Problem Solving

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

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?

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.

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!

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

thanks mogroth and cramya... for the explanation...quite clear now.