Question

the letters of of the word HOSTEL are arranged so that vowels occupy the end places then the number of arrangements is

Answer 1

In HOSTEL , there are 6 letters.

There are also 2 vowels: O and E.

It is said that the vowels will be at end positions.

So, the word can become:
O _ _ _ _ E
Or it can become:
E _ _ _ _ O

So, there are 2 possibilities for it.

So, now we have to arrange the remaining 4 letters in the remaining four places.

So, number of arrangements =
4P4
= 4!
= 24

Thus, the total number of possible permutations is = 2×24 = 48

Thus, the answer is 48.

Answer 2

Answer:

Step-by-step explanation:

the answer is 48

the number of constants are 4

so 4!

the number of vowels are 2

so 2!

we know the formula m*n

m=4!

n=2!

m*n

=4! * 2!

24 * 2

= 48