The number of words,which can be formed using | vedclass

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(C) The word $	ext{DAUGHTER}$ contains $8$ distinct letters: $D, A, U, G, H, T, E, R$. There are $3$ vowels: $A, U, E$ and $5$ consonants: $D, G, H,

Vedclass · Accepted Answer

$36000$

Vedclass · Answer

(C) The word $	ext{DAUGHTER}$ contains $8$ distinct letters: $D, A, U, G, H, T, E, R$. There are $3$ vowels: $A, U, E$ and $5$ consonants: $D, G, H, T, R$. Total number of words formed using all $8$ letters $= 8! = 40320$. To find the number of words where all vowels never come together,we subtract the number of words where all vowels are together from the total number of words. Treating the $3$ vowels $(A, U, E)$ as a single unit,we have $5$ consonants $+ 1$ unit $= 6$ entities. These $6$ entities can be arranged in $6!$ ways. The $3$ vowels within the unit can be arranged in $3!$ ways. Number of words where vowels are together $= 6! 	imes 3! = 720 	imes 6 = 4320$. Number of words where vowels never come together $= 8! - (6! 	imes 3!) = 40320 - 4320 = 36000$.

The number of words,which can be formed using all the letters of the word $\text{DAUGHTER}$,so that all the vowels never come together,is

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