Answer

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**Hint**: We require multiple words using the same letters as the given word so we prefer to use the concept of permutation here. Permutation is the process of organizing all the elements of a set into various sequences in mathematics. In other words, changing the order of elements in a given set is to be done here.

Formula used:

$ \Rightarrow {}^n{P_r} = \dfrac{{n!}}{{(n - r)!}} $

**Complete step-by-step answer**:

We are given a word ‘SOCIETY’, and we are required to find the total number of words that can be constructed by utilizing the letters in the given word.

For finding the total number of words, we must use the concept of permutations. In simple terms, when a set of a particular order is given, the method of rearranging or shuffling its elements is known as permuting and the process is termed as permutation.

So here when we permute the letters of ‘SOCIETY’ we can get the number of words that we can possibly make out of the letters of ‘SOCIETY’.

The words can either have two letters, three letters, four letters, five letters, six letters or seven letters.

Total number of words made by:

Two letters $ \Rightarrow {}^7{P_2} = \dfrac{{7!}}{{(7 - 2)!}} $

$ \Rightarrow \dfrac{{7!}}{{(7 - 2)!}} = \dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{5 \times 4 \times 3 \times 2 \times 1}} $

$ \Rightarrow \dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{5 \times 4 \times 3 \times 2 \times 1}} = 42 $

Three letters $ \Rightarrow {}^7{P_3} = \dfrac{{7!}}{{(7 - 3)!}} $

$ \Rightarrow \dfrac{{7!}}{{(7 - 3)!}} = \dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{4 \times 3 \times 2 \times 1}} $

$ \Rightarrow \dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{4 \times 3 \times 2 \times 1}} = 210 $

Four letters $ \Rightarrow {}^7{P_4} = \dfrac{{7!}}{{(7 - 4)!}} $

$ \Rightarrow \dfrac{{7!}}{{(7 - 4)!}} = \dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1}} $

$ \Rightarrow \dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1}} = 840 $

Five letters $ \Rightarrow {}^7{P_5} = \dfrac{{7!}}{{(7 - 5)!}} $

$ \Rightarrow \dfrac{{7!}}{{(7 - 5)!}} = \dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}} $

$ \Rightarrow \dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}} = 2520 $

Six letters $ \Rightarrow {}^7{P_6} = \dfrac{{7!}}{{(7 - 6)!}} $

$ \Rightarrow \dfrac{{7!}}{{(7 - 6)!}} = \dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{1} $

$ \Rightarrow \dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{1} = 5040 $

Seven letters $ \Rightarrow {}^7{P_7} = \dfrac{{7!}}{{(7 - 7)!}} $

$ \Rightarrow \dfrac{{7!}}{{(7 - 7)!}} = \dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{1} $

$ \Rightarrow \dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{1} = 5040 $

Total value $ = 42 + 210 + 840 + 2520 + 5040 + 5040 $

Total value $ = 13692 $

Therefore the total number of words that can be constructed with the letters of ‘SOCIETY’ is $ 13692 $ .

**So, the correct answer is “$ 13692 $”.**

**Note**: Similar to the concept of permutations, we can take a look at combinations. The combination is a method of choosing objects from a list in which the order of selection is irrelevant (different from permutations). A combination is the selection of r items from a collection of n items where no replacement takes place, and the order is irrelevant.

Formula: $ {}^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}} $

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