In how many ways can three people be accommodated in five rooms if they don't mind sharing?

Question

The answer is 125 if that helps

Answer 1

Surprisingly, it is a combined problem, permutations and combinations. This is because, if we have two persons A and B in the same room, it is the same two people if we place them in order B and then A.

 

For each case, we will use the Fundamental counting principle to find the total number of ways we can accommodate the three guests.

 

Case 1. All people are in one room. We have three people and 5 rooms.

 

The total number of ways is the product of in how many ways we can choose 1 room out of 5 ₅C₁ and

the number of ways we can choose 3 people out of 3 people ₃C₃ .

 

Case 1 = ₅C₁ • ₃C₃ = 5 • 1 =5  

 

Case 2. One room will take 2 people and one room will receive 1 person.

 

We need to multiply the total number of ways we can choose 2 rooms out of 5 ₅C₂ , the total number of ways we can choose 2 people out of 3 ₃C₂ and the number of ways we can mix the two groups of people ₂C₁.

 

Case 2=  ₅C₂ • ₃C_{2 • 2}C₁=10 • 3 • 2 = 60

 

Case 3. Each room will take 1 person. The product is the number of ways we can choose 3 rooms out of 5 ₅C₃ multiplied by the number of different ways we can place the three people in these three rooms ₃P₃.

 

Case 3 =₅C_{3 • 3}P₃ = 10 • 6 = 60.

 

We add Cases 1, 2 and 3:

(Case 1)+(Case 2)+(Case 3) = 5 +60+60 =125.