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
24
30
60
90
120
pls, help with this
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permutation, a good official question
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bluementor
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There are 5! = 120 different ways to finish the race.
Out of these 120 ways, half of them will be exact mirror opposites of the other half. For example:
Lets say the 5 participants are X, Y, Z, Meg and Bob. So one of the possibilites could be:
X, Y, Z, Meg, Bob (with Bob finishing last)
an opposite of this sequence would be:
Bob, Meg, Z, Y, X (B finishes first, ahead of Meg).
So in other words, you will have opposite pairs like the above in the 120 ways (so you will have 60 pairs). Since only one possibility in each pair will have Meg finishing ahead of Bob, the answer is 60 different ways.
Choose C.
-BM-
Out of these 120 ways, half of them will be exact mirror opposites of the other half. For example:
Lets say the 5 participants are X, Y, Z, Meg and Bob. So one of the possibilites could be:
X, Y, Z, Meg, Bob (with Bob finishing last)
an opposite of this sequence would be:
Bob, Meg, Z, Y, X (B finishes first, ahead of Meg).
So in other words, you will have opposite pairs like the above in the 120 ways (so you will have 60 pairs). Since only one possibility in each pair will have Meg finishing ahead of Bob, the answer is 60 different ways.
Choose C.
-BM-
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bluementor
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Lets reduce the problem to only 3 people: Bob, Meg and X.duongthang wrote:still not understand, pls, explain. best regards.
The number of ways the race can finish = 3! = 3x2x1 = 6
If you try listing the 6 different ways, you will get:
(1) BMX (Bob finishes first, Meg second, and X third)
(2) BXM
(3) XBM
(4) MBX
(5) MXB
(6) XMB
See that (1) and (6) are opposites to each other, in terms of arrangement. In (1) you have BMX, and if you reverse the sequence, you get (6) XMB. Between these two only (6) has the combination you want, i.e. Meg finishes before Bob.
The same can be done for the (2) and (5) pair, and the (3) and (4) pair. From each pair, only one combination will satisfy that Meg finishes before Bob. So in total here, you have 6 combinations, of which only 3 satisfies the criteria.
So if you generalize this further, if there are n participants in the race, then the number of ways in which Meg finishes before Bob will be:
n!/2 … this is what mike22629 used to solve this problem.
So for this problem, there are 5 participants in total. So the number of ways Meg can finish before Bob is:
5!/2 = 120/2 = 60.
Hope this helps.
-BM-
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maihuna
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I will prefer it positional way:
X X X B M : !4 = 24
X X X M Y : 3*3*2 = 18
X X M Y Z : 2*!3 = 12
B M X X X: !3 = 6
total = 60
prefer the positional one as it will be helpful for wider such questions rather than the pattern which will be limited in repeat and suitability.
X X X B M : !4 = 24
X X X M Y : 3*3*2 = 18
X X M Y Z : 2*!3 = 12
B M X X X: !3 = 6
total = 60
prefer the positional one as it will be helpful for wider such questions rather than the pattern which will be limited in repeat and suitability.
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Scott@TargetTestPrep
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We can create the following equation:tanviet wrote: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
24
30
60
90
120
total number of ways to finish the race = number of ways Meg finishes ahead of Bob + number of ways Meg does not finish ahead of Bob
Since the total number of ways to complete the race is 5! = 120, and since there are an equal number of ways for Meg to finish ahead Bob as there are for her not to finish ahead of Bob, Meg can finish ahead of Bob in 60 ways.
Alternate Solution:
If Meg comes in the first, then Bob can finish the race in any of the remaining positions, so there are 4! = 24 ways this could happen.
If Meg comes in the second, there are 3 choices for the first spot (since Bob can't finish ahead of Meg) and the last 3 spots can be filled in 3! ways; so there are 3 x 3! = 3 x 6 = 18 ways this could happen.
If Meg comes in the third, there are 3 choices for the first spot, 2 choices for the second spot, 2 choices for the fourth spot, and 1 choice for the last spot; so, this could happen in 3 x 2 x 2 = 12 ways.
Finally, if Meg comes in the fourth, then Bob must finish last and the first 3 spots can be filled in 3! = 6 ways.
Note that Meg cannot finish the race in the last position because then Bob will have finished the race ahead of Meg.
In total, there are 24 + 18 + 12 + 6 = 60 ways Meg can finish the race ahead of Bob.
Answer: C
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