Class 11

Math

Algebra

Permutations and Combinations

Find the number of words that can be formed by using all the letters of the word $_{′}DIFFERENTIATION_{′}$

- $15!$
- $4!.3!.2!_{4}$
- $3!×2!_{4}15! $
- none of these

Given word $DIFFERENTIATION=15$ letters.

We have repeated letters $−I$($3$ times ) $,F(2),E(2),N(2),T(2)$

Total permutations $=3!2!2!2!2115! $

$=3!(2!)_{4}15! $

Hence, the answer is $3!(2!)_{4}15! .$