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Are you sure that for (a) the answer is 5.346E^13 ?can someone please explain how to solve these questions:
-eight cards are selected with replacement from a standard pack of 52 playing cards with 12 picture carsd, 20 odd cards and 20 even cards.
a)how many different sequences of eight cards are possible?( ans:5.346 X 10 ^13
b)how many of the sequences in part (a) will contain three picture cards, three odd- numbered cards and two even numbered cards?(ans: 3.097 X10 ^12)
Choosing 8 from 52, sequence included, so that is permutation, 52P8, and I got 3.034E^13.
For (b) I also got a different result. The procedure is like choosing 3 from 12, 3 from 20 and 2 from 20, then put them in sequence. So it's first combination, 12C3 × 20C3 × 20C2, then permutation, 8P8 which is 8!.
I got 12C3 × 20C3 × 20C2 × 8! = 1.92E^12