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

1. $5$
2. $6$
3. $7$
4. $8$

Let's first give pens to all of them to satisfy their minimum requirement.
$$\begin{array}{} \underbrace{\text{Amal}}_{1(\geq 1)} & \underbrace{\text{Bimal}}_{2(\geq 2)} & \underbrace{\text{Kamal}}_{3(\geq 3)} \end{array}$$
After that, pens left $= 8-(1+2+3) = 8-6 = 2$ pens.

These $2$ pens can be distributed among  Amal, Bimal, and Kamal in the following ways:
$$\begin{array}{ccc} \underline{\text{Amal}} & \underline{\text{Bimal}} & \underline{\text{Kamal}} \\ 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \\ 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}$$
$\therefore$ The total number of distribution possible $=6$ ways.

$\textbf{Second Method:}$

Let's,

• Amal $=a$
• Bimal $=b$
• Kamal $=c$

$a+b+c=8; a\geq1, b\geq2, c\geq3$

$\Rightarrow a+b+c=8; a-1\geq 0, b-2\geq 0, c-3\geq 0 \quad \longrightarrow(1)$

Let,

• $a-1=x \implies a=x+1$
• $b-2=y \implies b=y+2$
• $c-3=z \implies c=z+3$

Now,  $x+1+y+2+z+3=8; x\geq 0, y\geq 0, z\geq 0$

$\Rightarrow x+y+z= 8-(1+2+3); x\geq 0, y\geq 0, z\geq 0$

$\Rightarrow x+y+z= 2; x\geq 0, y\geq 0, z \geq 0$

Number of distribution possible $= \;^{3+2-1}C_2 = \; ^4C_2 = \frac{4!}{2!2!} = \frac{4\times3\times2!}{2\times1\times2!}=6\;\text{ways}.$

Correct Answer $:\text{B}$

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