Question

Assume that a standard deck of 52 playing cards is randomly shuffled (13 cards of 4 types of suits - clubs, diamonds, hearts, and spades). If Alex draws a card from it 4 times with replacement, how many different combinations of suits can he get? (Suppose we further assume that the order doesn't matter: (clubs, diamonds, hearts, hearts) is equal to (hearts, clubs, diamonds, hearts). However, (clubs, diamonds, hearts, hearts) is different with (hearts, clubs, clubs, hearts).)

Answer 1

Alex has to draw 4 cards with replacement. So, on each possible draw 4 possible suit outcomes exist.
If order matters, then
For each draw we have 4 possible suit outcomes, so total number of different outcomes:
4 x 4 x 4 x 4 = 4⁴ = 256

Since the order doesn't matter, we divide the problem into following cases of outcome category:

Case 1: All different
No. of ways: 1 (Spades, Diamonds, Clubs, Hearts) in any order

Case 2: 3 same, 1 different
No. of ways:
Pick 1st suit (out of 4 which will be drawn 3 times) in  ⁴C₁ = 4
Pick 2nd suit out of the 3 left in 3 ways
No. of ways = 4 x 3 = 12

Case 3: 2 of one kind, 2 of another kind
Pick 1st suit (which will be drawn 2 times) in ⁴C₁ = 4 ways
Pick 2nd suit (which will also be drawn 2 times) in ³C₁ = 3 ways
No. of ways = 4 x 3 = 12
But this 12 will contain (Spades, Spades, Diamonds, Diamonds) and (Diamonds, Diamonds, Spades, Spades) both. Since the order doesn't matter, total no. of ways = 12/2 = 6

Case 4: All of one kind
No. of ways: 4 (all Spades, all Diamonds, all Hearts or all Clubs)

Case 5: 2 of one kind, 2 different
Pick 1st suit (out of 4 which will be drawn twice) in ⁴C₁ = 4 ways
Pick 2nd suit in 3 ways
Pick 3rd suit in 2 ways
Total no. of ways = 4 x 3 x 2 = 24, Each outcome will occur twice and since order doesn't matter,
total no. of ways = 24/2 = 12

So total number of different combination that can be drawn = 1 + 12 + 6 + 4 + 12 = 35