Problem Solving

vabhs192003 wrote:Q. In how many ways can 3 letters out of five distinct letters A,B,C,D & E be arranged in a straight line so that A and B are never together?

The question is asking us to arrange three out of five letters in a straight line, not to arrange all the five letters.

Also the question has some ambiguity in the word "together". If "never together" means A and B will not be placed side by side, then the solution is as follows...

Number of ways to select 3 letters from 5 = 5C3 = 10
Number of ways to arrange three letters in a straight line = 3! = 6
Hence, number of possible arrangements without any restriction = 6*10 = 60

Now, A and B will be together only if they are among the selected three letters.
Number of ways to select 3 letters from 5 such that A and B are always selected = Number of ways to select 1 letter from 3 = 3C1 = 3
Now, in each of the above selection, we can arrange the three letters in 3! = 6 ways.
However, A and B will not be together only if they are placed on the first and last place, which can be done in 2 ways. Hence, A and B will be together for the rest (6 - 2) = 4 ways.
Hence, A and B will be always together for 3*4 = 12 arrangements

Hence, A and B will never be together for (60 - 12) = 48 arrangements


However, if "never together" means A and B will not be selected together, then the answer will be (60 - 18) = 42