Question

How many permutations are there for the word ‘Numbers’?

Answer 1

We have to find the total number of permutations for the word ‘numbers’. This implies that we have to find the total number of words which can be formed by rearranging the word, ‘numbers’ in various different ways. Thus, we shall apply the formula of finding permutations and calculate the factorial of the total number of letters in the word ‘numbers’. Further we shall divide it by the factorial of the number of occurrences of each letter in this word.

Complete step-by-step answer:
Given the word, ‘numbers’.
The formula of permutations for finding the total permutations, $p$ of the word, numbers is given as
$p=\dfrac{n!}{{{m}_{a}}!.{{m}_{b}}!.....{{m}_{z}}!}$
Where,
$n=$total number of letters in the given word
${{m}_{a}},{{m}_{b}},.....,{{m}_{z}}=$ number of occurrences of the letters $a,b,.....,z$ in the given word
The total number of letters in this word are 7, thus, $n=7$.
Here, ${{m}_{n}},{{m}_{u}},{{m}_{m}},{{m}_{b}},{{m}_{e,}}{{m}_{r}}$and ${{m}_{s}}$ are the total number of occurrences of the letters n, u, m, b, e, r and s respectively.
We see that ${{m}_{n}},{{m}_{u}},{{m}_{m}},{{m}_{b}},{{m}_{e,}}{{m}_{r}}$and ${{m}_{s}}$ are all equal to 1.
Thus, we get number of permutations as
$p=\dfrac{n!}{{{m}_{n}}!.{{m}_{u}}!.{{m}_{m}}!.{{m}_{b}}!.{{m}_{e}}!.{{m}_{r}}!.{{m}_{s}}!}$
$\Rightarrow p=\dfrac{7!}{1!.1!.1!.1!.1!.1!.1!}$
We know that $7!=5040$and $1!=1$. Substituting these values, we get
$\Rightarrow p=\dfrac{5040}{1.1.1.1.1.1.1}$
$\Rightarrow p=\dfrac{5040}{1}$
$\Rightarrow p=5040$
Therefore, the total number of permutations for the word ‘numbers’ is 5040.

Note: Permutation is a one-to-one mapping from a finite set to itself or an ordering of a finite set of distinct elements. In other words, permutation is a transformation of a set’s prime form, by applying one or more of certain operations, specifically, transposition, inversion and retrograde.