Question

In how many ways $9$ Mathematics paper can be arranged so that the best and the worst
a. May come together
b. May not come together

Answer 1

In the question, we are asking to find out the ways the best & worst papers out of $9$ mathematics papers should be together and not. To solve this question, use the concept of permutation and combination. First take out the ways of arranging all $9$ papers. Then take out the ways in which best and worst papers are together and the differ the former by latter to get the number of ways in which best & worst will not be together.

Complete step-by-step answer:
Total number of papers are $9$.
$\therefore$Total number of ways in which all $9$ papers will arrange is $= 9!$ ways.
For case I:
Consider the best and worst paper as one.
Therefore, now no. of papers become $8$.
So, they can be arranged in $8!$ ways.
Now, the two papers (best & worst paper) can be arranged in themselves in $2!$ ways.
Therefore, total no. of ways in which best & worst can be arranged is $= \left( {8! \times 2!} \right)$ways.
For case II:
Since, the total no. of ways in which $9$ papers are arranged is $9!$ ways.
& total no. of ways in which the best & worst be together is $\left( {8! \times 2!} \right)$ ways
Therefore total no. of ways in which they will not together,
= \left\\{ {9! - } \right.\left. {\left( {8! \times 2!} \right)} \right\\} ways
$= 8!\left( {9 - 2} \right)$ ways
$= 7 \times 8!$ ways

Note: The question asked was from the topic permutation as when we arrange some objects in different ways along with some restrictions. So, we will have to read the question very carefully to understand what has been asked as an ultimate answer. Do the calculations attentively to avoid silly mistakes instead of knowing concepts & procedures to be applied.