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 :-).

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!:)

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.

Hello, @All.

In this question it means like how many ways (it means without repetition of letters) the "LEADER" letter can be arranged it supposed to be 6!=720. If the question is like how many different ways (with repetition of letters) can letter "LEADER" arranged then 6!/2! (2! is because of E letter repetition) = 360.

Hence the solution for this problem should be 720 as per my knowledge.

Hi @Thokie,

Q: There are 9 switches in a fuse box, how many different arrangements are there?
Ans: 2 2 2 2 2 2 2 2 2.

Each switch has two possible, on or off placing a 2 in each of the 9 position.
So, we have
2 raise to the power 9 = 512.

So it means there are 512 different arrangements can be done.

Hope this'll help.

Please See the question once carefully, it's just asked for number of ways so word repetition is not considered so the correct answer is 720 (6!) ,

If the question is asked like " How many DIFFERENT ways word Leader can be arranged" then it would have been 6!/2 for E-letter repeating 2 times.

I have a query.

LEADER= (LADR) +) (EE) = 4 + 1 = 5! = 120, and divided it by 2C2!, why can't be the answer 120.

The same way for CORPORATION problem is solved. All vowels are a set & consonants are of another set and as the vowels are scrambled it was divided by vowels combinations.

LEADER
It contains 6 letters i.e=6!
Now take repeated letters i.e E(2 times), so it can be divide by 6!
i.e LEADER =6!/2!.
= 6 * 5 * 4 * 3 * 2 * 1/2 * 1
= 360
So,360 is correct answer.

Take the example of the word EYE:

It can be arranged in 3 ways only , namely EYE , EEY, YEE as the letter E is repeated twice.

So the answer is not 3! but 3!/2!.

Applying the same for LEADER, we have the answer 6!/2!.