{2 \times 1} = 28 $$

$$ \binom{6}{2} = \frac{6!}{2!4!} = \frac{6 \times 5}{2 \times 1} = 15 $$

$$ \binom{8}{2} \binom{6}{2} = 28 \times 15 = 420 $$

7. Fill committee 3

Now we choose 4 members from the remaining 4 to be on committee 3. This can be done in $\binom{4}{4} = 1$ way.

8. Calculate the total arrangements (overcounting)

Multiply all the results together to get the total arrangements.

$$ \binom{10}{2} \times \binom{3}{2} \times \binom{3}{2} \times \binom{8}{2} \times \binom{6}{2} \times \binom{4}{4} = 45 \times 3 \times 3 \times 28 \times 15 \times 1 = 170100$$

9. Correct for overcounting

Since the order of the three committees does not matter, we must divide by the number of permutations of the first two committees we populated. The third committee is determined by the first two. The order of placement of the two members on two teams did not matter so we must divide by 2!. So, the overall correction for overcounting is dividing by $2! = 2$.

$$ \frac{170100}{2} = 85050 $$