Question

# Find how many $7$ letter words can be formed using the letters of the word ARIHANT.

Hint- Let’s consider a case which is similar to the condition of our question. Suppose we have $n$
number of balls ($r$ number of balls are identical) and n number of boxes each to be filled with one ball.
We will first select any r boxes out of the n boxes, in $C\left( {n,r} \right)$ ways. Now we will place the identical balls in the selected boxes, one in each. We are now left with $n - r$ boxes and $n - r$ balls, which can be placed in the boxes in $(n - r)!$ ways.
Therefore the number of ways of arranging $n$ objects in a row of which $r$ objects are identical is $C\left( {n,r} \right) \times (n - r)! = \dfrac{{n!}}{{r!(n - r)!}} \times (n - r)! = \dfrac{{n!}}{{r!}}$

Complete step by step solution:
We have to find number of $7$ letter words that can be formed using the letter of word ARIHANT
Set of the letter of the word ARIHANT =$\left\{ {A,R,I,H,A,N,T} \right\}$
Total number of letters = $7$
No. of letters identical = $2$
Here we have the similar condition as given below
The number of ways of arranging $n$ objects in a row of which $r$objects are identical =
$\dfrac{{n!}}{{r!}}$
On comparing we can assume that$n = 7$and $r = 2$
Therefore the number of ways of arranging $7$letters in a row of which $2$letters is identical =$\dfrac{{7!}}{{2!}}$
Since the factorial of a positive integer$n$, denoted by $n!$, is the product of all positive integers less than or equal to n:
$n! = n \times (n - 1) \times (n - 2) \times (n - 3) \times ......... \times 3 \times 2 \times 1$

$\therefore \dfrac{{7!}}{{2!}} = \dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times1}} = \dfrac{{5040}}{2} = 2520$
Hence the number of $7$ letter words that can be formed using the letters of word ARIHANT is $2520$.

Note:
The number of ways to arrange $n$ objects in a row, of which exactly $r$ objects are
identical, is$\dfrac{{n!}}{{r!}}$.
In a generalized form, if we wish to arrange a total no. of $n$objects, out of which $p$ are
of one type, $q$ of second type are alike, and $r$ of a third kind are same, then such a
computation is done by $\dfrac{{n!}}{{p!q!r!}}$