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]?

Question

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]?

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?

Answer 1

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 $$\binom{8}{2}\binom{8}{0}\binom{18}{1}\binom{18}{4} + \binom{8}{0}\binom{8}{2}\binom{18}{3}\binom{18}{2} + \binom{8}{1}\binom{8}{1}\binom{18}{2}\binom{18}{3}$$

Answer 2

If we have those $7$ card with the desired conditions then we can arrange them in $7!$ ways. So let's first see in how many ways we can select them. Base on the type (Power or Regular) and the color (Red or Black) of the cards we can partition our set of card into 4 disjoint sets $PR$ (which stands for Power and Red), $PB$, $RR$ or $RB$. Note that $|PR|=8=|PB|$ and $|RR|=18=|RB|$. Now we can select $7$ card only in on of the following ways:

Way #1: Sellecting $2\, PR$, $0\, PB$, $1\, RR$, and $4\, RB$, which can be done in $$ {8 \choose \color{red}2}\times{8 \choose 0}\times{18 \choose \color{red}1}\times{18 \choose 4} $$ different ways;

Way #2: Sellecting $1\, PR$, $1\, PB$, $2\, RR$, and $3\, RB$, which can be done in $$ {8 \choose \color{red}1}\times{8 \choose 1}\times{18 \choose \color{red}2}\times{18 \choose 3} $$ different ways;

Way #1: Sellecting $0\, PR$, $2\, PB$, $3\, RR$, and $2\, RB$, which can be done in $$ {8 \choose \color{red}0}\times{8 \choose 2}\times{18 \choose \color{red}3}\times{18 \choose 2} $$ different ways.

Thus the number of sellections is $$ {8 \choose \color{red}2}{8 \choose 0}{18 \choose \color{red}1}{18 \choose 4} + {8 \choose \color{red}1}{8 \choose 1}{18 \choose \color{red}2}{18 \choose 3} + {8 \choose \color{red}0}{8 \choose 2}{18 \choose \color{red}3}{18 \choose 2}, $$

and therefore the number of arrangements is $$ 7!\times \left( {8 \choose \color{red}2}{8 \choose 0}{18 \choose \color{red}1}{18 \choose 4} + {8 \choose \color{red}1}{8 \choose 1}{18 \choose \color{red}2}{18 \choose 3} + {8 \choose \color{red}0}{8 \choose 2}{18 \choose \color{red}3}{18 \choose 2} \right). $$