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
PS - gmatscore.com
This topic has expert replies
Source: Beat The GMAT — Problem Solving |
-
thumpin_termis
- Senior | Next Rank: 100 Posts
- Posts: 60
- Joined: Fri Jun 01, 2007 11:02 pm
-
jayhawk2001
- Community Manager
- Posts: 789
- Joined: Sun Jan 28, 2007 3:51 pm
- Location: Silicon valley, California
- Thanked: 30 times
- Followed by:1 members
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.
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
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
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
-
jayhawk2001
- Community Manager
- Posts: 789
- Joined: Sun Jan 28, 2007 3:51 pm
- Location: Silicon valley, California
- Thanked: 30 times
- Followed by:1 members
If you take "JB" as 1 entity, you are essentially missing out cases800GMAT 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
like JXYZB, JXYBZ etc.
-
jayhawk2001
- Community Manager
- Posts: 789
- Joined: Sun Jan 28, 2007 3:51 pm
- Location: Silicon valley, California
- Thanked: 30 times
- Followed by:1 members
Yes, if B immediately follows J, then your approach would 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 ....?
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
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
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
- Attachments
-
