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- Sep 9th 2006, 04:56 PMAfterShockCouple Counting Probs
Thanks

- Sep 10th 2006, 03:19 AMrgep
1a) I think you mean 4^200

1b) Yes, it's a multinomial.

1c) Yes: here's a way to see these. Write the A's, C's, G's and T's in order with bars between them: AA...A|CC...C|GG...G|TT...T and then see that you only need to specify the positions of the three bars in the 203 symbols you wrote down.

2a) 10 choices for Pr, 9 for Se, 8 for Tr gives 10*9*8

2b) If G is Se then 9*8 ways of choosing Pr then Tr from the other 8. If G is not an officer then 9*8*7 ways of choosing 3 officers from the other 9. Total 9*8 + 9*8*7

2c) Number of ways of choosing all male officers out of the 5 men is 5*4*3: hence number of ways with at least one female is 10*9*8 - 5*4*3

2d) Number of ways of choosing Pr, Se from the 5 female and Tr from the 5 males is 5*4 * 5; similarly for male Se and male Pr to give 5*4*5 * 3