Question

There are seven slots, where an interview can be
scheduled. Seven candidates are to be interviewed and
one slot can be assigned to one candidate.
Raman, Naman and Aman are three of the candidates.
In how many ways an interview schedule for all the
candidates can be made, such that Naman is
interviewed before Raman and Raman is interviewed
before Aman.​

Answer 1

Answer:

Total number of candidates = 9 The orders of Naman , Raman and Amon are fixed Therefore , Naman , Raman and Amon can be arranged only in 1 way Rest candidates = 6 If we take Naman , Raman and Amon as one entity , total entities to be arranged 7 Total arrangements = 7 ! = 5040

Answer 2

Given,

Number of slots = 7

Raman, Naman, and Aman are three of the candidates.

To find,

The number of ways an interview schedule for all the candidates can be made, such that Naman is interviewed before Raman and Raman is interviewed before Aman.​

Solution,

The number of ways the task can be performed is 5!

We can simply solve this mathematical problem by the following method.

We have to take care of the fact that, Naman is interviewed before Raman and Raman is interviewed before Aman.​

Thus,

We can first arrange Naman before Raman and Raman before Aman.

The order of these three would be then - Naman, Raman, Aman.

Now, the number of ways to interview the 7 candidates can be found by considering the three as one entity. This leaves us with only 5 entities to arrange, which can be done by 5!

Thus, the number of ways an interview for all can be scheduled is 5!