# Permutations.

• Feb 10th 2009, 09:24 PM
woollybull
Permutations.
9 cards are selected with replacement from a standard pack of 52 cards with 12 picture, 20 odd and 20 even cards.

a)How many different sequences of 9 cards are possible?
b)How many of the sequences in part (a) will contain 4 picture cards,3 odd and 2 even cards?

For part a) it will be 9 spaces all of which have 52 possibilities, so 52^9.
For part b) it will be 12^4 x 20^3 x 20^2. But the picture cards can be arranged so that the picture isnt just in the first 3 positions and the odd and even to be in other positions also.
So, the number of sequences will be 12^4 x 20^3 x 20^2 x 9!/(4!3!2!).

• Feb 12th 2009, 10:25 PM
Permutations
Hello woollybull
Quote:

Originally Posted by woollybull
9 cards are selected with replacement from a standard pack of 52 cards with 12 picture, 20 odd and 20 even cards.

a)How many different sequences of 9 cards are possible?
b)How many of the sequences in part (a) will contain 4 picture cards,3 odd and 2 even cards?

For part a) it will be 9 spaces all of which have 52 possibilities, so 52^9.
For part b) it will be 12^4 x 20^3 x 20^2. But the picture cards can be arranged so that the picture isnt just in the first 3 positions and the odd and even to be in other positions also.
So, the number of sequences will be 12^4 x 20^3 x 20^2 x 9!/(4!3!2!).

Choose the picture cards in $12^4$ ways and the positions they occupy in $^9C_4$ ways. Then choose the odd cards in $20^3$ ways, and their positions in $^5C_3$ ways. Finally choose the even cards in $20^2$ ways for the last two positions.
Total number of ways: $12^4 \times ^9C_4 \times 20^3 \times ^5C_3 \times 20^2$
$= 12^4 \times \frac{9!}{5!4!} \times 20^3 \times \frac{5!}{3!2!} \times 20^2$
$= \frac{12^4 \times 20^3 \times 20^2 \times 9!}{4!3!2!}$