Question

How many different words can be formed with the letters of the word ORDINATE so that
(i) The vowels occupy old places
(ii) Beginning with $O$
(iii) Beginning with $O$ and ending with $E$.

Answer 1

4 vowels and 4 consonants. Total 8 letters.
(i) No. of words $=4!\times 4!=24\times 24=576$
Because 4 vowels are to be in old places and the 4 consonants are to be adjusted in the remaining places.
(ii) $7!$ ways, $O$ being fixed.
(iii) $6!$ ways, $O$ fixed in $1^{st}$ and $E$ fixed in last.