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?

We need to distribute 8 identical pens among three people with specific minimum requirements:

Amal: at least 1 pen

Bimal: at least 2 pens

Kamal: at least 3 pens

The key insight is to satisfy the minimum requirements first, then distribute the remaining pens freely.

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Let's give each person exactly what they need at minimum:

Person	Minimum Required	Pens Given
Amal	1	1
Bimal	2	2
Kamal	3	3
Total	6	6

Remaining pens = 8 - 6 = 2 pens

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Now we need to find how many ways we can distribute 2 identical pens among 3 people with no restrictions.

If we give:

$x_1$ additional pens to Amal

$x_2$ additional pens to Bimal

$x_3$ additional pens to Kamal

Then: $x_1 + x_2 + x_3 = 2$ (where each $x_i \geq 0$)

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Let's list all possible ways to write 2 as a sum of three non-negative integers:

$x_1$	$x_2$	$x_3$	Distribution Pattern
2	0	0	(2,0,0)
1	1	0	(1,1,0)
1	0	1	(1,0,1)
0	2	0	(0,2,0)
0	1	1	(0,1,1)
0	0	2	(0,0,2)

Total ways = 6

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For advanced students: The number of ways to distribute n identical objects among k people is given by the combination formula:

${{}}^{n + k - 1}C_{k - 1}$ or ${{}}^{n + k - 1}C_n$

For our remaining 2 pens among 3 people:

${{}}^{2 + 3 - 1}C_2 = {{}}^4C_2 = \frac{4!}{2! \times 2!} = 6$

This confirms our answer!

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The "minimum requirements first" approach is a powerful problem-solving strategy:

Satisfy all constraints

Distribute remaining items freely

Count using systematic enumeration or combinations

Answer: 6 ways