Problem :
The number of ways in which four cards be selected from a pack of 52 cards such that there is exactly one pair ( pair has same number or alphabet)
What I tried :
Pairs may be : (2,2),(3,3) ….(K,K) etc.
$^{13}C_1 \times ^{13}C_1 \times ^{50}C_2$
But not getting the right answer.. please suggest..
Best Answer
There are a few problems with your solution.
For one, ${}^{13}C_1$ gives the number of ways to choose a card denomination without specifying a suit, but we have to specify a suit and a denomination to specify a card. To choose two cards of the same denomination, then, we must choose a denomination and must choose $2$ of the $4$ possible suits.
Now, to pick $2$ cards that are not a pair, and do not have the same denomination as the pair we've chosen, we must pick $2$ of the remaining cards in the deck, to be sure. However, ${}^{50}C_2$ gives the number of ways to choose any $2$ of the remaining cards, which allows for $4$ of a kind, $3$ of a kind, and $2$ pairs. We don't want any of these things to happen! Instead, we should pick $2$ of the denominations we haven't already picked from (How many such denominations are there? How many ways can we pick two of them?) and then pick a suit for each denomination.
Can you put it all together?