Question 1191618: From a deck of 52 playing cards, 5 cards are to be drawn.
a) How many 5-card hands are possible
b) How many 5-card hands are there consisting of 2 aces and 3 jacks?
c) How many 5-card hands are there consisting of all black suits?
d) How many 5-card hands are there consisting of all diamond suits?
e) How many 5-card hands are there consisting of 3 face-cards, 1 king, and 1 spade?
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
Part (a)
n = 52 cards total
r = 5 selections made
Use the nCr combination formula because order does not matter for any card hand.
n C r = (n!)/(r!(n-r)!)
52 C 5 = (52!)/(5!*(52-5)!)
52 C 5 = (52!)/(5!*47!)
52 C 5 = (52*51*50*49*48*47!)/(5!*47!)
52 C 5 = (52*51*50*49*48)/(5!) .... the factorial terms "47!" cancel
52 C 5 = (52*51*50*49*48)/(5*4*3*2*1)
52 C 5 = (311875200)/(120)
52 C 5 = 2598960
Answer is 2,598,960
This is a little under 2.6 million different five-card combinations.
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Part (b)
There are n = 4 aces total and we select r = 2 of them.
n C r = (n!)/(r!(n-r)!)
4 C 2 = (4!)/(2!*(4-2)!)
4 C 2 = (4!)/(2!*2!)
4 C 2 = (4*3*2*1)/(2*1*2*1)
4 C 2 = (24)/(4)
4 C 2 = 6
There are x = 6 ways to select the two aces in any order.
Then there are n = 4 jacks and r = 3 selections.
n C r = (n!)/(r!(n-r)!)
4 C 3 = (4!)/(3!*(4-3)!)
4 C 3 = (4*3*2*1)/(3*2*1*1)
4 C 3 = 24/6
4 C 3 = 4
There are y = 4 ways to select three jacks.
Or put simply, there are y = 4 ways to leave a certain jack card out.
Overall, there are x*y = 6*4 = 24 ways to select 2 aces and 3 jacks.
Answer: 24
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Part (c)
There are n = 26 black cards (13 spades + 13 clubs)
We make r = 5 selections
n C r = (n!)/(r!(n-r)!)
26 C 5 = (26!)/(5!*(26-5)!)
26 C 5 = (26!)/(5!*21!)
26 C 5 = (26*25*24*23*22*21!)/(5!*21!)
26 C 5 = (26*25*24*23*22)/(5!) ..... the 21! terms cancel
26 C 5 = (26*25*24*23*22)/(5*4*3*2*1)
26 C 5 = (7893600)/(120)
26 C 5 = 65780
Answer is 65,780
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Part (d)
There are n = 13 diamonds
We make r = 5 selections
n C r = (n!)/(r!(n-r)!)
13 C 5 = (13!)/(5!*(13-5)!)
13 C 5 = (13!)/(5!*8!)
13 C 5 = (13*12*11*10*9*8!)/(5!*8!)
13 C 5 = (13*12*11*10*9)/(5!)
13 C 5 = (13*12*11*10*9)/(5*4*3*2*1)
13 C 5 = (154440)/(120)
13 C 5 = 1287
Answer: 1287
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Part (e)
For any given suit, there are 3 face cards (J, Q, K).
That gives 4*3 = 12 face cards total.
n = 12
r = 3
n C r = (n!)/(r!(n-r)!)
12 C 3 = (12!)/(3!*(12-3)!)
12 C 3 = (12!)/(3!*9!)
12 C 3 = (12*11*10*9!)/(3!*9!)
12 C 3 = (12*11*10)/(3!)
12 C 3 = (12*11*10)/(3*2*1)
12 C 3 = (1320)/(6)
12 C 3 = 220
There are x = 220 ways to select the three face cards in any order.
There are four kings, so there are y = 4C1 = 4 ways to select one king card.
There are 13 spades, meaning there are z = 13C1 = 13 ways to pick a spade card.
The number of five card hands with 3 face-cards, 1 king and 1 spade is x*y*z = 220*4*13 = 11,440
Answer: 11,440
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