Given work DAUGHTER has 8 letters in which A, E,U are 3 vowels
We treat all vowels occur together as a single object and this single object together with 5 remaining objects (letters) will account for 6 object’*:’ These 6 objects can be arranged in 6P6 ways = 6! Ways corresponding to each of these arrangements, 3 vowels can be rearranged in 3! Ways ( 3 P3 ways)
∴ Required number of arrangements
= (6 x 5 x 4 x 3 x 2 x 1) x (3 x 2 x l)
= 720 x 6 = 4320
(ii) Required number of arrangements in which all vowels donot occur together = All possible arrangements of 8 letters taken all at time – number of arrangements in which the vowels occur together
= 8R8-6! x 3!
= 8! – 6! x 3!
= 61(8 x 7-3 x 2 x 1)
= 720 (50) = 36000
