Question

the no. of ways in which 3 persons can occupy 6 rooms separetly is​

Answer 1

Answer:

120 WAYS

Step-by-step explanation:

To understand how, first hand, we can make the question simpler: Let's assume that 2 people have to occupy 3 rooms.

Name these 3 rooms as R1, R2, R3 & 2 people occupy these rooms in the following ways…

First person can stay in either R1 or R2 or R3. Similarly, the second person can stay in either R1, or R2, or R3. So, when any 2 rooms are occupied by these 2 people, the number of ways will be counted as follows….

R1 R2

R1 R3

R2 R1

R2 R3

R3 R1

R3 R2

So, finally after they occupy the rooms… The above mentioned sample space will be the number of ways…

And here we have a total of 6 ways. Here, order is important, so we come to this conclusion that we don't have to find combinations. But we go for counting permutations.

Like, 1st person goes in R1, 2nd person goes in R2.

Then 1st person goes in R2, the 2nd goes in R1.

These will be 2 different ways…

Hence, we calculate permutation

ie, 3P2 = 3! / ( 3–2)! = 3 x 2 = 6 ways..

Now, our question is 3 people occupy 6 rooms…

Let those rooms be R1, R2, R3, R4, R5, R6

SO, different ways will be :

R1 R2 R3, R1 R2 R4, R1 R2 R5, R1 R2 R6

R1 R3 R4, R1 R3 R5, R1 R3 R6

R1 R4 R5, R1 R4 R6,

R1 R5 R6,

Now, Keeping R1 constant , we take remaining rooms in different order…..

R1 R3 R2, R1 R4 R2, R1 R5 R2, R1 R6 R2

R1 R4 R3, R1 R5 R3 , R1 R6 R3

R1 R5 R4 , R1 R6 R4

R1 R6 R5

So, this way we see, when 1st person is in R1, there are 20 ways, Then he goes to R2. There will be 20 ways.. Then he goes to R3, there will be 20 ways. This way total will be 20 * 6 = 120 ways.

Start naking sample space with R2 & continue the process.. as shown above . Then do with R3,R4,R5,R6

The above pattern shows that we need to count permutations…

nPr = n! / ( n- r) ! , where 0 < r <,= n

6P3 = 6! / (6–3)!

=6P3 =( 6 x 5 x 4 x 3 x 2 x 1) / ( 3 x 2 x 1 )

= 6 x 5 x 4

= 120 ways ………ANS