How many distinct permutations can be made from the letters of the word "infinity"?

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Answer 1 · 2017-01-25T14:47:14

$The Reqd. No. of Permutations= 3360 .$ $"The Reqd. No. of Permutations="3360.$

Suppose that, out of $n$ $n$ things, $r_{1}$ $r_1$ are of first type, $r_{2}$ $r_2$ are of

second type, $r_{3}$ $r_3$ are of third type,..., where $r_1+r_2+r_3+...=n.$

Then, no. of possible distinct permutations is given by

$(n!)/{(r_1!)(r_2!)(r_3!)...}$

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, $n=8, r_1=3, r_2=2, r_3=r_4=r_5=1$ .

$"The Reqd. No. of Permutations="(8!)/{(3!)(2!)(1!)(1!)(1!)}$

$=(8xx7xx6xx5xx4)/(2!)=3360.$

Enjoy Maths.!