Since we know, Permutation of n objects taking r at a time is ⁿP_r,and permutation of n objects taking all at a time is n!

And, we also know Permutation of n objects taking all at a time having p objects of the same type, q objects of another type, r objects of another type is . i.e. the, number of repeated objects of same type are in denominator multiplication with factorial.

Given, the word INDEPENDENCE. It has 12 letters and it has 3 repeated letters ‘N’, ‘D,’ ‘E.’ Of which, the letter N is repeated thrice, the letter D is repeated twice, and the letter E is repeated 4 times in that word. All other letters are distinct.

The problem can now be rephrased as to find total number of permutations of 12 objects (I, N, D, E, P, E, N, D, E, N, C, E) of which three objects are of same type (N, N, N), two objects are of another type (D, D), and four objects are of different type (E, E, E, E).

Total number of such permutations

= 1663200

Hence, a total number of permutations of the word INDEPENDENCE is 1663200.