Combination question(plz explain)

IMO D

It is a permutation question.

My approach is that I group A & E and treat them as a new letter, AE.

Now, the problem turns to be Total of available ways to arrange 4 letters AE B C D in a line =4!=24.

akpareek wrote:Arrange A, B, C, D, E in a line. How many ways are possible such that A always comes before E ?

A) 60 B) 48 C) 36 D)24 E) 18

hi all
i am novice, and my small thoughts about this problem
A in the first place

A,B ___ ____ ____ first case ABCDE keep the AB fixed and CDE will results in 3! outcomes
the same is valid for
A__ B__ ___ for example ACBED again A and B are fixed and for CDE 3! out comes
again
A __ __ B___ 3!-outcomes
A __ ___ __ B 3! outcomes

in total when A is first we have 3!*4 outcomes

__ AB__ __ 3!
__ A__ B__ 3!
__ A __ __ B 3!
when A is second 3!*3

when A is third
__ __ AB __ 3!
__ __ A __ B 3!

3!*2 for A the tird
and the last case
__ __ __ A B 3!

in total
3!*4+3!*3+3!*2+3!*1=3!(4+3+2+1)=60
so my answer A

some smart math students assert that total number of permutations is 5!
and exactly in half cases B will be behind
5!/2

maihan wrote:IMO D

It is a permutation question.

My approach is that I group A & E and treat them as a new letter, AE.

Now, the problem turns to be Total of available ways to arrange 4 letters AE B C D in a line =4!=24.

You have assumed that A always comes immediately before E. You have ignored all the arrangements where the other letters can intervene between A and E (A...C... E, etc).

This:

some smart math students assert that total number of permutations is 5!
and exactly in half cases B will be behind
5!/2

is certainly the best approach. A will precede E in half of the total number of arrangements while E will precede A in the other half.

Understand. Thanks for pointing out my mistake.