Question

5 chocolates need to be placed in 3 containers. Each container can hold all the 5 chocolates. In how
many ways can the chocolates be placed in the containers if no container can be empty and all
chocolates and containers are different?

Answer 1

Answer:

Total Number of ways = 60 + 90 = 150.(Ans.)

Step-by-step explanation:

Answer :-

According to the question, we have 5 chocolates to be placed in 3 containers where no container remains empty.

Hence, we can have the following kinds of distributions :

Case 1 :-

The distribution will be (3,1,1) that is, one container gets three chocolates and the remaining two containers get one chocolate each.

So,

→ Total Number of ways to choose the container which gets 3 chocolates = 3 (since, 3 containers are there).

→ Total Number of ways to select 3 chocolates out of 5 = (5)C(3) = {(5!)/(5-3)!(3!)} = {(5!) / (2!)(3!)} = (5*4*3!)/((2!)(3!) = (5*4)/2 = 10 .

→ Total Number of ways to distribute rest of the chocolates = 2.

Therefore,

→ Total Number of ways = 3 * 10 * 2 = 60 ways.

________

Case 2 :-

The distribution will be (1,2,2) that is, one container gets one chocolate and the remaining two containers get two chocolates each.

→ Total Number of ways to select 1 chocolate out of 5 = (5)C(1) = {(5!)/(5-1)!(1!)} = {(5!) / (4!)(1!)} = (5*4!)/((4!)(1!) = (5/1) = 5 .

→ Total Number of ways to choose the container which gets 1 chocolates = 3 (since, 3 containers are there).

→ Total Number of ways to select 2 chocolates out of remaining 4 for second container = (4)C(2) = {(4!)/(4-2)!(2!)} = {(4!) / (2!)(2!)} = (4*3*2!)/((2!)(2!) = (4*3)/2 = 6 .

Therefore,

→ Total number of ways = 5 * 3 * 6 = 90 ways.

_________

Hence,

→ Total Number of ways = 60 + 90 = 150.(Ans.)