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In how many ways can one choose 6 cards from a normal deck of cards so
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15 Nov 2022, 11:36
15 Nov 2022, 11:36
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In how many ways can one choose 6 cards from a normal deck of cards so as to have all suits present?
A. \(13^4*304\)
B. \(C^6_{48}\)
C. \(13^4*27*47\)
D. \(13^4*48*47\)
E. \(13^4*C^6_{48}\)
A. \(13^4*304\)
B. \(C^6_{48}\)
C. \(13^4*27*47\)
D. \(13^4*48*47\)
E. \(13^4*C^6_{48}\)
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Re: In how many ways can one choose 6 cards from a normal deck of cards so
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29 Nov 2022, 05:55
29 Nov 2022, 05:55
1
Expert Reply
CORRECT SOLUTION:
There are four suits: Clubs (♣), Hearts (♥), Spades (♠) and Diamonds (♦),. 13 each. We need to select 6 cards, so as to have each of C, H, S, D.
Well 6 chosen cards can have ONLY two following patterns:
{C, H, S, D, X, X};
OR
{C, H, S, D, X, Y}.
Where XX means that we have two same suits for the last two cards and XY means that we have two different suits for the last two cards.
Which means that from the 6 cards chosen: we could either have 3 cards of the same suit and the other 3 cards with the remaining 3 suits (for example, {C, C, C, H, S, D}) OR 2 cards of same suit and the other 2 cards with remaining 2 suits (for example, {C, C, H, H, S, D}).
Case 1 - {C, H, S, D, X, X}:
Select 1 suit that will have three cards from it = 4C1
The number of ways to select 3 cards from this suit = 13C3
Fourth card, 1 from 13 = 13C1
Fifth card, 1 from 13 = 13C1
Sixth card, 1 from 13 = 13C1
Total for this case = \(4C1*13C3*13C1*13C1*13C1=4*(11*2*13)*13*13*13=13^4*88\)
Case 2 - {C, H, S, D, X, Y}:
Select 2 suits that will have two cards from it = 4C2
The number of ways to select 2 cards from first two card suit = 13C2
The number of ways to select 2 cards from second two card suit = 13C2
Fifth card, 1 from 13 = 13C1
Sixth card, 1 from 13 = 13C1
Total for this case = \(4C2*13C2*13C2*13C1*13C1=6*(6*13)*(6*13)*13*13=13^4*6^3\)
GRAND TOTAL = \(13^4*88+13^4*6^3=13^4(88+216)=13^4*304\).
Answer: B.
There are four suits: Clubs (♣), Hearts (♥), Spades (♠) and Diamonds (♦),. 13 each. We need to select 6 cards, so as to have each of C, H, S, D.
Well 6 chosen cards can have ONLY two following patterns:
{C, H, S, D, X, X};
OR
{C, H, S, D, X, Y}.
Where XX means that we have two same suits for the last two cards and XY means that we have two different suits for the last two cards.
Which means that from the 6 cards chosen: we could either have 3 cards of the same suit and the other 3 cards with the remaining 3 suits (for example, {C, C, C, H, S, D}) OR 2 cards of same suit and the other 2 cards with remaining 2 suits (for example, {C, C, H, H, S, D}).
Case 1 - {C, H, S, D, X, X}:
Select 1 suit that will have three cards from it = 4C1
The number of ways to select 3 cards from this suit = 13C3
Fourth card, 1 from 13 = 13C1
Fifth card, 1 from 13 = 13C1
Sixth card, 1 from 13 = 13C1
Total for this case = \(4C1*13C3*13C1*13C1*13C1=4*(11*2*13)*13*13*13=13^4*88\)
Case 2 - {C, H, S, D, X, Y}:
Select 2 suits that will have two cards from it = 4C2
The number of ways to select 2 cards from first two card suit = 13C2
The number of ways to select 2 cards from second two card suit = 13C2
Fifth card, 1 from 13 = 13C1
Sixth card, 1 from 13 = 13C1
Total for this case = \(4C2*13C2*13C2*13C1*13C1=6*(6*13)*(6*13)*13*13=13^4*6^3\)
GRAND TOTAL = \(13^4*88+13^4*6^3=13^4(88+216)=13^4*304\).
Answer: B.
gmatclubot
Re: In how many ways can one choose 6 cards from a normal deck of cards so [#permalink]
29 Nov 2022, 05:55
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