Difficult Math Question #20 - Combinations

[800guy]

OA coming after some people answer..

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) x 48 x 47
b. (13^4) x 27 x 47
c. 48C6
d. 13^4
e. (13^4) x 48C6

[gamemaster]

i think its about time to see the answer

i think it should be < 13^4 but i didnt find an easy way to solve it

can you tell us how to solve this?

Thanks

[abby_g]

THe answer is A.
A spade is selected in 13C1 ways. such other 3 so total of (13C1)^4.

now we want 2 cards. one card i can get in 48C1 and the other in 47C1.

13^4 * 48 * 47.

the 2 cards can be picked using 48P2, since we want ways. ANYWAYS the answer is still the same.

i hope its right. Curiousity kills the cat Answer please.

[800guy]

OA:

Sol: 52 cards in a deck -13 cards per suit
First card - let us say from suit hearts = 13C1 =13
Second card - let us say from suit diamonds = 13C1 =13
Third card - let us say from suit spade = 13C1 =13
Fourth card - let us say from suit clubs = 13C1 =13
Remaining cards in the deck= 52 -4 = 48
Fifth card - any card in the deck = 48C1
Sixth card - any card in the deck = 47C1

Total number of ways = 13 * 13 * 13 * 13 * 48 * 47 = 13^4 *48*47 ---> choice A

[gamemaster]

but arent you repeating on options like this?

this is what's confusing, i dont think you can just multiply (13^4) by 48*47

lets take 1,2,3 of type hart and 4,5,6 of 3 different types, so:

{1,4,5,6} {2,3}
{2,4,5,6} {1,3}
{3,4,5,6} {2,1}

are the same choises, you have here a very large number of repetitions

please elaborate on this if you understand what i mean

Thanks

[gamemaster]

i think i got it ...

the solution applies only for cases where the order of picking matters, so in this case:

{1,2,3,4} {5,6}

is different than

{2,2,3,4} {3,1}

This should be clarified in the question!, otherwise the solution should be different ... (i think )