# Permutation and Combination

Suppose that you are dealt a hand of 8 cards from a standard deck of 52 playing cards.

a) What is the probability that you are dealt no diamonds and exactly 1 jack?

b) What is the probability that you are dealt no more than 1 spade?

c) What is the probability that you are dealt at least 2 sevens and at least 3 fives?

https://brainmass.com/math/combinatorics/permutation-combination-437042

## SOLUTION This solution is **FREE** courtesy of BrainMass!

Hi,

a. There are 4 jacks in a deck but since there will be no diamonds, we are to work on 3 jacks only. Then on the remaining cards there are 39 total removing all diamonds, but we also have to remove 3 jacks, having 36 to play around with the 7 remaining cards. So we have:

P = (3/52)(36/51)(35/50)(34/49)(33/48)(32/47)(31/46)(30/45)

P = 0.00416

b. For 1 card, there are 13 spades. For the remaining cards, we have 39 to play around. So we have:

P = (13/52)(39/51)(38/50)(37/49)(36/48)(35/47)(34/46)(33/45)

P = 0.03606

c. For 2 cards, there are 4 sevens to play around. For the three cards, we have 4 fives to play around. For the remaining 3 cards, we have 44 cards to play around. we have:

P = (4/52)(3/51)(4/50)(3/49)(2/48)(44/47)(43/46)(42/45)

P = 7.54 x 10^-7

https://brainmass.com/math/combinatorics/permutation-combination-437042