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Bunuel
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Bunuel
A "standard" deck of playing cards consists of 52 cards in each of the 4 suits of Spades, Hearts, Diamonds, and Clubs. Each suit contains 13 cards: Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King. In how many ways 5 cards can be selected from a "standard" deck of playing cards, without a replacement, so that all 4 suits appear?

A. 1,287
B. 4,056
C. 52,728
D. 405,646
E. 685,464

Are You Up For the Challenge: 700 Level Questions
Conditions to keep in mind:
1. Each suit is represented in the 5 cards
2. Cards are to be replaced

Because our target is to make sure that all the suits appear, we have to consider each of them.
Ways in which 5 cards can appear -
2S H D C
S 2H D C
S H 2D C
S H D 2C
Total 4 ways.
As cards had to be replaced at any moment there are 52 cards in the deck i.e. 13 cards in each suit.
So, required answer = 4* 13C2*13C1*13C1*13C1 = 685,464

Answer E.­
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Can someone please help me out, I did the following calculations:

13*13*13*13*48

The first 4 '13' represent a card from each suit and as 4 cards are gone, the 5th card could be any of the remaining cards, therefore 48 (52-4)
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AabhishekGrover
Can someone please help me out, I did the following calculations:

13*13*13*13*48

The first 4 '13' represent a card from each suit and as 4 cards are gone, the 5th card could be any of the remaining cards, therefore 48 (52-4)


Yes the approach is okay...but you need to divide by 2...since there will be 2 same suite.....13x13x13x13x48/2 = 685464

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Let's be honest here.

It's not expected that the full multiplication should be done and people are cheating themselves by performing the multiplication via calculator while practicing.

The thing to notice is that the multiplication chain results in a one's digit of 4 and therefore answer E is the only viable option.
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