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Question: A box contains \(n\) pairs of shoes and \(2r\) shoes are selected \[(r < n)\]. The probability that ...

A box contains nn pairs of shoes and 2r2r shoes are selected (r<n)(r < n). The probability that there is exactly one pair is
A.n−1Cr−12nC2r\dfrac{{{}^{n - 1}{C_{r - 1}}}}{{{}^{2n}{C_{2r}}}}
B.n.n−1Cr−12nC2r\dfrac{{n.{}^{n - 1}{C_{r - 1}}}}{{{}^{2n}{C_{2r}}}}
C.(nC1.n−1Cr−1)2r−12nC2r\dfrac{{({}^n{C_1}.{}^{n - 1}{C_{r - 1}}){2^{r - 1}}}}{{{}^{2n}{C_{2r}}}}
D.None of these

Explanation

Solution

We shall use the concept of combinations to solve the given problem. We will first find the total number of outcomes using combination. Then we will find the number of favourable outcomes by finding the different number of pairs of getting one pair of shoes. Finally, we will use the formula of probability to find the correct option.

Formula used:
We will use the following formulas:
1.If A is an event, then P(A)=P(A) = Number of favourable outcomes ÷\div Total number of outcomes
2.The number of ways to select rr objects out of nn objects is C(n,r)=nCrC(n,r) = {}^n{C_r}

Complete step-by-step answer:
It is given that the box contains nn pairs of shoes. We know that each pair will contain 2 shoes. So, the box contains 2n2n shoes in all.
We have to select 2r2r out of these 2n2n shoes. This can be done in 2nC2r{}^{2n}{C_{2r}} ways.
Hence, the total number of outcomes is 2nC2r{}^{2n}{C_{2r}}.
We need to find the probability that there is exactly one pair.
Out of nn pairs, one pair can be selected in nC1{}^n{C_1} ways.
The remaining number of pairs is n−1n - 1. From these, we have to select r−1r - 1 pairs which are mismatched. This can be done in n−1Cr−1{}^{n - 1}{C_{r - 1}} ways.
The r−1r - 1 mismatched pairs that we select have to consist of single mismatched shoes. The number of ways in which these can be selected is 2r−1{2^{r - 1}}.
Now multiplying all the number of ways of selecting a pair of shoes we will get the number of favourable outcomes as (nC1.n−1Cr−1)2r−1({}^n{C_1}.{}^{n - 1}{C_{r - 1}}){2^{r - 1}}.
If A is the event of selecting exactly one pair, then using the formula of probability, P(A)=P(A) = Number of favourable outcomes ÷\div Total number of outcomes, we get
P(A)=(nC1.n−1Cr−1)2r−12nC2rP(A) = \dfrac{{({}^n{C_1}.{}^{n - 1}{C_{r - 1}}){2^{r - 1}}}}{{{}^{2n}{C_{2r}}}}.
Hence, the correct option is C.

Note: We should not get confused between combinations and permutations. In permutations, the order of elements is to be taken care of, whereas in combinations, the order of elements does not matter. Here, the order in which we draw the cards is not a considerable factor, and this is the reason we have used combinations.