Ten people are to be divided into 3 committees in such a way that every committee must have at least one member and no person can serve on all three committees. In how many ways ca... Ten people are to be divided into 3 committees in such a way that every committee must have at least one member and no person can serve on all three committees. In how many ways can this be done? Read more

Steps to Solve

1. Total ways without restrictions

First, consider the number of ways to assign each of the 10 people to one of the three committees without any restrictions. Each person can be assigned to any of the 3 committees, so there are $3^{10}$ ways.

2. Enforce the non-empty condition

We need to ensure that each committee has at least one member. We use the Principle of Inclusion-Exclusion (PIE).

• Subtract the cases where at least one committee is empty: There are $\binom{3}{1} \cdot 2^{10}$ ways to have at least one committee empty.
• Add back the cases where at least two committees are empty: There are $\binom{3}{2} \cdot 1^{10}$ ways to have at least two committees empty.

So, by PIE, the number of ways to have each committee with at least one person is: $3^{10} - \binom{3}{1}2^{10} + \binom{3}{2}1^{10} = 3^{10} - 3 \cdot 2^{10} + 3 = 59049 - 3 \cdot 1024 + 3 = 59049 - 3072 + 3 = 56080$

3. Enforce the condition that no person serves on all three committees

Now we need to subtract the cases where at least one person serves on all three committees. Since no person can serve on all three committees, we want to subtract the arrangements where at least one person is in all three committees. For each person, they can either be on all three committees or not. Thus, each person has $2^3 -1 = 7$ possibilities excluding no committee, but since they can not be assigned to all three, we should subtract one more possibility, so each person has $2^3 - 2=6$ possibilities. So each committee must have at least one member, we have $3^{10}$ ways. Since no person can serve on all three committees, each person can be assigned to any combination of the committees, except all three. Each person only has the option of belonging to subsets of {committee 1, committee 2, committee 3} without the subset {committee 1, committee 2, committee 3}. The number of such assignments is $2^3 - 1 = 7$. Since no person can serve on all three committees, subtract cases where one or more people serve on all three committees. The number of subsets, not including the empty set, is $2^3 -1 = 7$. The number of options for committees for each person is $2^3-1=7$. But we must exclude all the empty committees.

We want to subtract the cases where at least one person is in all three committees from the 56080 arrangements we got from the previous step. The complement is where each person can not be assigned to all three committees. This means that each of the 10 people cannot be in all three committees. For each person there are $2^3 - 1 = 7$ possible non-empty subsets of committees that person can be in. Since we also disallow membership in all three committees, there are $2^3 - 2 = 6$ options for each of the 10 people. Without the non-empty requirement, the number of ways is $6^{10}$.

4. Apply Principle of Inclusion-Exclusion (PIE) again

Consider the number of ways of assigning each person to the 3 committees such that no one is in all three committees. For each person, there are seven options. So initially, we have $7^{10}$. Now we need to ensure each committee has at least one member.

Using PIE: $6^{10} - \binom{3}{1}5^{10} + \binom{3}{2}4^{10} - \binom{3}{3}3^{10}$ But here the number of ways to place a person is $2^3 - 2 = 6$, because we do not want all three, and we can't have empty sets.

$6^{10} - 3(5^{10})+ 3(4^{10})- 3^{10}$ $60466176 - 3(9765625) + 3(1048576)-59049= 60466176 - 29296875 + 3145728 - 59049 = 34255980$

5. Final Answer

Thus, the number of ways to divide ten people into three committees such that each committee has at least one member and no person serves on all three committees is 34255980.