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

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Yathin13
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Six children, Austin, Ben, Carlos, Dave, Edmund, Frank are running in [#permalink] New post  08 Sep 2021, 21:42
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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

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KarishmaB
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Re: Six children, Austin, Ben, Carlos, Dave, Edmund, Frank are running in [#permalink] New post  08 Sep 2021, 21:56
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Yathin13 wrote:
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. 72


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/
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100mitra
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Re: Six children, Austin, Ben, Carlos, Dave, Edmund, Frank are running in [#permalink] New post  08 Sep 2021, 22:53
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

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Re: Six children, Austin, Ben, Carlos, Dave, Edmund, Frank are running in [#permalink] New post  Updated on: 09 Sep 2021, 05:57
VeritasKarishma wrote:
Yathin13 wrote:
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. 72


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/


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

Originally posted by sherxon on 08 Sep 2021, 23:24.
Last edited by sherxon on 09 Sep 2021, 05:57, edited 1 time in total.
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antoniogmat
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Re: Six children, Austin, Ben, Carlos, Dave, Edmund, Frank are running in [#permalink] New post  09 Sep 2021, 02:13
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!
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KarishmaB
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Re: Six children, Austin, Ben, Carlos, Dave, Edmund, Frank are running in [#permalink] New post  09 Sep 2021, 21:42
Expert Reply
sherxon wrote:
VeritasKarishma wrote:
Yathin13 wrote:
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. 72


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/


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


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.
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chetan2u
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Re: Six children, Austin, Ben, Carlos, Dave, Edmund, Frank are running in [#permalink] New post  12 Sep 2021, 04:39
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antoniogmat wrote:
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!



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.
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antoniogmat
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Re: Six children, Austin, Ben, Carlos, Dave, Edmund, Frank are running in [#permalink] New post  12 Sep 2021, 10:36
chetan2u wrote:
antoniogmat wrote:
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!



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.


Thank you very much for your help, I have it much clearer now. :thumbsup:
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Re: Six children, Austin, Ben, Carlos, Dave, Edmund, Frank are running in [#permalink] New post  12 Sep 2021, 23:52
Total combinations = 6! = 720

In half, Austin will finish before Dave

360

D
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Bunuel
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Re: Six children, Austin, Ben, Carlos, Dave, Edmund, Frank are running in [#permalink] New post  15 Sep 2021, 00:30
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Yathin13 wrote:
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


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