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
