Combinatorics Doubt

ashesh.rajhans wrote: 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

Case 1: 3 cards of one suit, the other 3 cards of the remaining 3 suits
Number of suit options for the 3 cards in red = 4. (Any of the 4 suits.)
Number of ways to choose 3 cards of this suit = 13C3 = (13*12*11)/(3*2*1) = 13*2*11.
Number of card options for the second suit = 13.
Number of card options for the third suit = 13.
Number of card options for the fourth suit = 13.
To combine these options, we multiply:
4*13*2*11*13*13*13 = 13�(4*2*11) = 13�(88).

Case 2: a pair of one suit, a pair of another suit, the other 2 cards of the two remaining suits
Number of ways to choose two suits for the two pairs in red = 4C2 = (4*3)/(2*1) = 6. (Any 2 of the 4 suits.)
Number of ways to choose 2 cards for the first pair in red = (13*12)/(2*1) = 13*6.
Number of ways to choose 2 cards for the second pair in red = (13*12)/(2*1) = 13*6.
Number of card options for the third suit = 13.
Number of card options for the fourth suit = 13.
To combine these options, we multiply:
6*13*6*13*6*13*13 = 13�(6*6*6) = 13�(216).

Total ways = Case 1 + Case 2 = 13�(88) + 13�(216) = 13�(88 + 216) = [spoiler]13�(304[/spoiler]).

None of the answer choices is correct.
Ignore this problem, which is far too complex for the GMAT.
What is the source?

ashesh.rajhans wrote:Guys,

Can you help me out with this ?

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

My Ans : (13^4) * 48C2
Ans Given : (13^4) x 48C1 x 47C1

My doubt is since we are only choosing - why double the selections ?
Isn't selecting AB same as BA here ?

Any help would be appreciated,
[email protected]

All given Answer choices are definitely Incorrect

Case 1: 1, 1, 2, 2 cards from 4 different suits respectively
Select 2 suits from which 2 cards need to be chosen = 4C2
Select 2 cards from each of selected two suits = 13C2 * 13C2
Select 1 card from each of remaining two suits = 13 * 13
Total Ways = 4C2 * (13C2 * 13C2) * (13 * 13) = 13^4 *6*6*6 = 13^4 * 216


Case 2: 1, 1, 1, 3 cards from 4 different suits respectively
Select 1 suits from which 3 cards need to be chosen = 4C1 = 4
Select 3 cards from selected suit = 13C3 = 13*22
Select 1 card from each of remaining two suits = 13 * 13 * 13
Total Ways = 4 * (13 * 22) * (13 * 13 * 13) = 13^4 * 88

Total ways = 13^4 * (88+216) = 13^4 * 304

Thanks GMATinsight and GMATGuruNY for the solution.

@GMATinsight, The source is https://www.gmating.com/ .
It's an old website, was being managed by a student.

Another approach:

Since we need all four suits, we only need to consider the suits of the fifth and sixth cards.

If the fifth and sixth cards have different suits, we can choose these suits in (4 choose 2) = 6 ways. Once we choose them, we have (13 choose 2) ways to choose 2 cards from each of the two overrepresented suits, and (13 choose 1) ways to choose 1 card from each of the other suits. This gives

6 * (13 choose 2)² * (13 choose 1)²

If the fifth and sixth cards have the same suit, we can choose this suit in (4 choose 1) = 4 ways. From here, the logic is similar

4 * (13 choose 3) * (13 choose 1)³

The sum of these two values is

6 * 78² * 13² + 4 * 286 * 13³

or

2³ * 13� * (3³ + 11)

or

13� * 304