# Q.2.    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?

S seema garhwal

In the word EQUATION, we have

vowels = 5(A,E,I,O,U)

consonants = 3(Q,T,N)

Since all the vowels and consonants occur together so (AEIOU) and (QTN) can be assumed as single objects.

Then, permutations of these two objects taken at a time $=^2P_2=2!=2$

Corresponding to each of these permutations, there are $5!$ permutations for vowels and $3!$ permutations for consonants.

Thus, by multiplication principle, required the number of different words $= 2\times 5!\times 3!=2\times 120\times 6=1440$

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