Question

How many permutations of the word “spell” are there?

Answer 1

Hint: A permutation is a sequenced selection of items from some collection. When we are given a collection of digits or letters, we have to find the number of ways we can arrange them in a sequential order to calculate the number of permutations. If we have \[n\] similar items in a given collection, then the number of permutations will be \[{{\left( \dfrac{1}{n} \right)}^{th}}\] of the total number of possible permutations.

Complete step by step answer:
As per the given question, we are given a word “spell”. We need to find the number of permutations we have with the letters present in the word “spell”.
In the given word “spell”, we have the letter ‘l’ repeated two times. So, let us rename one of the lowercase l to uppercase L.
Now we have a set of 5 letters. They are \[\left\{ s,p,e,l,L \right\}\]. It is the same as the case where there are 5 chairs in which 5 persons are arranged to sit. Let the number of ways be ‘N’.
So, while making the first choice we have 5 persons in choice. That is 5 ways. --(1)
After the first arrangement, we have 4 people to choose from. That is 4 ways. ---(2)
For the third arrangement, there are 3 persons. That is, 3 ways of arrangement --(3)
Similarly, we can make the fourth arrangement in 2 ways and the last one in just 1 way. ---(4)
From point 1 to 4, it is a continuous process after which it gets completed. So, we can write \[N=5\times 4\times 3\times 2\times 1\] which is equal to 120.
Actually, we have the letter ‘L’ repeated two times. Hence, the number of permutations N becomes half.
\[\, therefore, \] There are 60 permutations of the word “spell”.

Note:
For a \[n\] letter word or \[n\] digit number, the number of possible permutations can be calculated by the factorial of \[n\]. That is, \[n!\]. We know that \[n!=n\times (n-1)\times ......\times 3\times 2\times 1\]. If we have \[m\] letters or digits in common where \[m\le n\], then the number of permutations decreases to \[\dfrac{n!}{m!}\].