Hi I am really having trouble trying to work out:
A group of 30 people consists of 15 women and 15 men, How many ways to:
-
form 10 pairs from the group?
-
divide the group into two groups (group 1 and group 2) of equal size?
-
divide the group into 2 equal groups, where each group in its own has as many men as women in it?
-
divide the groups into two groups of equal size such that group 1 contains at least 4 men?
-
divide the group into two groups, each having size at least one?
My answers: say $p_1, p_2,…,p_{30} \in$ group of 30 people
1)Not sure at all about this one, but I know that it is not ${30\choose 2}*{28\choose 2}*…*{10\choose 2}$ as there will be double counting.
2) ${30 \choose 15}$
Reasoning: Form groups of size 15. thus if group1 is $p_1,p_2,…,p_{15}$ then group2 would be $p_{16},p_{17},…,p_{30}$. Thus effectively dividing the group into 2 groups of equal size. I believe that each group is NOT arbitrary? so if another case: if group1 is $p_{16},p_{17},…,p_{30}$ then group2 would be $p_1,p_2,…,p_{15}$ this what we want and not double counting. Is this what they are asking?
3)Not to sure about this one, as we can make 15 packs of 1 man and 1 woman, but 15/2 does not make sense here, thus I suppose the best we can do is have 2 groups with 7 men and 7 women: ${15 \choose 7} * {15\choose 7}$ but this is not fully dividing the group.
Reasoning: pick 7 men from 15 and 7 women form 15. $m_1,m_2,…,m_7,w_1,w_2,…,w_7$. Are the groups arbitrary? and thus answer is: ${15 \choose 7} * {15\choose 7}/2$
4)${30\choose 15} – {18\choose 15}$
Reasoning: get all the ways to divide the group in half, then remove all the cases where the is not at least 4 men in group
5)if arbitrary groups: ${30\choose 1}+{30\choose 2}+…+{30\choose 15}$if not arbitrary groups:${30\choose 1}+{30\choose 2}+…+{30\choose 29}$
Reasoning: add the number of ways to form a group of size 1, to the number of ways to form a group of size 2, to…. I only do to ${30\choose 15}$ if they are asking for arbitrary groups or there would be double counting,
This is all the information that they give regarding the questions, and I am so lost. like for question 3, what determines the group size, if they want 15: 15, then it not possible.
I would greatly appreciate any help that you can offer to help me to understand how to answer these questions
Best Answer
$$\binom{30}{2}\binom{28}{2}\binom{26}{2}\binom{24}{2}\binom{22}{2}\binom{20}{2}\binom{18}{2}\binom{16}{2}\binom{14}{2}\binom{12}{2}$$ is the number of ways of selecting ten labeled pairs of two people from the group. Since the groups are not labeled, we must divide by the $10!$ ways we could select the same ten pairs of people, so the number of ways $10$ pairs could be formed from the group is $$\frac{1}{10!}\binom{30}{2}\binom{28}{2}\binom{26}{2}\binom{24}{2}\binom{22}{2}\binom{20}{2}\binom{18}{2}\binom{16}{2}\binom{14}{2}\binom{12}{2}$$
Your answer $$\binom{30}{15}$$ is correct if the groups are labeled, which I believe is the author's intent since the question refers to group 1 and group 2.
If the groups were unlabeled, we would have to divide by the $2!$ ways we could select the same two groups of $15$ people, so there would be $$\frac{1}{2!}\binom{30}{15}$$ ways to divide the group of $30$ people into two unlabeled groups of $15$ people.
Since there are an odd number of men and an odd number of women, this is impossible.
The problem indicates that the groups are labeled, so there would be $\binom{30}{15}$ ways of selecting the members of group 1 if there were no restrictions. From these selections, we must subtract those cases in which there are fewer than four men in group 1. A group of $15$ people with exactly $k$ men can be selected from $15$ men and $15$ women in $$\binom{15}{k}\binom{15}{15 - k}$$ ways. Hence, the number of ways we could select a group with fewer than four men is $$\binom{15}{0}\binom{15}{15} + \binom{15}{1}\binom{15}{14} + \binom{15}{2}\binom{15}{13} + \binom{15}{3}\binom{15}{12}$$ Therefore, the number of ways of dividing the group of $30$ people into two labeled groups of $15$ people if group 1 has at least four men is $$\binom{30}{15} - \binom{15}{0}\binom{15}{15} - \binom{15}{1}\binom{15}{14} - \binom{15}{2}\binom{15}{13} - \binom{15}{3}\binom{15}{12}$$
Observe that the groups are not labeled.
Suppose Amy is one of the women. For each of the remaining $29$ people, we have two choices: We can either place a person in Amy's group or the other group. That gives us $2^{29}$ divisions into two groups. However, we are prohibited from placing all $29$ of the other people in the same group as Amy since then the other group would be empty. Hence, there are $2^{29} - 1$ ways to divide the groups into two groups, each with size at least one.