Six children, Austin, Ben, Carlos, Dave, Edmund, Frank are running in

Question

Six children, Austin, Ben, Carlos, Dave, Edmund, Frank are running in the race. In how many ways Austin finishes the race before Dave if there are no ties?

A. 60
B. 120
C. 240
D. 360
E. 720

KarishmaB · Answer

With no ties, 6 children can finish the race in 6! ways. In half of these, Austin will finish before Dave and in the other half, Dave will finish before Austin. Hence, in 6!/2 = 360 ways, Austin finishes before Dave. Answer (D) Check: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/1 ... s-part-ii/

100mitra · Answer

Correct option : D (360)

Memebrs : 6
Range : 4

Their are 4 condition out of 6, 
when Austin finishes before Dave, remaining 2 is either equal or after.

Permutations : nPr = 6P4 = 360

Posted from my mobile device

sherxon · Answer

Hi  VeritasKarishma ,
If ties were possible, would we divide it by 3?

antoniogmat · Answer

I can't understand the result. This is my approach:

If Dave finishes 6th, Austin can finish in 5 positions (1st to 5th)= 5
If Dave finishes 5th, Austin can finish in 4 positions (1st to 4th)= 4
.
.
.
If Dave finishes 1st, Austin cannot finish in a better position.
Therefore, 5*4*3*2*1=5!=120

Could please someone explain where I am mistaken and why answer b is not correct?
Thank you!

KarishmaB · Answer

If ties were possible, the question would have to tell you whether 2 way ties, 3 way ties etc are possible. Also, can all ranks be tied etc. You would need to consider those cases too. The question may become too cumbersome.

chetan2u · Answer

You have gone wrong in your approach. 
1) If D is 6th and A is 5th, remaining 4 can be in 4! ways. As A can be in 4, 3, 2 or 1st position, we can take it as with D in 6th position, A can be in any of 5 positions and remaining 4 in 4! ways. => 5*4!
2) Similarly when D is in 5th position, A can take any of the top 4 positions and remaining 4 will take 4! ways. => 4*4!
3) When D is in 4th position => 3*4!

Total : 5!+4*4!+3*4!+2*4!+1*4!=4!(5+4+3+2+1)=24*15=360

Hope it clears your query.

antoniogmat · Answer

Thank you very much for your help, I have it much clearer now.

rvgmat12 · Answer

Total combinations = 6! = 720

In half, Austin will finish before Dave

360

D

Bunuel · Answer

Discussed here: https://gmatclub.com/forum/six-mobsters ... ml#p645860

bumpbot · Answer

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Six children, Austin, Ben, Carlos, Dave, Edmund, Frank are running in

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