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P and C

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P and C

by nikhilagrawal » Tue Jul 22, 2008 3:50 am
65. Goldenrod and No Hope are in a horse race with 6 contestants. How many different arrangements of finishes are there if No Hope always finishes before Goldenrod and if all of the horses finish the race?
(A) 720
(B) 360
(C) 120
(D) 24
(E) 21

Pls explain with soln ..
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by newera » Tue Jul 22, 2008 4:19 am
I'm still learning permutation and combination, but will give it a shot.

I think the OA is C.

There are 6 contestants. If No Hope is before Goldenrod, then there are 5 arrangements how that is possible.

Either: NGXXXX, XNGXXX, XXNGXX, XXXNGXX, XXXXNG

Now you have to tackle the different arrangements for the other horses. There are 4! ways the four remaining horses can be arranged.

So, 4! * 5 = 24 * 5 = 120

I could be missing a step or very wrong...let me know if this is right. Thanks!
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by airan » Tue Jul 22, 2008 5:48 am
Solution 1:
There can be 6! different possibilities alltogether. Now half of them will be where GoldenRod will be in first 3 and No hope in second 3 i.e. in half of them, one will be ahead of another.
So total combinations is 360.


Solution 2: ( Found on internet)

if A finishes 1... B could be in any of other 5 positions in 5 ways and other horses finish in 4! ways, so total ways 5*4!

if A finishes 2... B could be in any of the last 4 positions in 4 ways. but the other positions could be filled in 4! ways, so the total ways 4*4!

if A finishes 3rd... B could be in any of last 3 positions in 3 ways, but the other positions could be filled in 4! ways, so total ways 3*4!

if A finishes 4th... B could be in any of last 2 positions in 2 ways, but the other positions could be filled in 4! ways, so total ways... 2 * 4!

if A finishes 5th .. B has to be 6th and the top 4 positions could be filled in 4! ways..

A cannot finish 6th, since he has to be ahead of B

therefore total number of ways

5*4! + 4*4! + 3*4! + 2*4! + 4! = 120 + 96 + 72 + 48 + 24 = 360



I dont think either of them are the best approches..Can anyone provide a better and simpler approach ?
Thanks
Airan
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by gupta.amit3 » Tue Jul 22, 2008 6:21 am
I will go with C: 120.

Explanation as given by newera.
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by newera » Tue Jul 22, 2008 6:33 am
Ok, i think my answer is wrong because i assume that No Hope has to be right behind Goldenrod. But Airan pointed out that doesn't have to be the case. i.e. NXXGXX is very possible as well

Someone please explain for me too lol. :)
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by pepeprepa » Tue Jul 22, 2008 6:34 am
For 6 contestants including Goldenrod and No hope
I agree with 5*4!=120

But we are false, Goldenrod can be anywhere after No Hope not just after. We did not read the question precisely.
It's 360 as Airan explained in his two methods.
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by arorag » Tue Jul 22, 2008 8:08 am
Lets look at this way
6 H so they can arrange in 6!
Now this is uniform disturibution so 1/2 time G will be ahead of N and 1/2 N will be ahead of G.
Ans is 6!-6!/2 = 6! 1/2= 360
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by beeparoo » Wed Jul 23, 2008 7:30 pm
arorag wrote:Lets look at this way
6 H so they can arrange in 6!
Now this is uniform disturibution so 1/2 time G will be ahead of N and 1/2 N will be ahead of G.
Ans is 6!-6!/2 = 6! 1/2= 360
Totally agree with Arorag except you don't need this redundant step: 6! - 6!/2.

(1/2)6! is enough based on logic.
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