Q. 184.0( 3 Votes )

In how many ways

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) = = = = 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! = = × 3! = × 6 = 720.


Hence total number of arrangements possible is 720.


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