Question

In how many ways can 8 identical pens be distributed among Amal, Bimal, and Kamal so that Amal gets at least 1 pen, Bimal gets at least 2 pens, and Kamal gets at least 3 pens?​

Answer 1

Let us assume Amal gets 1 + x pens and Bimal gets 2 + y pens and Kamal gets 3 + z pens

1 + x + 2 + y + 3 + z = 8

x + y + z = 2 (x , y , z being whole numbers) or

x + (1 + y) + (3 + z )

x + y + z = 8 – 3 = 5 (x , y, z being naturals numbers)

Take 5 ones 1 _ 1 _ 1 _ 1 _ 1 , + symbol should be added in any 2 slots. (There are 4 empty slots)

This is done in 4C2 ways = 6 ways.

Answer

6 ways

Answer 2

Step-by-step explanation:

Given, Amal, Bimal, and Kamal should have at least 1, 2, 3 pens respectively.

After Amal, Bimal and Kamal are given their minimum required pens, the pens left are 8 - (1 + 2 + 3) = 2 pens

Now, The remaining 2 pens can go to different people (3 ways – 1,1,0; 0,1,1; 1,0,1) or the same person (3 ways – 2,0,0; 0,2,0; 0,0,2)

So total no. ways to divide the pens among three = 3 + 3 = 6 ways