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# Statistics Problem Permutations watch

1. I am only able to do part A (720 I hope). For part c I think it may be 6 choose 2 which is 15. Could I have some help on the other parts(and part a and c if these are the wrong answers)?
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2. (Original post by John Flores)
I am only able to do part A (720 I hope). For part c I think it may be 6 choose 2 which is 15. Could I have some help on the other parts(and part a and c if these are the wrong answers)?
Anyone?
3. Part b- what is the probability he gets ferrari? As each race is independent then u can do this probability to the power of 10
4. (Original post by John Flores)
I am only able to do part A (720 I hope). For part c I think it may be 6 choose 2 which is 15. Could I have some help on the other parts(and part a and c if these are the wrong answers)?
For part c, the answer is 15, but it is not obtained through using 6C2; it just happens that for this case that the answer is equal to 6C2.

There are 6C2 ways of picking the first pair, and then 4C2 ways of picking the second pair, and then 2C2 ways of picking the third pair. Note however that when multiplying these all together, the same teams in different orders, namely 3! times, will occur. So we actually get 6C2* 4C2 * 2C2 / 3! = 15

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