Question

Two teachers are interviewing top 6 students in the FTRE exam, in two different subjects starting at the same time. Each teacher interview for 15 minutes; the number of ways in which interview can be scheduled is \[\]
A.$6!$ \[\]
B. $5!$\[\]
C.$44\times 6!$\[\]
D.$265\times 6!$\[\]

Answer 1

Hint: We find the number of ways $t$ the 6 students can be arranged in time slots for first teacher as ${{S}_{1}},{{S}_{2}},{{S}_{3}}$ and the time slots for the second teacher as ${{S}_{4}},{{S}_{5}},{{S}_{6}}$. We see that one student can be allotted only one slot and his or her time slot cannot be allotted any other students. So we find the number of derangements fro slot shuffling ${{D}_{6}}$ and the total number of ways for scheduling the interview as $t\times {{D}_{6}}$.\[\]

Complete step-by-step answer:
We know from combinatorics that derangement is the permutation of elements of a set such that no element appears in the original position. The number of derangements of a set of size $n$ is known as the subfactorial of $n$ or the ${{n}^{\text{th}}}$ derangement number. It is denote by ${{D}_{n}}$ and given by the expression
\[{{D}_{n}}=n!\left( 1-\dfrac{1}{1!}+\dfrac{1}{2!}-...+\dfrac{1}{n!} \right)=n!\sum\limits_{k=0}^{n}{{{\left( -1 \right)}^{k}}\dfrac{1}{k!}}\]

We are given the question that two teachers are interviewing top 6 students in FTRE exam, in two different subjects starting at the same time. Let us denote the time slots first teacher as ${{S}_{1}},{{S}_{2}},{{S}_{3}}$ and the time slots for second teacher as ${{S}_{4}},{{S}_{5}},{{S}_{6}}$. \[\]
We observe that if any of the 6 students can appear in the interview on the time slots${{S}_{1}},{{S}_{2}},{{S}_{3}},{{S}_{4}},{{S}_{5}},{{S}_{6}}$.So the number of ways we can allot the time slot is $t=6!$.\[\]
 One student can appear only once in one time slot allotted to him or her irrespective whether the interviewer is the first teacher or the second teacher. The time slot allotted to him or her cannot be given to any other student. So the problem is a derangement problem of a set of size $n=6$ where the elements are students and positions are time slots. So the number of derangements is the number of ways time slot can be shuffled which is,
\[\begin{align}
  & {{D}_{6}}=6!\left( 1-\dfrac{1}{1!}+\dfrac{1}{2!}-...+\dfrac{1}{6!} \right) \\
 & \Rightarrow {{D}_{6}}=6!\left( 1-1+\dfrac{1}{2}-\dfrac{1}{6}+\dfrac{1}{24}-\dfrac{1}{120}+\dfrac{1}{720} \right) \\
 & \Rightarrow {{D}_{6}}=720\times \left( \dfrac{360-120+30-6+1}{720} \right)=265 \\
\end{align}\]
So we use the rule of product find the total number of ways to schedule the interviews
\[t\times {{D}_{6}}=6!\times 265\]

So, the correct answer is “Option D”.

Note: We note that the positions or slots are not fixed derangement problems. We can alternatively solve by finding the number of one-one map that can be made from the set $\left\{ {{S}_{1}},{{S}_{2}},{{S}_{3}},{{S}_{4}},{{S}_{5}},{{S}_{6}} \right\}$. The other type of derangement problems are letter-envelopes, unique digits on places etc. We note that the derangement follows the recursion formula $n!=n!\left( n-1 \right)+{{\left( -1 \right)}^{n}}$.