**Given : word EXAMINATION**

number of letters = 11

the dictionary words before starting letter E will be starting with A

**Case(1)** first letter A and second letter also A

remaining 9 letters have two I and two N

so these are permutations with repetition,

so number of ways = 9!/(2!2!) = 9!/4

**Case(2)** first letter A and second letter also I

remaining 9 letters have two N

so these are permutations with repetition,

so number of ways = 9!/(2!) = 9!/2

**Case(3)** first letter A and second letter is N

remaining 9 letters have two I

so these are permutations with repetition,

so number of ways = 9!/(2!) = 9!/2

**Case(4)** first letter A and second letter is any of the 5 letters E, M, T, O, X

these 5 letters can be chosen in 5 ways

remaining 9 letters have two I and two N

so these are permutations with repetition,

so number of ways = 9!/(2!2!) = 5(9!/4)

The total number of ways from all the 4 cases

= (9!/4) + (9!/2) + (9!/2) + 5(9!/4)

= 9!(1/4 + 1/2 + 1/2 + 5/4)

=9! (5/2)