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horses and positions - Tough math

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by gmatrant » Sat Oct 18, 2008 9:26 am
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

Solution:

_ _ _ _ _ _ (6 dashes to be equated to 6 positions)

Case 1: No hope comes first, then number of arrangements with Goldenrod coming second is 5!
Case 2: No hope comes second, , then number of arrangements with Goldenrod coming second is 4!
Case 3 :No hope comes third then number of arrangements with Goldenrod coming second is 3!
Case 4 :No hope comes fourth, , then number of arrangements with Goldenrod coming second is 2!
Case 5 :No hope comes fifth, , then number of arrangements with Goldenrod coming second is 1 way
5! + 4! + 3! +2! +1 = 153, but answer is 360.

Can anyone tell me why my solution is wrong. What have i missed calculating??

Thanks
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by Morgoth » Sat Oct 18, 2008 9:44 am
Exactly half of the times one horse will be ahead of the other.

total arrangements = 6! = 720

No Hope always finishes before Goldenrod = 720/2 = 360.

OA?
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Re: horses and positions - Tough math

by Morgoth » Sat Oct 18, 2008 9:49 am
gmatrant wrote:
Solution:

_ _ _ _ _ _ (6 dashes to be equated to 6 positions)

Case 1: No hope comes first, then number of arrangements with Goldenrod coming second is 5!
Case 2: No hope comes second, , then number of arrangements with Goldenrod coming second is 4!
Case 3 :No hope comes third then number of arrangements with Goldenrod coming second is 3!
Case 4 :No hope comes fourth, , then number of arrangements with Goldenrod coming second is 2!
Case 5 :No hope comes fifth, , then number of arrangements with Goldenrod coming second is 1 way
5! + 4! + 3! +2! +1 = 153, but answer is 360.


Thanks
Case 1: No hope comes first, then number of arrangements with Goldenrod coming second is 5 * 4!
Case 2: No hope comes second, , then number of arrangements with Goldenrod coming second is 4* 4!
Case 3 :No hope comes third then number of arrangements with Goldenrod coming second is 3*4!
Case 4 :No hope comes fourth, , then number of arrangements with Goldenrod coming second is 2*4!
Case 5 :No hope comes fifth, , then number of arrangements with Goldenrod coming second is 4!

5*4! + 4*4! + 3*4! + 2*4! + 4! = 4! * 15 = 24 * 15 = 360


Hope this helps.
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by cramya » Sat Oct 18, 2008 4:35 pm
If you were to draw slots it would look like this:


NH - No Hope (any slot wiht NH has value 1 since we are fixing its spot)

Case1:
NH COMES FIRST

NH 5 4 3 2 1 = 120
---- ---- ---- ---- ---- ----

Case2:
NH COMES SECOND

4 NH 4 3 2 1 = 96
---- ---- ---- ---- ---- ----


Case3:
NH COMES THIRD

4 3 NH 3 2 1 = 72
---- ---- ---- ---- ---- ----


Case4:
NH COMES FOURTH

4 3 2 NH 2 1 = 48
---- ---- ---- ---- ---- ----


Case5:
NH COMES FIFTH

4 3 2 1 NH 1 = 24
---- ---- ---- ---- ---- ----

Add all you would get 360

Hope this helps!
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by cramya » Sat Oct 18, 2008 6:13 pm
NH - No Hope (any slot wiht NH has value 1 since we are fixing its spot)

What I emant by this there is only 1 way to fix NH in a slot i.e. 1
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by gmatrant » Sat Oct 18, 2008 10:20 pm
thanks mogroth and cramya... for the explanation...quite clear now.
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