How many words can be formed using the letter `A` thrice, the letter `B` twice and the letter `C` thrice?

Options

A. `(8!)/((3!)xx(2!)xx(3!))`
B. `8!`
C. `((8xx2xx3xx3)!)/((3!)xx(2!)xx(3!))`
D. None

Answer

A

Explanation

We are given 8 letters, namely AAABBCCC. Out of these letters three are alike of one kind (3 A's), two are alike of second kind (2 B's) and three are alike of third kind (3 C's).
Hence, the required number of permutations
`=(8!)/((3!)xx(2!)xx(3!))=560.`