Question

A box contain $n$ pairs of shoes and $2r$ shoes are selected, $(r<n)$. The probability that there is exactly one pair is

• A. $\frac{{}^{n-1}{C}_{r-1}}{{}^{2n}{C}_{2r}}$
• B. $\frac{n{.}^{n-1}{C}_{r-1}}{{}^{2n}{C}_{2r}}$
• C. $\frac{({}^n{C}_1.{}^{n-1}{C}_{r-1})2^{r-1}}{{}^{2n}{C}_{2r}}$
• D. None of these

Answer 1

The correct option is C $\frac{({}^n{C}_1.{}^{n-1}{C}_{r-1})2^{r-1}}{{}^{2n}{C}_{2r}}$

The box contains $2n$ shoes. We can choose $2r$ shoes out of $2n$ shoes in ${}^{2n}{C}_{2r}$ ways.
We can choose one pair out of $n$ pairs in ${}^n{C}_1$ ways.
Now we have to avoid a complete pair.
While choosing $(2r-2)$ shoes out of remaining $(n-1)$ pairs of shoes, we first choose $(r-1)$ pairs out of $(n-1)$ pairs.
This can be done in ${}^{n-1}{C}_{r-1}$ ways.
From each of these $(r-1)$ pairs, choose $(r-1)$ single (Unmatching) shoes.
This can be done in $2^{r-1}$ ways.
Thus the number of favorable way is $\left({}^n{C}_1{.}^{n-1}{C}_{r-1}\right)2^{r-1}$
Hence, the probability of the required event $=\frac{\left({}^n{C}_1{.}^{n-1}{C}_{r-1}\right)2^{r-1}}{{}^{2n}{C}_{2r}}$