Aptitude - Permutation and Combination - Discussion

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

This problem is combination or permutation ?

360 is the correct answer because "C" is repeated

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

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

Hi @Pavithra the word LEADER has repeated letter 'E' twice so 6!/2! = 6*5*4*3 = 360.

In how many ways ELECTION can be arranged so that the vowels occupy the odd places.

There are 9 switches in a fuse box, how many different arrangements are there?

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.

In this letter e repeated twice.
So that in permutation swapping of e letter.
Does not form new word and so they are counted.
Once at every case so divided by 2!.

Why not sometime division and sometime by considering vowel?