DS quesiton . please help

[silvia928]

4.) 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

The answer is : C

[getneonow]

(1+2+3+4) * 3! = 60


Neo

[kausis]

A slightly more detailed answer would be, since Meg finishes ahead of Bob, Meg can be at the 1st, 2nd, 3rd or 4th position (and not the last!).

So, if Meg is 1st, Bob can be in any one of the other 4 positions
-OR- (+)
if Meg is 2nd, Bob can be in any of the 3 next positions
-OR- (+)
---'' ---- 3rd, -------------------------'' --- 2 next positions
-OR- (+)
---'' ---- 4th, then Bob can be only at the last position (1 way)
-AND- (x)
the other three cyclists can be arranged in 3! ways.

Hence the final answer would be (4 + 3 + 2 + 1)x3! = 60 [C]

[Stuart@KaplanGMAT]

4.) 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

The answer is : C

A great first step would be figuring out that this is a problem solving question, not a data sufficiency one - please label your posts properly and put them in the proper section.

[silvia928]

thanks for all your help.