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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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.
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.
Naveenan Ramachandran
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Case 1: Jen is at 1st place.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.
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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Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 201112  will stay active as time permits
18005664043 (USA)
+9199201 32411 (India)