Meg and Bob are among the 5 participants in a cycling race. If each participant finishes the race and no two participants finish at the same time, in how many different possible orders can the participants finish the race so that Meg finishes ahead of Bob?
A. 24
B. 30
C. 60
D. 90
E. 120
can anyone help??
ps 16
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- jayhawk2001
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Is it 60 ?
I believe this was discussed in a recent thread. Total number of
combinations = 5! = 120. Out of this, half the number of times, M will
be before B and half the number of times B will be before M.
So, total = 120 / 2 = 60.
Alternatively, you can look at positions to compute this
When M is in first place, B can finish in 4 different places (i.e. MB---, M-B--
etc.). For each such case, we have 3! permutations for the other set of 3
people to finish the race.
So, when M is in first place, we have 4 * 3! possibilities.
Similarly, computing for M in second place, third place etc. we get
3! * (4 + 3 + 2 + 1) = 6 * 10 = 60 possibilities
I believe this was discussed in a recent thread. Total number of
combinations = 5! = 120. Out of this, half the number of times, M will
be before B and half the number of times B will be before M.
So, total = 120 / 2 = 60.
Alternatively, you can look at positions to compute this
When M is in first place, B can finish in 4 different places (i.e. MB---, M-B--
etc.). For each such case, we have 3! permutations for the other set of 3
people to finish the race.
So, when M is in first place, we have 4 * 3! possibilities.
Similarly, computing for M in second place, third place etc. we get
3! * (4 + 3 + 2 + 1) = 6 * 10 = 60 possibilities
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Jay's method is the best...Still
Select 2 position from the total positions = 5C2
Assume that in the above position always Meg is ahead of the other
and the rest position can be filled in 3*2 ways
so total no of ways in which Meg finished before Bob
5C2 * 3* 2
and hence = 60
Select 2 position from the total positions = 5C2
Assume that in the above position always Meg is ahead of the other
and the rest position can be filled in 3*2 ways
so total no of ways in which Meg finished before Bob
5C2 * 3* 2
and hence = 60