Aptitude - Permutation and Combination - Discussion

6!/2! = 6*5*4*3*2*1/2.

= 720/2.

= 360.

Hi factorial value is 1*2*36*4*5*6* = 720.

So 720/2 = 360.

The word as 6 letters, the letter e is repeated twice, so:

npr = 6!/2! = 360.

Whether we need to divide by 2 or 2!. As in this, it doesn't matter but if "e" was repeated thrice, then do we need to divide numerator by 3 or 3!

Can anybody explain me when permutation & when combination used?

It is asking how many way not how many different ways.

So, answer should be 6! = 720.

Lets look at it this way:

In LEADER, there are 2 Es and 4 Unique alphabets (LADR).

Now selection of 2 Es in any of the 6 positions is a combination problem. Hence E can be selected in 6C2 ways. Which is 6x5/2! = 15.

For each of these 15 choices of E, there are 4 remaining slots for 4 unique alphabets. Hence this is a permutation problem. The 4 unique alphabets can be arranged in 4! ways i.e. 4 x 3 x 2 x 1= 24.

Therefore total ways of arranging is 15x24 = 360.

Yes friends answer 360 is correct because E is repeated 2 times.

It is a permutation question, in which some letters are repeated so it can be solved as, ans = 6!/2! =720/2 = 360.

Please explain following question I am not understand what is mean by 6! = 360?
(1!) (2!) (1!) (1!) (1!).

!- what is this?