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Question

Question: How many distinguishable ways can the letter of the word SASSAFRAS be arranged using all the letters...

How many distinguishable ways can the letter of the word SASSAFRAS be arranged using all the letters?

Explanation

Solution

We have to find the total number of distinguishable ways in which the letter of the word SASSAFRAS can be arranged using all the letters. This implies that we have to find the total number of permutations for the word ‘SASSAFRAS’. Thus, we shall apply the formula of finding permutations and calculate the factorial of the total number of letters in the word ‘SASSAFRAS’. 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, ‘SASSAFRAS’.
The formula of permutations for finding the total permutations, pp of the word, numbers is given as
p=n!ma!.mb!.....mz!p=\dfrac{n!}{{{m}_{a}}!.{{m}_{b}}!.....{{m}_{z}}!}
Where,
n=n=total number of letters in the given word
ma,mb,.....,mz={{m}_{a}},{{m}_{b}},.....,{{m}_{z}}= number of occurrences of the letters a,b,.....,za,b,.....,z in the given word
The total number of letters in this word are 9, thus, n=9n=9.
Here, mS,mA,mF,{{m}_{S}},{{m}_{A}},{{m}_{F}},and mR{{m}_{R}} are the total number of occurrences of the letters S, A, F, and R respectively.
We see that mS=4{{m}_{S}}=4, mA=3{{m}_{A}}=3, mF=1{{m}_{F}}=1 and mR=1{{m}_{R}}=1,
Thus, we get number of permutations as
p=n!mS!.mA!.mF!.mR!p=\dfrac{n!}{{{m}_{S}}!.{{m}_{A}}!.{{m}_{F}}!.{{m}_{R}}!}
p=9!4!.3!.1!.1!\Rightarrow p=\dfrac{9!}{4!.3!.1!.1!}
We know that 9!=362880, 4!=24, 3!=69!=362880,\text{ }4!=24,\text{ }3!=6and 1!=11!=1. Substituting these values, we get
p=36288024.6.1.1\Rightarrow p=\dfrac{362880}{24.6.1.1}
p=362880144\Rightarrow p=\dfrac{362880}{144}
p=2520\Rightarrow p=2520
Therefore, the total number of permutations for the word ‘SASSAFRAS’ is 2520.

Note: One possible mistake we can make while solving these problems related to these permutations of a word is that we can make a mistake while counting the number of repetitive letters in the particular word. Thus, this counting of words must be done carefully.