Question 1191612: From a standard deck of 52 cards, how many six-hand cards are possible if there are at least three aces and three other cards?
Answer by math_tutor2020(3817)   (Show Source):
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If there are exactly 3 aces in the hand, then there are x = 4 ways to pick these three cards.
This is equivalent to having x = 4 ways to not pick a particular ace.
We have n = 52-4 = 48 non-ace cards to pick from and r = 3 slots to fill.
Use the nCr combination formula.
n C r = (n!)/(r!(n-r)!)
48 C 3 = (48!)/(3!*(48-3)!)
48 C 3 = (48!)/(3!*45!)
48 C 3 = (48*47*46*45!)/(3!*45!)
48 C 3 = (48*47*46)/(3!)
48 C 3 = (48*47*46)/(3*2*1)
48 C 3 = (103776)/(6)
48 C 3 = 17296
We have y = 17296 ways to select the three other non-ace cards.
In total, there are x*y = 4*17296 = 69,184 ways to select 6 cards such that exactly 3 are aces.
Let P = 69184
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Now consider the six card hand having exactly 4 aces. There's only one way to select all four aces.
There are 48 C 2 = 1128 ways to pick the other cards.
That gives 1*1128 = 1128 ways to have a 6-card hand with exactly 4 aces.
Let Q = 1128
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Add the previous results:
P+Q = 69184+1128 = 70,312
This is the number of six-card hands with at least 3 aces.
In the event of having 4 aces, we can't have "3 other cards" as your teacher mentioned because that would mean having a 7-card hand.
Answer: 70,312
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