How many ways are there to select 3 red cards and 4 black cards so that there are 2 power cards [jack to ace] and 5 regular cards [2 to 10]

combinatoricsdiscrete mathematics

How many ways are there to select 3 red cards and 4 black cards from a deck of 52 playing cards so that there are 2 power cards [jack to ace] and 5 regular cards [2 to 10]?

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My attempt:

There are $$\binom{26}{3}\times\binom{26}{4}$$ ways to pick 3 red cards and 4 black cards from a standard deck of 52 cards. Moreover, there are $$\binom{36}{5}\times\binom{16}{2}$$ ways to pick 5 regular cards and 2 power cards from a standard deck of 52 cards. Now, the required answer is an overlap between the two selections, so that among 7 cards there are 3 red cards and 4 black cards and there are 5 regular cards and 2 power cards. How do I get the desired answer?

Best Answer

There are four cards of each rank, so there are $4 \cdot 4 = 16$ power cards as there are four power cards of each of the ranks J, Q, K, A. Of these, $8$ are red and $8$ are black. Of the other $52 - 16 = 36$ cards in the deck, $18$ are red and $18$ are black.

The number of ways of selecting exactly $a$ red power cards, $b$ black power cards, $c$ red regular cards, and $d$ black regular cards is $$\binom{8}{a}\binom{8}{b}\binom{18}{c}\binom{18}{d}$$

For the power cards, there are three possibilities:

  • Both are red, in which case one regular red card and four regular black cards must be selected, which can be done in $$\binom{8}{2}\binom{8}{0}\binom{18}{1}\binom{18}{4}$$ ways.
  • Both are black, in which case three regular red cards and two regular black cards must be selected, which can be done in $$\binom{8}{0}\binom{8}{2}\binom{18}{3}\binom{18}{2}$$ ways.
  • One is red and the other is black, in which case two regular red cards and three regular black cards must select, which can be done in $$\binom{8}{1}\binom{8}{1}\binom{18}{2}\binom{18}{3}$$ ways.

Hence, the number of ways to select $3$ red cards and $4$ black cards from a standard deck so that there are $2$ power cards and $5$ regular cards is $$\binom{8}{2}\binom{8}{0}\binom{18}{1}\binom{18}{4} + \binom{8}{0}\binom{8}{2}\binom{18}{3}\binom{18}{2} + \binom{8}{1}\binom{8}{1}\binom{18}{2}\binom{18}{3}$$

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