Question

There are 20 pairs of shoes in a closet. Four shoes are selected at random. The probability that there is exactly one pair is

• A. $\frac{{}^{20}{c}_1}{{}^{40}{c}_4}$
• B. $\frac{{}^{20}{C}_1\times {(}^{38}{C}_2-19)}{{}^{40}{C}_4}$
• C. $\frac{{}^{20}{C}_1{\times }^{38}{C}_2}{{}^{40}{C}_4}$
• D. $\frac{{}^{20}{C}_1{\times }^{19}{C}_2}{{}^{40}{C}_4}$

Answer 1

Answer: (B)

$\frac{{}^{20}{C}_1\times {(}^{38}{C}_2-19)}{{}^{40}{C}_4}$

Out of 20 pairs, we need to select 1 pair and the other 2 shoes should not form a pair.
Thus, this can be achieved in: ${}_1^{20}C\ast {(}_2^{19}C{\ast }_1^{2}C{\ast }_1^{2}C)$
(i.e. select one pair, then out of the remaining 19 pairs, select 2 pairs and from each pair select one shoe)
We can also say that: select one pair, then out of the remaining 38 shoes, select 2 shoes and eliminate the 19 cases where these 2 shoes form a pair, i.e. ${}_1^{20}C\ast {(}_2^{38}C-19)$
Therefore, probability $=\frac{{}_1^{20}C\ast {(}_2^{38}C-19)}{{}_4^{40}C}$