Question

If the different permutations of all the letters of the word BHASKARA are listed as in a dictionary, how many strings are there in this list before the first word starting with B?

Answer 1

Hint: To solve the given question, we will first make use of the fact that the words which are there in the list before the first word starting with B will start from A. So, we will find out what the total number of repeating letters of A is. Then we will fix A in the starting position and we will do permutation for the rest 7 letters with the help of the formula \[\dfrac{n!}{r!}\] where n is the total letters and r is the number of repeating letters.

Complete step-by-step answer:
Before solving the question, we must know what a permutation is. A permutation is an arrangement of entities into a sequence or a linear order. Now, the total number of letters in the word BHASKARA is 8 out of which there are three As. Now, we will make use of the fact that the words which are present in the dictionary before the words starting with B, we will start from letter A. Thus, we have to find the total number of words in which the starting letter is A. To do this, we will fix one A in the starting position and we will find the total permutations of the remaining 7 letters. Now, in the remaining letters, we have two As and five other letters.
The number of permutations of ‘n’ entities having ‘r’ entities common of one kind is given by
\[\text{Number of permutations}=\dfrac{n!}{r!}\]
Thus, we have,
\[\Rightarrow \text{Number of permutations}=\dfrac{7!}{2!}\]
\[\Rightarrow \text{Number of permutations}=\dfrac{7\times 6\times 5\times 4\times 3\times 2\times 1}{2\times 1}\]
\[\Rightarrow \text{Number of permutations}=\dfrac{5040}{2}\]
\[\Rightarrow \text{Number of permutations}=2520\]
Thus, there are 2520 words in the dictionary before the first word starting with B.

Note: One might be confused as to why the answer is not \[\dfrac{8!}{3!}\] where there are a total of 8 letters and 3 As. This is wrong because we have already fixed one A so we will not apply permutation for it. We will apply permutation only for the remaining 7 letters which consist of 2 As. Thus the answer is \[\dfrac{7!}{2!}.\]