We have $20$ individuals consisting of $10$ married couples. We pick $5$ of them. In how many ways can we pick them so that the 5 chosen are unrelated?? (that is, no two are married to each other)?
Try:
Let $m_i$ be the males such that and $f_i$ be females and we say they are related if $f_i=m_i$ We want to have a list $(a_1,a_2,a_3,a_4,a_5)$ such that such that we dont have a situation in which $f_i=m_i$
$a_1$ can be any of them 20 people, say it is $m_1$ so $a_2$ may not be $f_1$ or $m_1$( since it has been chosen already) and so for $a_2$ we have $18$ choices, say we pick $m_2$ for this case, then by same reasoning, we have $16$ choices for $a_3$ and so on. Thus,
in total we have $20 \times 18 \times 16 \times 14 \times 12 $ ways to pick those 5 individuals.
The answer in my books is ${ 10 \choose 5 }2^5$
What is mistake in here? My answer seems to be reasonable.
The justification for the book's answer is that we can choose five of the ten married couples in $\binom{10}{5}$ ways, then choose one of the two spouses from each selected couple, giving $\binom{10}{5}2^5$ possible selections.
The reason your answer is incorrect is that the order in which we choose the people does not matter. For simplicity, let's assume that each of the $20$ people has a different first name. Notice that choosing Peter, Amy, Xavier, Elisabeth, and Melissa in that order results in the same group of people as choosing Elisabeth, Amy, Peter, Melissa, and Xavier in that order. Consequently, we must divide your answer by the $5!$ orders in which the same group of five people can be selected. Observe that $$\frac{20 \cdot 18 \cdot 16 \cdot 14 \cdot 12}{5!} = \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 2^5}{5!} = \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5!}{5!5!} \cdot 2^5 = \frac{10!}{5!5!} \cdot 2^5 = \binom{10}{5}2^5$$