**Question:**

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

**Solution:**

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

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

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

The vowels in the word are $\mathrm{E}, \mathrm{A}, \mathrm{O}$.

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 $\mathrm{P}(5,5)$

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

$P(5,5) \times 3 !=\frac{5 !}{(5-5) !} \times 3 !=\frac{5 !}{0 !} \times 3 !=\frac{120}{1} \times 6=720$

Hence total number of arrangements possible is 720.