Combinatorics

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