Question

Four cards are drawn at random from a pack of $52$ playing cards. The probability that the draw contains exactly one pair (i.e., having the same number) is

• A. $\frac{13{\times }^{12}{C}_2\times 4^2}{{}^{52}{C}_4}$
• B. $\frac{13{\times }^4{C}_2{\times }^{12}{C}_2\times 4^2}{{}^{52}{C}_4}$
• C. $\frac{13{\times }^4{C}_2{\times }^{48}{C}_2}{{}^{52}{C}_4}$
• D. none of these

Answer 1

Answer: (B)

$\frac{13{\times }^4{C}_2{\times }^{12}{C}_2\times 4^2}{{}^{52}{C}_4}$

There are 13 sets of the same card. We can choose 1 of the 13 in $C_1^{13}=13$ ways. In that set, there are 4 cards. We can select 2 of the 4 in $C_2^4=6$ ways.The other 2 cards should not form a pair. Hence, we cannot select any card from the set we have selected the first 2 cards. Thus, there are 12 sets left. We need to select 2 sets from the 12 and after that 1 card out of 4 cards from each set. This can be done in $C_2^{12}\times C_1^4\times C_1^4$ ways.
Hence, probability $=\frac{13{\times }^4{C}_2{\times }^{12}{C}_2{\times }^4{C}_1{\times }^4{C}_1}{{{}^{52}C}_4}$