The number of permutations that can be made out of the letters of the word “DEFENCE" which start with a consonant and end with a consonant is:

Concept:

Suppose a set of n objects has n₁ of one kind of object, n₂ of a second kind, n₃ of a third kind, and so on with n = n₁ + n₂ + n₃  +...+ n_k then the number of distinguishable permutations of the n objects is  = \(\frac{{{\rm{n}}!}}{{{{\rm{n}}_1}! \times {{\rm{n}}_2}! \times {{\rm{n}}_3}! \ldots \ldots \ldots {{\rm{n}}_{\rm{k}}}!}}\)

Calculation:

The word 'DEFENCE' has 3 vowels “E, E, E" and 4 consonants” D, F, N, C". 

The consonant for the first place can be selected in 4 ways.

First placed occupied by consonant now remaining consonant are three for the last place.

Then, the consonant for the last place can be selected in 3 ways.

The other 5 places in between can be occupied by the remaining 5 letters in (5!/3!) ways = 20 ways.   (Out of 5 letters, 3 letters are repeated)

Hence, the total number of ways = 4 × 3 × 20 = 240 ways