0
$\begingroup$

How to find the number of ways that two pairs of card and one card be chosen to choose a total of five cards from a standard deck of 52 cards? Ex: AA228, KKJJ3, 77885 are some examples of such choosing. Please provide the logic and not just the solution for this. And also, even though it's impossible to verify this by making a large list and counting them, but, how do we verify that the answer is correct?

  • 0
    It's not impossible to verify it by making a large list and counting them. There are only 52C5 or 2,598,960 hands of five cards. This is easily checked by making a list on a computer. – 2017-01-07

1 Answers 1

1

This is not really very difficult, and the proof is basically that the rules of combinations and choices are well understood.

To choose the two ranks of the pairs, you have $\binom {13}2$ options. This is a binomial coefficient representing the number of ways $2$ items can be chosen from a pool of $13$, often said "$13$ choose $2$". The actual number is calculated as: $$\binom {13}2 = \frac{13!}{(13-2)!2!} = \frac{13\cdot 12}{2\cdot 1} = 78$$

Then to chose the suits for each pair, you have $\binom {4}2 = 6$ options each. Finally to choose the remaining card there are $11$ remaining ranks and $4$ suits in each so there are $44$ options.

Multiply these together to reflect that the choices are independent: $$\binom {13}2\binom {4}2\binom {4}2\binom {44}1 = 78\cdot 6\cdot 6\cdot 44 = 123552 $$

  • 0
    why do we only consider one suit of cards? Aren't there 4.13C2(i.e 4 times 13 choose 2) ways for two cards to be chosen? – 2017-01-07
  • 1
    We do consider all suits of cards, but first, to get the pairs, we consider only rank. Then the $\binom 42$ gives us the chosen suits for each pair. – 2017-01-07