Question

There are nine slots, where an interview can be scheduled. Nine candidates are to be interviewed and one slot con be assigned to one candidate. Ramon. Noman and Amon are three of the candidates. in how many ways on interview schedule for all the candidates can be made, such that Naman is interviewed before Roman and Raman is interviewed before Amon​

Answer 1

The number of ways = 5040

Step-by-step explanation:

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

Hope this answer is helpful.

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Q: Out of 9 people waiting for their turns for an interview, in how many ways can a selection for 4 be made if 1 particular person is always selected?

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