Using a standard deck of 52 cards how many 5 card hands contain
A. 2 queens?
B. 3 fives?
C. At most 2 sevens?
D. One pair and one three of a kind?
To find the number of 5-card hands that satisfy each condition, we can use combinations. The formula for combinations is nCr = n! / (r! * (n-r)!), where n is the total number of items and r is the number of items being chosen at a time.
A. Number of 5-card hands with 2 queens:
There are 4 queens in a deck, so we need to choose 2 of them. The other 3 cards can be any of the remaining 50 cards (52 cards - 2 queens). The formula for combinations gives us:
nCr = 4C2 * 50C3 = (4! / (2! * (4-2)!)) * (50! / (3! * (50-3)!))
= (4 * 3 / (2 * 1)) * (50 * 49 * 48 / (3 * 2 * 1))
= 6 * (19,600)
= 117,600
Therefore, there are 117,600 5-card hands that contain exactly 2 queens.
B. Number of 5-card hands with 3 fives:
Similarly, there are 4 fives in a deck, and we need to choose 3 of them. The other 2 cards can be any of the remaining 48 cards (52 cards - 3 fives). Applying the formula again:
nCr = 4C3 * 48C2 = (4! / (3! * (4-3)!)) * (48! / (2! * (48-2)!))
= (4 / 1) * (48 * 47 / (2 * 1))
= 4 * (48 * 47)
= 8,928
So, there are 8,928 5-card hands that contain exactly 3 fives.
C. Number of 5-card hands with at most 2 sevens:
We need to count the number of hands that have either 0, 1, or 2 sevens. Let's calculate each case separately:
Number of hands with 0 sevens:
Choosing all 5 cards from the remaining 50 cards (52 cards - 2 sevens):
nCr = 50C5 = 50! / (5! * (50-5)!)
= 2,118,760
Number of hands with 1 seven:
Choosing 1 seven and the other 4 cards from the remaining 50 cards:
nCr = 2C1 * 50C4 = (2! / (1! * (2-1)!)) * (50! / (4! * (50-4)!))
= 2 * (50 * 49 * 48 * 47 / (4 * 3 * 2 * 1))
= 2 * 230,300
= 460,600
Number of hands with 2 sevens:
Choosing 2 sevens and the other 3 cards from the remaining 50 cards:
nCr = 2C2 * 50C3 = 1 * (50! / (3! * (50-3)!))
= 50 * 49 * 48 / (3 * 2 * 1)
= 19,600
The total number of hands with at most 2 sevens is the sum of the above three cases:
2,118,760 + 460,600 + 19,600 = 2,599,960
Therefore, there are 2,599,960 5-card hands that contain at most 2 sevens.
D. Number of 5-card hands with one pair and one three of a kind:
To form one pair and one three of a kind, we need to choose one rank (e.g., 6, 10) for the pair and one rank for the three of a kind (e.g., 5, K). There are 13 ranks in total, so we have to count the combinations for each case:
Number of hands with one pair:
Choosing 2 cards of the chosen rank and 3 other cards from the remaining 48 cards:
nCr = 13C1 * 4C2 * 48C3 = 13 * (4! / (2! * (4-2)!)) * (48! / (3! * (48-3)!))
= 13 * 6 * (48 * 47 * 46 / (3 * 2 * 1))
= 13 * 6 * (22,176)
= 1,775,872
Number of hands with three of a kind:
Choosing 3 cards of the chosen rank and 2 other cards from the remaining 48 cards:
nCr = 13C1 * 4C3 * 48C2 = 13 * (4! / (3! * (4-3)!)) * (48! / (2! * (48-2)!))
= 13 * 4 * (48 * 47 / (2 * 1))
= 13 * 4 * (1,128)
= 58,368
The product of the two cases gives the total number of hands:
1,775,872 * 58,368 = 103,259,040,896
Therefore, there are 103,259,040,896 5-card hands that contain one pair and one three of a kind.