Question

There is a deck with 42 cards, with 4, 5, 6, 7, 8, 9, A. There are three suits spades, hearts, and clovers. In each suit there are two of each card. So there are two 7 of spades, 7 of hearts, 7 of clovers and so on. Each hand dealt consists of 5 cards.

1. How many hands contains exactly a pair that consists of two cards with the same value?

2. How many hands contains exactly one pair with the same value but different suits?

3. How many hands contain exactly one 9 of spades?

Discrete Math.

Answer 1

{Using notation, selection or r objects/elements out of n distinct objects/elements is

Spades	Hearts	Clovers
7*2	7*2	7*2

1> Same value card is 6 with two same kind.

We have 7 different cards.

Selecting type of pair (exactly one) out of 7 pair = 7 ways =

Next in 6 same value cards selecting 2 cards:

• both are same suits in ways
• both are different suits in ways.

Total number of ways = 7*(6+6) = 42 is selecting two same value cards.

We need to select other 3 cards too of different value. We are left with 6 different value cards.

Number of ways to select 3 cards out of 6 is ways. Then there 3 more option within same value.

Thus total number of ways for 5 cards selection with exactly a pair that consists of two cards of same value = 42*20*3 = 2520 ways

2> Same as above, we just need to remove counting of selection of same suits.

Thus total number of ways for 5 cards selection with exactly a pair that consists of two cards of same value but different suits = 21*20*3 = 1260 ways

3> 9 of spades is already selected. We can not select one more 9 of spades. So we are left with 42-2 = 40 cards. And we still need to select 4 cards without any restriction.

Thus total number of required ways : ways. answer