Of the 2,598,960 different five card hands from a 52 card deck, how many would contain 2 black cards and 3 red cards?

Question

Savannah Anderson · Accepted Answer

First we take the cards in order, and then we divide by the number of orders for the five cards, as order doesn't matter.

1st black card: 26 choices 2nd black card: 25 choices 1st red card: 26 choices 2nd red card: 25 choices 3rd red card: 24 choices A total of #26xx25xx26xx25xx24=10,140,000# But since all orders are equal, we divide by the number of orders for a five card hand: #5xx4xx3xx2xx1=5! =120#, so: Answer: #(10,140,000)/120=84,500#

Isaiah Anthony · Answer

undefined To find the number of five-card hands containing 2 black cards and 3 red cards from a standard 52-card deck, you can calculate it using combinations.

Number of ways to choose 2 black cards from 26 black cards: $ C(26, 2) $

Number of ways to choose 3 red cards from 26 red cards: $ C(26, 3) $

Multiply these two values to get the total number of five-card hands with 2 black cards and 3 red cards:

$$ C(26, 2) \times C(26, 3) = \frac{26!}{2! \times (26-2)!} \times \frac{26!}{3! \times (26-3)!} $$

$$ = \frac{26 \times 25}{2 \times 1} \times \frac{26 \times 25 \times 24}{3 \times 2 \times 1} $$

$$ = 325 \times 2600 $$

$$ = 845,000 $$

So, there are 845,000 different five-card hands containing 2 black cards and 3 red cards from a 52-card deck.