Probability help?

Question

There are 10 contestants on a game show. The longer you stay the momre money you get.

A. how many combinations can the first four be voted off?
B. In how many can the final four of the ten be chosen?

Answer 1

A. Multiply these fractions 1/10, 1/9, 1/8, 1/7. For the first person off, it can be anybody. So the prob is 1/10. For the next person it can be anyone out of the remaning 9. And so on. So there are 10 ties 9 times 8 times 7 combinations or the first 4. It seems like a lot, i know.

B. it is the total combinations for being voted off devided by the combinations of the top 6 being voted off Minus the four people combination for possible selected.. So 10 factorial devided by 6 factorial minus 4 factorial, if you will.

A. 10x9x8x7
B. 10!/ (6!-4!)

I would check with other people to see if this is right. I am only an 8th grader, but I got 26th place in a State wide math compitition with problems like these all the time.=D

Hope that helps.=D

Answer 2

A. how many combinations can the first four be voted off?

You didn't say if order is important.

If order matters 10*9*8*7 = 5040
If it does not matter C(10,4) = 210

B. In how many can the final four of the ten be chosen?

Chosen at what point? Before the game starts? In that case the answers are the same as in A.

If order matters 10*9*8*7 = 5040
If it does not matter C(10,4) = 210

Or did you mean among the remaining six persons once they are known. In that case:

If order matters 6*5*4*3 = 360
If it does not matter C(6,4) = 15

The aswerer above me is not necessarily wrong. He just made some assumptions about the vague question that were different than mine.