Question

Eight chairs are numbered 1 to 8. Two women and three men wish to occupy one chair each. First, the women choose the chairs from amongst the chairs marked 1 to 4, and then the men select the chairs from amongst the remaining. The number of possible arrangements is,
A. ${}^{6}{{C}_{3}}\times {}^{4}{{C}_{2}}$
B. ${}^{4}{{P}_{2}}\times {}^{4}{{P}_{3}}$
C. \[{}^{4}{{C}_{2}}+{}^{4}{{P}_{3}}\]
D. None of these

Answer 1

Hint: We will first start by using the method of selecting r objects out of n objects that is ${}^{n}{{C}_{r}}$ for finding the ways in which we can select two chairs for women and three for men. Then we will permute the men and women among themselves.

Complete step-by-step answer:
Now, we have been given 8 chairs which are numbered from 1 to 8. Also, it has been given that women choose the chairs from amongst the chairs marked 1 to 4, and then men select from remaining chairs. In total there are 2 women and three men who wish to occupy one chair each.
Now, we know the number of ways of selecting r objects among n is ${}^{n}{{C}_{r}}$. So, we have the ways in which we can choose two chairs among four numbered 1 to 4 is ${}^{4}{{C}_{2}}$ and we can arrange the women then in 2! ways. Also, we have the ways of selecting 3 chairs among the rest 6 chairs is ${}^{6}{{C}_{3}}$ and in them we can permute the men in 3! ways.
So, in total we have number of possible arrangements as,
${}^{4}{{C}_{2}}\times 2!\times {}^{6}{{C}_{3}}\times 3!$
Now, we know that ${}^{n}{{C}_{r}}={}^{n}{{P}_{r}}\times r!$.
Therefore, we have,
Total ways $={}^{4}{{P}_{2}}\times {}^{6}{{P}_{2}}$
Hence, the correct option is (D).

Note: It is important to note that we have used a fact that ${}^{n}{{C}_{r}}={}^{n}{{P}_{r}}\times r!$. This can be understood as we know that${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!}$ and ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$. So, substituting this we have ${}^{n}{{C}_{r}}={}^{n}{{P}_{r}}\times r!$.