Question

The difference between the different permutation of the word BANANA and rank of the word BANANA is
(A) 60
(B) 35
(C) 24
(D) None of these

Answer 1

Hint: First of all, we have to find the total number of words formed by BANANA. We know the formula that the total number of permutations of n items in which m items are the same is equal to $${\displaystyle \frac{n!}{m!}}$$. At first, we have the words starting with the letter A. We have to form a 6 letter word and A is fixed at the first position, so now we have 5 letters remaining that are B, N, A, N, and A. We have the letter A and letter N repeated twice. Using the given formula we will get total number of words starting from A. Now, we have the words starting from the letter BAA. We have to put the letters ANN at three different positions and we have the letter N repeated twice. Use this formula and get the total number of words starting from BAA. Now, we have the words starting with the letters BAN. The possible words are BANAAN, BANANA, and BANNAA. Now, get the rank of the word BANANA. Now, calculate the difference between the different permutations of the word BANANA and rank of the word BANANA.

Complete answer:

According to the question, it is given that we have the word BANANA. We have to find the difference between the different permutations of the word BANANA and rank of the word BANANA.

First of all, we have to find the total number of permutations of the word BANANA.

We know the formula that the total number of permutations of n items in which m items are the same is equal to $${\displaystyle \frac{n!}{m!}}$$ .

In the word BANANA, we have six letters. 

The total number of permutations of 6 letters = $$6!$$

But in the letter BANANA, we have the letter A repeated thrice and the letter N repeated twice.

So, the total number of permutations of 6 letters = $${\displaystyle \frac{6!}{3!\times 2!}}={\displaystyle \frac{1\times 2\times 3\times 4\times 5\times 6}{(1\times 2\times 3)(1\times 2)}}={\displaystyle \frac{4\times 5\times 6}{2}}=60$$ ……………………………..(1)

In the word BANANA, we have the letters B, A, and N. We know the orders of the letters. The letter A comes first then B, and then N.

At first, we have the words starting with the letter A.

Since we have to form a 6 letter word and A is fixed at the first position, now we have 5 letters remaining that are B, N, A, N, and A. 

We have to put these five letters at five different positions and we have the letter A and letter N repeated twice.

So, the total number of words starting from the letter A = $${\displaystyle \frac{5!}{2!2!}}={\displaystyle \frac{1\times 2\times 3\times 4\times 5}{(1\times 2)(1\times 2)}}=2\times 3\times 5=30$$ ……………………………(2)

Now, we have the words starting with the letters BAA.

Since we have to form 6 letter words and the letters B, A, and A are fixed at the first three positions so now, we have 3 letters remaining that are N, N, and A. 

We have to put these three letters at three different positions and we have the letter N repeated twice.

So, the total number of words whose first three letters are B, A, and A = $${\displaystyle \frac{3!}{2!}}={\displaystyle \frac{1\times 2\times 3}{(1\times 2)}}=3$$  ……………………………(3)

Now, we have the words starting with the letters BAN.

Since we have to form 6 letter words and the letters B, A, N are fixed at the first three positions so, now we have 3 letters remaining that are A,A and N. 

The possible words are BANAAN, BANANA, BANNAA

The word “BANANA” is the 2nd letter. 

The rank of the word BANANA = 30 + 3 +2 = 35 ……………….(4)

So, the rank of the word BANANA is 35.

The difference between the different permutations of the word BANANA and rank of the word BANANA = 60 -35 = 25.

Answer is none of these.

Hence, the correct option is D.

Note: In this question, one might think that the total number of permutations for the word BANANA is $$6!$$ . This is wrong because in the word BANANA we have the letter A repeated thrice and the letter N repeated twice. So, we have to divide $$6!$$ by $$3!$$ and $$2!$$ . Therefore, the total number of possible permutations is $${\displaystyle \frac{6!}{3!\times 2!}}$$.