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

All 6 horses can finish the race in 6! way (assuming no tie).

If no tie is possible between No Hope and Goldenrod, then in half of these cases No Hope will be before Goldenrod and in half of these cases after (not necessarily right before or right after). How else? So, there are 6!/2=360 different arrangements of finishes where No Hope always finishes before Goldenrod.

Answer: B.

Similar question: six-mobsters-have-arrived-at-the-theater-for-the-premiere-of-the-126151.html
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combinations are symmetrical, so in half combinations goldenrod will be ahead and in exactly half it will be behind.

ans: 6!/2 = 360

just think of it like this

THere are 6 possible outcomes for person A to finish the race
and 5 for person B and so on

so u should multpily them all 6X5X4X...

and u get 720

I have solved this in below fashion



So, When 'N' finishes first then there are 5 places where G can take (i.e 5 ways)
When 'N' finishes second then there are 4 places where G can take (i.e 4 ways)
When 'N' finishes third then there are 3 places where G can take (i.e 3 ways)
...

No of ways is 5*4*3*2 = 120.

please explain flaw in my explanation.

Thanks in Advance

The most effective way to solve this is Bunuel's. However if we wanna take your approach more calculus are needed:

first of all 6 spots to fill _ _ _ _ _ _
If 'N' finishes first then the other 5 spots can be filled in \(5!\) ways N 5 4 3 2 1

And now it gets complicated...
If 'N' second then the other 5 spots can be filled in \(4*4!\) ways 4 N 4 3 2 1. Why this?
N is in second position and there are 5 horses left. Of those only 4 can occupy the first position (every horse EXCEPT g), so write 4 on the first line.
We have 4 slots left and 4 horses => 4 3 2 1 for the remaining spots

With the same method if N finished 3rd, we get \(4*3*3!\) ways 4 3 N 3 2 1
N on the third slot, 5 horses left. The first spot can be occupied by 4 horses (every horse EXCEPT g), the second can be occupied by three horses (every horse EXCEPT g and the previusly chosen one); then for the others spots we have 3 horses => 3! ways.
if N finished 4th, we get \(4*3*2*2!\) ways 4 3 2 N 2 1
if N finished 5th, we get \(4*3*2*1\) ways 4 3 2 1 N 1

Sum them up and you get 360. This approach however is really long

Hope it's clear

I will also apply the brunel specified method.

I just wanted to know where I went wrong and correct my understanding.

If N is on the first place G can take ANY of the remaining 5 places and the remaining 4 horses can be arranged in 4! number of ways: 5*4!=120;
If N is on the second place G can take 4 places and the remaining 4 horses can be arranged in 4! number of ways: 4*4!=96;
If N is on the third place G can take 3 places and the remaining 4 horses can be arranged in 4! number of ways: 3*4!=72;
If N is on the fourth place G can take 2 places and the remaining 4 horses can be arranged in 4! number of ways: 2*4!=48;
If N is on the fifth place G can take only 1 place and the remaining 4 horses can be arranged in 4! number of ways: 1*4!=24;

120+96+72+48+24=360.

Answer: B.

Hope it's clear.
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Questions about the same concept to practice:
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meg-and-bob-are-among-the-5-participants-in-a-cycling-race-58095.html
in-how-many-different-ways-can-the-letters-a-a-b-91460.html
mary-and-joe-are-to-throw-three-dice-each-the-score-is-the-126407.html

Hope it helps.
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@Bunuel,I used the method you explained later but i want to use the first method(All cases by 2) next time..How do i come to know in which questions i can use this approach..that in half cases one thing must have happened and in the other half cases the other thing?

Here's my thinking:
order (from 2nd-8th place) of 6 horses does't matter. So we have 6!=720 choices
720 choices include 2 cases: No Hope comes first and Goldenrod comes first.
Final answer is 720/2=360. Please correct me if there is anything wrong. Thanks

Hi Veterans,
I have a query, Since Glen and No Hope are always fixed wr to each other, cant we just combine them into an entity and thus have 5 entities left with us which can be arranged in 5! ways ? However, this gives wrong answer.
Appreciate a quick reply.

Thanks a ton!

No Hope always finishes before Goldenrod does NOT mean that No Hope always finishes right before Goldenrod, there might be some other contestants between them.
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Thanks Bunuel, clear as a crystal!

Hi Veterans,

I have a very embarrassing doubt,

The question stem reads "Goldenrod and No Hope are in a horse race with 6 contestants" doesn't that means we have 8 horses in the race ? (By calculating 6! and dividing it by 2 will only give stats about the 6 horses !!)

There are 6 horses and hence 6! outcomes are possible and in that 1/2 the time No Hope will be ahead of Goldenrod and 1/2 the times lag.
So it's \(\frac{6!}{2}\) = 360 (B)

Please correct me if I'm wrong, but since the question said Goldenrod and No Hope are in a race with 6 contestants, doesn't it mean that there are total 8 horses? Why is the correct answer 6!/2 rather than 8!/2? Thank you!

I think that the question means no that Goldenrod and No Hope are in a race with 6 OTHER contestants but rather that Goldenrod and No Hope are in a race WHICH HAS 6 contestants (so "a race with 6 contestants").
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Arrangements = 6!/2 = 360 Since 1 in 2 cases, No Hope finishes before Goldenrod.

IMO B
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This has been highlighted already but I would like to indicate that the question is ambiguous. I was under the impression that there were 8 total horses not 6.
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