Randolph has a deck of 12 playing cards made up of only 2

Assuming the question means "only one pair" (not a valid assumption though)

You have ordered the cards. Note that you need to select two kinds of cards - one a pair and another a set of two cards which are different.
You can select the pair in 6 ways (pair of 1 or pair of 2 or pair of 3 etc). Another way of saying this is: select the first card of the pair in 12 ways and then the second card in 1 way only. But since there is no first-second arrangement, divide 12 by 2! to get 6.
You select the dissimilar cards in 10*8 ways - 10 ways for the first one and 8 ways for the second one. Again, there is no first-second so you divide this by 2 to get 10*8/2 = 40 ways.

Total you get 6*40 = 240 ways

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

First, pick the pair with the same number. Since there are 6 pairs, this can be accomplished

6 ways

Next, pick two different numbers that will represent the remaining two cards. There are 5 sets of 2

5!/2!3! = 10

For each of these sets of 2 there are 2 suits to pick among, so

2*2= 4

Total ways: 6*10*4 = 240

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I have a different approach for this :

lets assume we have to fill 4 blanks with digits 1 to 6 :
exactly one pair
first blank - 12 ways to fill
second blank - 10 ways (11 digits left but cannot choose a pair as it is reserved for eg for first blank)
third blank - 8 ways (10 digits left but cannot choose a pair as it is reserved for eg for first blank )
fourth blank - paired digit - 1 way

12*10*8*1 = 960
since order does not matter divide by 4
960/4 = 120

Total possible combination is 12C4= 495
Total possible combination when cards are not same = (12*10*8*6)/4!= 240
So 2 cards are same combination is 495-240= 255.
Answer is 255 which is not in options.

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