Question

a standard deck of 52 playing cards has 4 suits with 13 different cards in each suit.

1. if the order in which the cards are dealt is not important, how many different 5 - card hands are possible?
2. how many different 5 - card hands are of the same color?

Explanation:

Step1: Calculate the number of ways to choose 5 cards from 52

We use the combination formula \(C(n,r)=\frac{n!}{r!(n - r)!}\), where \(n = 52\) and \(r=5\).
\(C(52,5)=\frac{52!}{5!(52 - 5)!}=\frac{52\times51\times50\times49\times48}{5\times4\times3\times2\times1}=2598960\)

Step2: Calculate the number of ways to choose 5 - card hands of the same color

There are 2 colors (red and black), and each color has 26 cards.
For each color, using the combination formula \(C(n,r)\) with \(n = 26\) and \(r = 5\)
\(C(26,5)=\frac{26!}{5!(26-5)!}=\frac{26\times25\times24\times23\times22}{5\times4\times3\times2\times1}=65780\)
Since there are 2 colors, the total number of 5 - card hands of the same color is \(2\times C(26,5)=2\times65780 = 131560\)

Answer:

1. The number of different 5 - card hands (when order doesn't matter) is \(2598960\)
2. The number of 5 - card hands of the same color is \(131560\)