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