M7MBA wrote: ↑

Sun Jun 14, 2020 1:42 pm

Randolph has a deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, there are 2 cards in the deck that have the same value.

Randolph likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for a pair of cards that have the same value. How many such combinations are possible?

A. 240
B. 960
C. 120
D. 40
E. 5760

[spoiler]OA=A[/spoiler]

Solution:

Since there are 12 cards, the number of ways one can choose 4 cards is:

12C4 = (12 x 11 x 10 x 9)/(4 x 3 x 2) = 11 x 5 x 9 = 495

When 4 cards are chosen, in terms of the number of pairs of cards that have the same value, there could be 0, 1, or 2 pairs. If we can determine the number of ways the 4 cards have 0 pairs and 2 pairs, then we can subtract those two results from 495 to obtain the number of ways the 4 cards would have (exactly) 1 pair.

The number of ways 4 cards have no pair is:

(12 x 10 x 8 x 6) / (4 x 3 x 2) = 5 x 8 x 6 = 240

(Note: in the above calculation, 12 is the number of ways one can choose the first card, 10 the second card, 8 the third card, and 6 the fourth card. However, since the order of the 4 cards doesn’t matter, we need to divide by 4! or 4 x 3 x 2.)

The number of ways 4 cards have 2 pairs is:

6C2 = (6 x 5)/2 = 15

(Note: Since we are really choosing 2 pairs from the available 6 pairs where order doesn’t matter.)

Therefore, the number of ways 4 cards have exactly 1 pair is 495 - 240 - 15 = 240.

Answer: A