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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are u sure???

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which class r u Nawaznutss...???

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  1. sorry my above answer was my mistake my new answer is5!*6!*56
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vowels are E,U,A,I,O

Consonants are Q,T,N

Take all vowels as one alphbet and similar for consonants

5 vowels can be arranged in (factorial 5) ways. Similarly 3 consonants can be arranged in (factorial 3 ) way. But we can exchange the position of vowel and consonant so it can be done in 2 ways.

So total number of ways=(facorial5)(factorial 3)x2=240x6x2=2880

So total 2880 words can be formed.

Hope you got it



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In the word EQUATION, there are 5 vowels, namely, A, E, I, O, and U, and 3 consonants, namely, Q, T, and N.

Since all the vowels and consonants have to occur together, both (AEIOU) and (QTN) can be assumed as single objects. Then, the permutations of these 2 objects taken all at a time are counted. This number would be

Corresponding to each of these permutations, there are 5! permutations of the five vowels taken all at a time and 3! permutations of the 3 consonants taken all at a time.

Hence, by multiplication principle, required number of words = 2! × 5! × 3!

= 1440

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The answer is 1440 as when we take factorial we are only considering both ways and hence we won't multiply it by 2
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