Question

In a hotel, four rooms are available. Six persons are to be accommodated in these four rooms in such a way that each of these rooms contains at least one person and at most two persons. Then the number of all possible ways in which this can be done is _____

Answer 1

we have to find the number of all possible ways in which Six persons are to be accommodated in these four rooms in such a way that each of these rooms contains at least one person and at most two persons.

solution : This question is based on grouping.let's solve it.

let R₁ , R₂ , R₃ and R₄ be the four rooms in a hotel. six persons are arranged in these four rooms in such a way that each of these rooms contains at least one person and at most two persons.

so, arrangements of persons in the room can be 1, 1, 2, 2

using formula, number of ways = p ! × q! /(r₁! r₂ ! r₃ !.... × n! s₁ ! s₂ ! 2! ...)

here p = 6 , q = 4

so, number of possible ways = (6! × 4!)/{(1! × 1! × 2!) × (2! × 2! × 2!)} = (6 × 5 × 4 × 3 × 2 × 24/16)

= 6 × 5 × 4 × 3 × 3

= 1080

Therefore the number of all possible ways in which this can be done is 1080