Question

Find the number of 5 letter words that can be formed using the letters S, A, R, A, N, A and M.

Answer 1

Now to solve this question we will take 3 cases. First we will find the number of words formed by taking all letters different. Next we will find the number of words formed when there are two A’s in the word. Lastly we will find the number of words formed when there are 3A’s in the word.

Complete step by step solution:
Consider the given letters S, A, R, A, N, A and M.
Here we can see that there are 3A’s and rest all letters are different.
Now we want to form a 5 letter word from these letters.
Hence either the letter will contain only 1 A’s that are all distinct letters or the letter contains just 2 A’s or the letter has 3 A’s.
We will solve this problem using cases.
Case 1: All the 5 letters are different.
Now let's consider that all the 5 letters in the word are different. Now hence the choice of letters are S, A, R, N, M. Hence the total number of words formed in this case is 5! = 120.
Now consider Case 2: There are 2 A’s in the word.
Now let us take out 2 A’s. Now we want to select 3 letters from the remaining 4 letters which are S, R, N and M.
The number of ways 2 select 3 letters from 4 letters is $^{4}{{C}_{3}}=\dfrac{4!}{3!1!}=4$
Now we can again arrange the letters in 5! Ways
Also since we have 2A’s we will divide the total number by 2!
Hence the number of words formed in this case is $\dfrac{5!\times 4}{2!}=120\times 2=240$
Now consider Case 3: There are 3 A’s in the word.
Now we will first fix 3 A’s in the word Now we want to select 2 letters from 4 available letters.
The number of ways to do so is $^{4}{{C}_{2}}=\dfrac{4!}{2!2!}=6$
Now again we can arrange the letters in 5! Ways.
Now since we have 3A’s we will divide by 3!.
Hence the number of words formed in this case is $\dfrac{6\times 5!}{3!}=120$
Hence the total number of words formed in 3 cases is 240 + 120 + 120 = 480.
Hence the total number of words formed is 480.

Note: Now note that here since we have 3A’s in the letter set we cannot directly use the formula $^{7}{{C}_{5}}$ as according to this counting AASRNM and AASRNM are same words as it counts every permutation. Hence to solve these types of questions we take cases.