Question

All the six letters of the more SACHIN are arranged to & fro. Different words repeating any letter in any one word. The words so formed are then arranged as in a dictionary. What will be the position of the word SACHIN in that sequence?
\[\begin{array}{*{20}{l}}
  {\left( A \right){\text{ }}436.0} \\
  {\left( B \right){\text{ }}570.0} \\
  {\left( C \right){\text{ }}601.0} \\
  {\left( D \right){\text{ }}751.0}
\end{array}\]

Answer 1

Hint:We have to read the question very carefully to understand what information has been given & what has been asked as an answer. We can solve this type of questions by applying the concept of factorial and permutation. As per question , S is the first letter so it is fixed & we know that if we take letters A, C, H, I , N from word SACHIN at the first position then the five letters can make new words by putting each letter in 5 different ways , so find no. of ways it can be done .Then, as only one word will be formed by taking 5 at first position so, add it to position of word in the sequence.

Complete step-by-step answer:
Here, we asked to find the position of the word SACHIN in the sequence given in question.
 If we take letters A, C, H, I, N from word SACHIN at the first position then the five letters can make new words in \[5!\] ways for each letter.
So total no. of ways will be 5 times the possibilities of making each word i.e,
\[5! \times 5\]$ = 5 \times \left( {5 \times 4 \times 3 \times 2 \times 1} \right) = 120 \times 5 = 600$
Elaborately, we can also solve it in another way that is –
As ‘S’ is fixed being the starting letter,
No. of letters that can start with ‘A’: \[5!\] $ = \left( {5 \times 4 \times 3 \times 2 \times 1} \right) = 120$
Similarly, no. of letters that can start with ‘C’: \[5!\] $ = \left( {5 \times 4 \times 3 \times 2 \times 1} \right) = 120$
Similarly, no. of letters that can start with ‘H’: \[5!\] $ = \left( {5 \times 4 \times 3 \times 2 \times 1} \right) = 120$
Similarly, no. of letters that can start with ‘I’: \[5!\] $ = \left( {5 \times 4 \times 3 \times 2 \times 1} \right) = 120$
Again, no. of letters that can start with ‘N’: \[5!\] $ = \left( {5 \times 4 \times 3 \times 2 \times 1} \right) = 120$
Hence total no. of words can be sum of all possibilities: $120 + 120 + 120 + 120 + 120 = 600$
Again, if we take another 5 letters at their position as in the word ‘SACHIN’, then only one word will be formed & that will be in $600 + 1 = {601^{th}}$position in the sequence.

So, the correct answer is “Option C”.

Note:First of all, the question should be read very carefully to understand what information is given & what has been asked as an answer in the question. To solve this problem, we should have the concept of Permutation crystal clear in our mind, that is applied for numerical related arrangements. It can also be solved by applying concepts of arithmetic series & sequence as mentioned in the solution part.