Aptitude - Permutation and Combination - Discussion

@Tanaya: both permutation and combination are used in this section.

In some problems we arrange the letters means they are of type permutation..and in some cases like selection group we used combination :-).

@Abhishek: let me explain you with the example

Suppose your name is : AAB which contain A 2 times (in above question E repeate 2 times so i consider name with one character repeat twice)

AAB can be arranged in

AAB ( 1)
ABA (2)
BAA (3)

Means 3 ways..

This ans can be calculate directly using formula

Formula:


(number of characters)!
------------------------
(number of repeat characters) !


In AAB there are 3 characters and one character repeat twice

So above formula become

3! 3*2*1
-- = ------ = 3 which we got above by doing manually
2! 2*1

similarly in above question

LEADER contains 6 character and E repeat twice so using formula

6!
--- = 360.
2!

Hope you got this :-).

Why not 5! * 2!? because E is repeated so now consider 5 letters so 5! and E is repeated twice so 2! ??

I have query that all these problems belong to combination and not permutation section?

Helo frndz in this word LEADER thre r 6 leters in which 2 leters are repeted thats why we divide it by 2! and the ans is 6!/2!

Why we take 6!/2! Whay we have not take vowel 3 and consonent 3. like (3+1)4
4!= 4*3*2*1= 24
3!/2!=3
24*3= 72

Hi poornima,

E is repeated in two times so we divided
leader -l,e,a,d,r ( e is repeted 2 times so considered single time)
6!=6x5x4x3x2x1/2x1=360

Can anybody explain me when permutation & when combination used?

How did 360 come?

What is (1!)? What is the meant of this symbol? And can anyone provide the easy method for this type of problem?

@Savitri.

It is;

L = 1.
E = 2.
A = 1.
D = 1.
R = 1.

(1!)(2!)(1!)(1!)(1!)

How it is? Explain me.