In how many ways can one choose 6 cards from a normal deck of cards so

First let's deal with the solution provided, which is NOT correct. Let's use the same exact approach for simpler but essentially the same exact problem:

In how many ways can one choose 3 letters from A1, A2, A3, B1, B2, B3 so as to have an A and B present?

(I copied EXACTLY the solution, just changed the numbers and cards by letters)

If you simply write down all the cases, then you'd see that there are only 18 ways:

A1, A2, B1;
A1, A2, B2;
A1, A2, B3;

A1, A3, B1;
A1, A3, B2;
A1, A3, B3;

A2, A3, B1;
A2, A3, B2;
A2, A3, B3;

A1, B1, B2;
A2, B1, B2;
A3, B1, B2;

A1, B1, B3;
A2, B1, B3;
A3, B1, B3;

A1, B2, B3;
A2, B2, B3;
A3, B2, B3.

So, there are duplicates in answer provided as correct.

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.

P.S. This is NOT a GMAT question, so you can ignore it.