Question

three cards are drawn from a pack of 52 cards.how many selections will contain
1)all red cards
2)two face cards and one ace card
3)one is club and the other two are diamond cards
4)all cards have same suit


ans this question in detail....!!​

Answer 1

Answer:There are 26 red(13 hearts +13 diamonds) cards in a pack of 52 cards . Now, 3 cards out of total 52 cards can be selected in C(52, 3) = 52!/3! 49! =22100 no. of ways and 3 red cards out of 26 red cards can be selected in C(26, 3)= 26!/3! 23! =2600 no. of ways. Hence the required probability = 2600/22100 = 26/221 = 2/17.

There are 4C1 * 48C2 = 4512 ways to have exactly one ace, 4C2 * 48C1 = 288 ways to have exactly two aces and 4C3 * 48C0 = 4 ways to have three aces, for a total of 4804 ways.

You also could have taken the total number of three card hands (52C3 = 22,100) and subtracted the number of hands with no aces (48C3 = 17,296) for a difference of 4804.

Step-by-step explanation:please mark me brainliest???