Total number of ways 2 people can finish the race

nPr = 5P2 = 5!/(5-2)! = 5!/3! = 20

Since there are 20 possible ways that these two can finish the race in various order, half of them will be John finishing ahead of Bill, and the other half will be Bill finishing head of John. So simply half the the number from above.

20/2 = 10

Another vote for 10.

f2001290 -- if there's no explanation given for these questions, I'd
recommend more authentic sources like MGMAT, kaplan and PR.

Let there be 5 runners
J B X Y Z


So number of ways would be

With J at 1st place and B behind him @ any other 4 places = 4
With J at 2nd place and B behind him @ any other 3 places = 3
With J at 3rd place and B behind him @ any other 2 places = 2
With J at 4th place and B behind him @ any other 1 place = 1

Total = 4+3+2+1= 10

I am getting 12.
if we consider john and bill as one entity then :

JB PQR
The above group can be rearranged in 4! ways.
However the arrangement of JB and BJ is reduced to just JB

So 4!/2! = 12...


Is there something wrong in my approach..pls help

800GMAT wrote:I am getting 12.
if we consider john and bill as one entity then :

JB PQR
The above group can be rearranged in 4! ways.
However the arrangement of JB and BJ is reduced to just JB

So 4!/2! = 12...


Is there something wrong in my approach..pls help

If you take "JB" as 1 entity, you are essentially missing out cases
like JXYZB, JXYBZ etc.

Oh I see.......I guess, i misunderstood the question to mean that B is immediately after J

By the way, if the question did state that B is immediately after J, then would my approach be correct ....?

800GMAT wrote:Oh I see.......I guess, i misunderstood the question to mean that B is immediately after J

By the way, if the question did state that B is immediately after J, then would my approach be correct ....?

Yes, if B immediately follows J, then your approach would be correct.

thankx jayhawk

I think gmatscore.com chose D using the following reasoning- total ways for 5 people to finish the race = 5! = 120. In half of the races John will finish the race ahead of Bill....hence 60

One observation- I tried this reasoning on set containing 3 runners
ABC can finish a race in 3! = 6 ways
ABC
ACB
BAC
BCA
CAB
CBA
In the above possibilites, A finishes ahead of B in half of the possibilites....i.e. in 3 ways

using similar reasoning, I think, D is the right answer

jayhawk2001 wrote:
If you take "JB" as 1 entity, you are essentially missing out cases
like JXYZB, JXYBZ etc.

Hi jayhawk....also one more observation....if I were missing out cases dont u think my answer should be less than 10.....thnkx

60 is the right answer

John is in the first place = 1*4! = 24
John is in the second place = 3C1 * 1 *3! =18
John is in the third place = 3P2 * 2!= 12
John is in the fourth place = 3! * 1 * 1 =6

24+18+12+6 = 60