Problem Solving

Bill has a small 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.

Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?

8/33
62/165
17/33
103/165
25/33

Can someone please solve this using 'the number of ways' method ?

P(At least one pair) = P(1 - none pair)

Total Combinations = 12C4 [=(12*11*10*9)/(4*3*2*1)] ....A
None pair = (12*10*8*6)/(4*3*2*1) .....B

In "none pair", 12 because first card can be selected in 12 ways, then 10 because second card can be taken in 10 ways (out of 11 remaining, so that it is different from first selected)...... and so on.
Order doesn't matter, so we divided by 4*3*2*1

Now, P(none pair) = B/A = 16/33

P(At least one pair) = 1 - 16/33 = 17/33 (Answer)

Trader AK wrote:P(At least one pair) = P(1 - none pair)

Total Combinations = 12C4 [=(12*11*10*9)/(4*3*2*1)] ....A
None pair = (12*10*8*6)/(4*3*2*1) .....B

In "none pair", 12 because first card can be selected in 12 ways, then 10 because second card can be taken in 10 ways (out of 11 remaining, so that it is different from first selected)...... and so on.
Order doesn't matter, so we divided by 4*3*2*1

Now, P(none pair) = B/A = 16/33

P(At least one pair) = 1 - 16/33 = 17/33 (Answer)

Perfect..Thank you. I couldnt realize that order doesnt matter , hence was getting the wrong answer