Aptitude - Permutation and Combination - Discussion

Why we should take like 6*5*4*3?

360 is the correct answer because "C" is repeated

6!/2! = 6*5*4*3 = 360.

This problem is combination or permutation ?

In how many different ways can the letters of the word 'ORANGE' be arranged so that the three vowels never come together ?

Answer :

The total number of ways of arranging ORANGE = 6!
The total number of ways o,a and e can be arranged = 3!
The total number of groups when the vowels are one group and the rest are individuals= 4!
= 6! - 4! x 3!
= 576

In how many different ways can the letters of the word 'EXTRA' be arranged so that the two vowels never come together ?

Answer:

The total number of ways of arranging EXTRA = 5!
The total number of ways e and a can be arranged = 2!
The total number of groups when the vowels are one group and the rest are individuals XTR(EA) = 4!
= 5! - 4! x 2!
= 72

LEADER

One way of arranging this is
"EADERL" (The first 'E' and the second 'E')
If I want to put the second 'E' in place of the first 'E' to make another arrangement, I would get
"EADERL" which is the same as the previous one.

POINT: we have shown that eventhough I interchange the first 'E' and the second 'E', we came up with the same arrangement.
MEANING... if there are TWO same letters, there will be DUPLICATION...
That is the reason why we DIVIDE by TWO!

It's not 5! because still, there are six letters we are arranging!

Hope this insight can help you!:)

@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