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The Reqd. No. of Permutations=3360.
Suppose that, out of n things, r1 are of first type, r2 are of
second type, r3 are of third type,..., where r1+r2+r3+...=n.
Then, no. of possible distinct permutations is given by
n!(r1!)(r2!)(r3!)...
In our Example, there are total 8 letters in the word INFINITY ,
out of which, 3 letters are of one type (i.e., the letter I ), 2
are of second type (i.e., the letter N ) and the remaining 3 are
(i.e., the letters F,T and Y) are each of 1 type.
Thus, n=8,r1=3,r2=2,r3=r4=r5=1.
The Reqd. No. of Permutations=8!(3!)(2!)(1!)(1!)(1!)
=8×7×6×5×42!=3360.
i think this is the answer
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