Question

Consider a standard deck of 52 cards. How many 5 card hands have 2 pairs in them?

Answer 1

Hint: This question is from the topic permutation and combination. In this question, first we will find out the number of ways of selecting two cards from 13 different cards and one card from the 11 different cards. After that, we will find the number of ways of selecting cars according to four different suits. After that, we will solve the question and find the answer.

Complete step-by-step solution:
Let us solve this question.
In this question, we have given a deck of 52 cards. We have to find the number of ways if we pick up 5 cards having two pairs of the same rank in them from the deck of 52 cards.
So, we can say that we will take 2 cards of the same rank, two other cards of the same rank (but different from the first), and one other card of a different rank.
First of all, we will find the ways of selecting cards of the same rank. As we know that a deck of 52 cards have 13 different ranks that are Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, king, queen, and jack. So, we will select two cards of different ranks from 13 different rank cards.
The number of ways of selecting 2 cards from 13 different cards will be \[{}^{13}{{C}_{2}}\].
Now, for selecting the third card from 11 different ranks (we have left 11 from 13, because we have already chosen the two cards), the number of ways of selecting will be \[{}^{11}{{C}_{1}}\].
So, now we have selected 5 cards in which two pairs of different rank and one other card of any different rank. The number of ways of selecting will be \[{}^{13}{{C}_{2}}\times {}^{11}{{C}_{1}}\]
As we know that in a deck of 52 cards, there are 13 cards of different ranks and each card has a different suit that are spade, heart, diamond, and club. So, there are 4 different suits. We will have to choose 2 suits from 4 different suits.
So, the number of ways of selecting for 2 different suits for the first pair of card of same rank will be \[{}^{4}{{C}_{2}}\]
 And, the number of ways of selecting for 2 different suits for the second pair of card of same rank will be\[{}^{4}{{C}_{2}}\]
For the fifth card, we will choose one suit from the 4 suits.
So, the number of ways of selecting the 5th card from the four suits will be \[{}^{4}{{C}_{1}}\].
Hence, the number of ways of selecting 5 cards having 2 pairs in them will be \[{}^{13}{{C}_{2}}\times {}^{11}{{C}_{1}}\times {}^{4}{{C}_{2}}\times {}^{4}{{C}_{2}}\times {}^{4}{{C}_{1}}\]
That is equal to
\[\dfrac{13!}{2!\left( 13-2 \right)!}\times \dfrac{11!}{1!\left( 11-1 \right)!}\times \dfrac{4!}{2!\left( 4-2 \right)!}\times \dfrac{4!}{2!\left( 4-2 \right)!}\times \dfrac{4!}{1!\left( 4-1 \right)!}=78\times 11\times 6\times 6\times 4=123552\]

Note: As this question is from the topic of permutation and combination and binomial theorem, so we should have a better knowledge in those topics to solve this type of question easily. We should know the formula from the topic binomial theorem that is \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\], where n and r are integers and n should always be greater than r. We should also know that\[a!=a\times \left( a-1 \right)\times \left( a-2 \right)\times ........\times 3\times 2\times 1\]. Don’t forget these formulas. They are very helpful in this type of question.