Given $16$ pens to be distributed between $4$ people ,then find the number of distributions with the given assumptions.

Question

Given $16$ pens to be distributed between $4$ people assuming the fourth person gets at least $3$ pens and the three other ones should not get more than $5$ pencils,in how many ways it's possible?

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We are looking for the integral solutions to the equation $b_1+b_2+b_3+b_3+b_4=16$ such that $b_4 \ge 3$ and $b_1,b_2,b_3 \le 5$,this is given by :

$$[x^{16}]\left[\sum_{3 \le k \le 16}^{}x^k \right]\left[\sum_{0 \le k \le 13}^{}x^k \right]^3$$ $$=[x^{13}]\left[\sum_{0 \le k \le 13}^{}x^k \right]\left[\sum_{0 \le k \le 13}^{}x^k \right]^3$$ $$=[x^{13}](1-x^{14})^4(1-x)^{-4}$$ $$=[x^{13}]\sum_{k=0}^{4}\binom{4}{k}(-1)^kx^{14k}\sum_{k \ge 0}^{}\binom{k+3}{k}x^{k}$$ $$=\binom{16}{13}$$

However the answer is not this,so where was I wrong?

Answer 1

$b_1 + b_2 + b_3 + b_4=16$

As $b_4 \geq 3$, the number of ways to distribute $16$ pens will be equivalent to

$b_1 + b_2 + b_3 + b_4 = 13 \,$ without a restriction on $b_4$. The number of ways -

$\displaystyle = {16 \choose 3}$

Now $b_1, b_2, b_3 \leq 5$ so we should exclude number of ways in which one of them is $\geq 6$ which will be given by -

$b_1 + b_2 + b_3 + b_4 = 7$

Number of ways $\displaystyle = {3 \choose 1} {10 \choose 3} \,$ as there are ${3 \choose 1}$ ways to choose a number from $b_1, b_2, b_3$ that can be $\geq 6$.

But this will overcount cases where $2$ numbers are $6$ or greater. So we need to add those back.

$b_1 + b_2 + b_3 + b_4 = 1$ (two numbers are $6$ or more).

Number of ways $\displaystyle = {3 \choose 2} {4 \choose 3} \,$ as there are ${3 \choose 2}$ ways to choose two numbers from $b_1, b_2, b_3$ that can be $\geq 6$.

Now $3$ numbers cannot be all $\geq 6$ at the same time.

So applying P.I.E, our final answer is

${16 \choose 3} - {3 \choose 1} {10 \choose 3} + {3 \choose 2} {4 \choose 3} = 212$

Answer 2

I'm not sure where you went wrong, but here is a solution using a generating function. $$\begin{align} f(x) &= (x^3+x^4+x^5+\dots)(1+x+x^2+x^3+x^4+x^5)^3 \\ &= x^3 \cdot \frac{1}{1-x} \cdot \left( \frac{1-x^6}{1-x} \right)^3 \\ &= x^3 (1-x^6)^3 (1-x)^{-4} \\ &= x^3 (1-3x^6+3x^{12}-x^{18}) \sum_{i=0}^{\infty}\binom{4+i-1}{i} x^i \end{align}$$ So $$[x^{16}]f(x) = \binom{4+13-1}{13} -3\cdot \binom{4+7-1}{7} + 3 \cdot \binom{4+1-1}{1} = \boxed{212}$$