Problem Solving

If Mike and Ella are two of 4 participants in a race, how many different ways can the race finish where Ella always finishes in front of Mike?

A. 6
B. 12
C. 16
D. 18
E. 20

Night reader wrote:If Mike and Ella are two of 4 participants in a race, how many different ways can the race finish where Ella always finishes in front of Mike?

A. 6
B. 12
C. 16
D. 18
E. 20

There are 4! = 24 possible arrangements of the 4 participants.

In 1/2 of the arrangements, Mike will be in front of Ella; in the other 1/2 of the arrangements, Ella will be in front of Mike.

Thus, Ella will finish in front of Mike in 1/2 * 24 = 12 of the arrangements.

The correct answer is B.

my approach is conservative

when Ella finishes first => Mike can take any of the other three places, hence 3 => other two participants take the remaining 2 places in 2! => 3*2!

--- Ella comes second => Mike is in any two places, 2*2!

--- Ella is third => Mike is in one place, 1*2!

3*2! + 2*2! +1*2! = 12

There are 4! ways for all the contestants to finish the race - 24 ways

The condition is Ella should always finish in front of mike.

So , total number of ways - no of ways that Mike finishes before Ella = No of ways that Ella finishes before Mike

Mike finishes in 1 st place - so there are 3! ways for ella and other 2 participants to finish
Mike finishes in 2 nd place - 2! ways for Ella and the other participant to finish ( we are not bothered about who comes first as the condition is only between Mike and Ella)
Mike finishes in 3rd place - So there is just 1! way for Ella to finish.

Hence , 24 - (3!+2!+1!) = 24 - 9 = 15

Where am I going wrong??

What if the question actually meant this..
2nd place
Mike
1st Place
Ella

and
3rd place
Mike
2nd place
Ella

and

4th place
Mike
3rd Place
Ella.

What if i needed to find out where Ella and Mike donot finish in any other position.. other than being in (1 and 2 or 2 and 3 or 3 and 4)..

In this case i would go with E M A B...(A and B being other participants)
So, we would have to consider Ella and mike to be together...which means we are left with 3! and it comes to 6...
Isn't this question ambiguous...
Shouldn't the question be telling about other assumptions as well???

junegmat221 wrote:What if the question actually meant this..
2nd place
Ella
1st Place
Mike

and
3rd place
Ella
2nd place
Mike

and

4th place
Ella
3rd Place
Mike.

What if i needed to find out where Ella and Mike donot finish in any other position.. other than being in (1 and 2 or 2 and 3 or 3 and 4)..

In this case i would go with E M A B...(A and B being other participants)
So, we would have to consider Ella and mike to be together...which means we are left with 3! and it comes to 6...
Isn't this question ambiguous...
Shouldn't the question be telling about other assumptions as well???

the question does not appear to be any ambiguous. The combination of 3! is possible only if Ella finishes first and Mike can take any of the other three places => 3 arrangements, the other two participants can be placed in 2! ways => in total 3!

For this problem to have Ella and Mike together plus combination of 3!, the problem should have introduced one more participant provided Ella finished in front of Mike or
state that Ella and Mike together will finish first at any time and the other two participants can take their places 2!/(2-3)! as only two participants are left and three places are available... this is impossible => the factorial of negative is undefined.

So, we would have to consider Ella and mike to be together...which means we are left with 3! and it comes to 6...

I am sorry with this explanation...
Actually it should have been 4!/ (2!)= 12 ways.

2!->since we have taken Ella and Mike together...

Anyway,
Coming back to the question again

where Ella always finishes in front of Mike

Ella always finishes in front of mike.

Mike Back Ella At the front..
I shall put the order of the races where Mike Finishes at the back of Ella.
I am sorry about my previous post where i mentioned Mike to be finishing first which i will edit now.
The Race combination might be.

1. E M A B
2. E M B A
3. A E M B
4. B E M A
5. A B E M
6. B A E M


So i possibly assumed they have to be together..
Isn't this a possible assumption?
Night Reader, i guess u are getting my assumptions right this time...

junegmat221 wrote: The Race combination might be.

1. E M A B
2. E M B A
3. A E M B
4. B E M A
5. A B E M
6. B A E M

I am not sure what you are trying to do here. We have only one constraint - Ella is in front of Mike. If we had another constraint as understood by you - Ella is right in front of Mike then it would be 2! arrangements - and that's all. Cancel AB and BA for it is the same arrangement for us (no constraint).

Did you see 2! or 2 in the answer choices? The questions consist of both the stimuli and the questions stems. You've probably solved the quadratic equations in GMAT; you might know that these equations have two answers, and usually only one answer is sought in GMAT question stem.

Simply look at this problem this way,

if we have Ella coming first, we have only the set of three ___ ___ ___ to be filled in

since Mike can be the second, third and fourth..... M=3... A(orB)....B(orA) =
E M....A(orB)....B(orA)
E A(orB)....M....B(orA)
E A(orB)....B(orA)....M


Ella can be the second too - Mike can not be the first, so

___ ___ ___ ___ A(orB)....E....M....B(orA)

or A(orB)....E....B(orA)....M => it doesn't matter whether A or B is the first or fourth 2 * 2!

and finally Ella is the third then Mike is the last only => 1* 2!

in total 3*2! & 2*2! & 1*2!

sum up these permutations, for these are separate events - only one race may be started and ended at once => three sets of places are possible in each scenario

get 12 and select the right answer.

p.s. I tried to break down my answer to every tiny element of this permutation problem. If it's still not clear to you, read the permutations concept from the source I've read it - total GMAT math