In how many ways can the letters of the word ‘HEXAGON’ be permuted

1.	In how many ways can the letters of the word ‘HEXAGON’ be permuted? In how many words will the vowels be together?
Answer» There are 7 letters in the word HEXAGON. Formula: Number of permutations of n distinct objects among r different places, where repetition is not allowed, is P(n,r) = n!/(n-r)! Therefore, a permutation of 7 different objects in 7 places is P(7,7) = \(\frac{7!}{(7-7)!}\) = \(\frac{7!}{0!}\) = \(\frac{5040}{1}\) = 5040. They can be permuted in P (7,7) = 5040 ways. The vowels in the word are E, A, O. Consider this as a single group. Now considering vowels as a single group, there are total 5 groups (4 letters and 1 vowel group) can be permuted in P (5,5) Now vowel can be arranged in 3! Ways. Formula: Number of permutations of n distinct objects among r different places, where repetition is not allowed, is P(n,r) = n!/(n-r)! Therefore, the arrangement of 5 groups and vowel group is P(5,5)×3! =\(\frac{5!}{(5-5)!}\) × 3! = \(\frac{5!}{0!}\) x 3! = \(\frac{120}{1}\) x 6 = 720. Hence total number of arrangements possible is 720.

In how many ways can the letters of the word ‘HEXAGON’ be permuted? In how many words will the vowels be together?

Answer»

There are 7 letters in the word HEXAGON.

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, a permutation of 7 different objects in 7 places is

P(7,7) = \(\frac{7!}{(7-7)!}\) = \(\frac{7!}{0!}\) = \(\frac{5040}{1}\) = 5040.

They can be permuted in P (7,7) = 5040 ways.

The vowels in the word are E, A, O.

Consider this as a single group.

Now considering vowels as a single group, there are total 5 groups (4 letters and 1 vowel group) can be permuted in P (5,5)

Now vowel can be arranged in 3! Ways.

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, the arrangement of 5 groups and vowel group is

P(5,5)×3! =\(\frac{5!}{(5-5)!}\) × 3! = \(\frac{5!}{0!}\) x 3! = \(\frac{120}{1}\) x 6 = 720.

Hence total number of arrangements possible is 720.