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by mba1986 » Tue Aug 24, 2010 8:28 pm
If Bob and Jen are two of 5 participants in a race, how many different ways can the race finish where Jen always finishes in front of Bob?

Sadly no answers sorry.

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by 4GMAT_Mumbai » Tue Aug 24, 2010 9:12 pm
Hi,

Brute force method:

Case 1: Bob finishes last. Jen has 4 places to finish

Case 2: Bob finishes fourth. Jen has 3 places to finish

Case 3: Bob finishes third. Jen has 2 places to finish

Case 4: Bob finishes second. Jen has only 1 place to finish

Thus, total # of ways = 10 ways.

For each of these 10 ways, the other 3 people can finish in 3! ways. So, overall 10 times 6 = 60 ways.

A cleaner method:

# of overall ways for Jen to finish - 5

# of overall ways for Bob to finish - 4 (as one place has been occupied by Jen)

# of overall ways = 5 times 4 = 20 ways.

Half of these - Jen finishes before Bob = 20 / 2 = 10 ways.

For each of these 10 ways, the other 3 people can finish in 3! ways. So, overall 10 times 6 = 60 ways.

Hope this helps. Thanks.
Last edited by 4GMAT_Mumbai on Thu Aug 26, 2010 9:11 pm, edited 1 time in total.
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by Rahul@gurome » Tue Aug 24, 2010 9:31 pm
arjunsekhar wrote:If Bob and Jen are two of 5 participants in a race, how many different ways can the race finish where Jen always finishes in front of Bob?

Sadly no answers sorry.
Case 1: Jen is at 1st place.

Bob can be at any other 4 places, which means Bob can be arranged in 4 ways.
And the remaining 3 participants can be arranged in 3! ways.
Hence, total number of ways = 4 * 3! = 4 * 3 * 2 = 24 ways

Case 2: Jen is at 2nd place.

Since Jen always finishes in front of Bob, so Bob can be arranged in 3 ways.
And remaining 3 participants can be placed in 3! = 3 * 2 = 6 ways
Hence, total number of ways = 3 * 3! = 3 * 3 * 2 = 18 ways

Case 3: Jen is at 3rd place.

Since Jen always finishes in front of Bob, so Bob can be arranged in 2 ways.
And remaining 3 participants can be placed in 3! = 3 * 2 = 6 ways
Hence, total number of ways = 2 * 3! = 2 * 3 * 2 = 12 ways

Case 4: Jen is at 4th place.

Since Jen always finishes in front of Bob, so Bob can be arranged in 1 way.
And remaining 3 participants can be placed in 3! = 3 * 2 = 6 ways
Hence, total number of ways = 1 * 3! = 3 * 2 = 6 ways

Therefore, Adding the 4 cases, we get required answer = 24 + 18 +12 + 6 = 60 ways
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