Question

This exercise refers to a standard deck of playing cards. Assume that 5 cards are randomly chosen from the deck. How many hands contain exactly two 3s and two 7s?

Answer 1

Answer:

Total number of cards in hand contain exactly two 3s and two 7s is 56.

Step-by-step explanation:

Given : This exercise refers to a standard deck of playing cards. Assume that 5 cards are randomly chosen from the deck.

To find : How many hands contain exactly two 3s and two 7s?

Solution :

In a standard deck of card has 52 cards distributed in 4 sets.

Each set has {2,3,4,5,6,7,8,9,10,J,Q,K,A} has 13 cards.

Number of 3s in 52 cards are 4.

Number of 7s in 52 cards are 4.

So, Getting exactly two 3s out of 4 is $^4C_2$

Getting exactly two 7s out of 4 is $^4C_2$

The fifth card is from remaining cards i.e. 52-8=44

Getting fifth card is $^{44}C_1$

Total number of cards in hand contain exactly two 3s and two 7s is

$T=^4C_2+^4C_2+^{44}C_1$

$T=\frac{4!}{2!(4-2)!}+\frac{4!}{2!(4-2)!}+\frac{44!}{1!(44-1)!}$

$T=\frac{4\times3\times2!}{2!\times 2}+\frac{4\times3\times2!}{2!\times 2}+\frac{44\times43!}{1\times 43!}$

$T=6+6+44$

$T=56$

Therefore, Total number of cards in hand contain exactly two 3s and two 7s is 56.