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question of nCr (choosing)

helenli89

New member
Joined
Oct 1, 2009
Messages
48
Q: In a race, the 15 runners are randomly assigned the numbers 1,2,....15. Find the probability that the fifth runner to finish is the 3rd finisher with a single digit number.

my answer is: ((9 nCr 3)(6 nCr 2)7)/(15 nCr 5) But the answer is ((9 nCr 3)(6 nCr 2)7)/((15 nCr 4)*11) it says that there are (15 nCr 4) ways of choosing the first 4 and 11 ways to choose the 5th. But I don't understand this explaination, I think these two denominator should give the same number, but they don't.

Thanks.
 

arthur ohlsten

Full Member
Joined
Feb 20, 2005
Messages
854
I have a slightly different derivation, and I disagree with the answer posted.

look at the first 4 finishers. What is the probability that 4 will be single digit runners and 2 multidigit runners, from the total of 15 ?
(9 C 2)(6 C 2) / (15 C 4)

What is the probability that the 5 finisher is one of the remaing 7 single digit runners, from the total of 11 remaining runners?
7/11

What is the probability that the first 5 finishers have 2 single digit and 2 double digit finisher in the first 4 finishers and the 5th finisher is a single digit runner?

{(9C2)(6C2)7 }/{(15C4)11}

I disagree with the first term 9C3 in the answer.
I think my derivation is correct, but I have been wrong many times in my math career .

Arthur
 

helenli89

New member
Joined
Oct 1, 2009
Messages
48
arthur ohlsten said:
I have a slightly different derivation, and I disagree with the answer posted.

look at the first 4 finishers. What is the probability that 4 will be single digit runners and 2 multidigit runners, from the total of 15 ?
(9 C 2)(6 C 2) / (15 C 4)

What is the probability that the 5 finisher is one of the remaing 7 single digit runners, from the total of 11 remaining runners?
7/11

What is the probability that the first 5 finishers have 2 single digit and 2 double digit finisher in the first 4 finishers and the 5th finisher is a single digit runner?

{(9C2)(6C2)7 }/{(15C4)11}

I disagree with the first term 9C3 in the answer.
I think my derivation is correct, but I have been wrong many times in my math career .

Arthur

Yes you're right, I made a typo on my posting the answer is exactly what you have.
Thanks, now I understand the answer now.
 

arthur ohlsten

Full Member
Joined
Feb 20, 2005
Messages
854
Your welcome. I dislike the answer you say you were given for the number 11 in the posting.
I think considering the 5th place runner as having a probability of 7/11 explains the 7 in the numerator and the 11 in the denominator, is a much clearer explanation.

Good luck in the course

Arthur
 
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