How many words,with or without meaning,can be | vedclass

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Question

(A) The word $EQUATION$ consists of $8$ distinct letters: $5$ vowels $(A, E, I, O, U)$ and $3$ consonants $(Q, T, N)$.Since the vowels must occur toge

Vedclass · Accepted Answer

$1440$

Vedclass · Answer

(A) The word $EQUATION$ consists of $8$ distinct letters: $5$ vowels $(A, E, I, O, U)$ and $3$ consonants $(Q, T, N)$.Since the vowels must occur together and the consonants must occur together,we treat the group of vowels $(AEIOU)$ as $1$ unit and the group of consonants $(QTN)$ as $1$ unit.These $2$ units can be arranged in $2!$ ways.Within the vowel group,the $5$ vowels can be arranged in $5!$ ways.Within the consonant group,the $3$ consonants can be arranged in $3!$ ways.By the multiplication principle,the total number of words is $2! 	imes 5! 	imes 3! = 2 	imes 120 	imes 6 = 1440$.

How many words,with or without meaning,can be formed using all the letters of the word $EQUATION$ at a time so that the vowels and consonants occur together?

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