stne wrote:
Just edited my original post
Meant to take 5C2 not 10C2,
so my doubt
one pair
6C1*5C2(4!/2!)= 720
6C1 = ways to select the one card from 6 which will form the pair
5C2 = select two different cards from the 5 remaining to form the singles * This I think this is not correct , what would be correct *
4!/2! = permutations of 4 letters where two are identical
for two pairs
6C2 * 4!/2!2! = 90
6C2 = ways to select the two cards which will form the pair .
4!/2!2! = ways to arrange the 4 letters where 2 are same kind and another 2 are same kind .
Responding to a pm:
There is a difference between this question and the other one you mentioned.
In that question, you were using one digit twice which made them identical. Here the two cards that form the pair are not identical. They are of different suits. Also, here, you don't need to arrange them. You can assume to just take a selection while calculating the cases in the numerator and the denominator. The probability will not get affected. In the other question, you needed to find the number of arrangements to make the passwords/numbers and hence you needed to arrange the digits.
Hence, in this step, 6C1*5C2(4!/2!), it should be 6C1*5C2*2*2 instead.
6C1 = ways to select the one card from 6 which will form the pair
5C2 = select two different values from the 5 remaining to form the singles *This is correct *
2*2 = For each of the two values, you can select a card in 2 ways (since you have 2 suits)
Similarly, 6C2 * 4!/2!2! should be 6C2 only to select 2 pairs out of 6.
Probability = 255/495 = 17/33
Note: You can arrange the cards too and will still get the same probability. Just ensure that you arrange in numerator as well as denominator.
Only one pair = 6C1*5C2*2*2 * 4! (you multiply by 4! because all the cards are distinct)
Both pairs = 6C2 * 4! (you multiply by 4! because all the cards are distinct)
Select 4 cards out of 12 = 12C4 * 4! (you multiply by 4! because all the cards are distinct)
Probability = 255*4!/495*4! = 17/33
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