Number of ways in which the letters of the word 'ABBCABBC' can be arranged such that the word ABBC does not appear is any word is

Options

A. `256`
B. `391`
C. `361`
D. `498`

Answer

C

Explanation

`(c )` `A's=2`, `B's=4`, `C's=2`
Total words formed `=(8!)/(4!2!2!)=420` ……..`(i)`
Let `ABBC='xx'`
Number of ways in which `xxABBC` can be arranged `=(5!)/(2!)=60`
But this includes `xxABBC` and `ABBCxx`.
The word `ABBCABBC` is counted twice in `60`m hence it should be `59`.
Hence, required number of ways `=420-59=361`