In how many ways can 2N identical items be distributed among N people, so everyone gets at least 1 items and at most 3 items I think I need to use Inclusion-Exclusion principle here but I do not know how.
Also I was thinking that when everyone must get at least 1, then I can distribute 1 to everyone and have N items remaining to distrubute among N people so that some get 1 more and some get 2 more(3 items at max per person)
 A: Your thoughts are a good start. Indeed you can distribute $1$ item to everyone and then you have $N$ more items to distribute to $N$ people, with everyone getting at most $2$ items. You can do this by summing over the number of people who get $2$ items. If $k$ people get $2$ items, then $N-2k$ people must get $1$ item and $N-k-(N-2k)=k$ people must get $0$ items. There are $\binom N{k,k,N-2k}=\frac{N!}{(N-2k)!k!^2}$ ways to make those choices, so in total there are
$$
\sum_{k=0}^{\left\lfloor\frac n2\right\rfloor}\frac{N!}{(N-2k)!k!^2}
$$
different distributions. I don’t think there's a closed form for this; Wolfram|Alpha just returns hypergeometric series.
A: Denote a person who gets $1,2,3$ items by $-,0,+$ respectively. Then the problem reduces to finding the number of ways to assign signs to $n$ people such that there are as many $+$ people as there are $-$ people. As a sum:
$$\sum_{k=0}^{\lfloor n/2\rfloor}\binom n{2k}\binom{2k}k$$
These are the central trinomial coefficients, OEIS A002426.
A: Generating Function Approach
The number we want is
$$
\begin{align}
\underbrace{\left[x^{2n}\right]\overbrace{\left(x+x^2+x^3\right)^n}^{\substack{\text{each of $n$ people get}\\\text{$1$, $2$, or $3$ items}}}}_{\substack{\text{number of ways}\\\text{to distribute $2n$ items}}}
&=\left[x^n\right]\left(1+x+x^2\right)^n\tag1\\
&=\left[x^n\right]\left(1-x^3\right)^n(1-x)^{-n}\tag2\\[12pt]
&=\left[x^n\right]\sum_k\binom{n}{k}(-1)^kx^{3k}\sum_j\binom{-n}{j}(-1)^jx^j\tag3\\[3pt]
&=\sum_k(-1)^k\binom{n}{k}\binom{2n-3k-1}{n-3k}\tag4\\
&=\sum_{k=0}^{\lfloor n/3\rfloor}(-1)^k\binom{n}{k}\binom{2n-3k-1}{n-1}\tag5
\end{align}
$$
Explanation:
$(1)$: $\left[x^{n+k}\right]x^kP(x)=\left[x^n\right]P(x)$
$(2)$: $\left(1+x+x^2\right)(1-x)=1-x^3$
$(3)$: apply the Generalized Binomial Theorem
$(4)$: apply negative binomial coefficients
$\phantom{\text{(4):}}$ extract the coefficient of $x^n$ by setting $j=n-3k$
$(5)$: the terms in $(4)$ are non-zero for $3k\le n$, then apply $\binom{n}{k}=\binom{n}{n-k}$

Inclusion-Exclusion Approach
The number of ways to distribute $2n$ identical balls to $n$ people where each person gets $1$, $2$, or $3$ balls, is the same as the number of ways to distribute $n$ identical balls to $n$ people where each person gets $0$, $1$, or $2$ balls (just start each person with $1$ ball).
Let $S(j)$ be the arrangements where person $j$ has $3$ or more balls. We want to count the number of arrangements in none of the $S(j)$. We will apply the Generalized Principle of Inclusion-Exclusion, where the sum of the intersection of $k$ of the $S(j)$ is
$$
N(k)=\overbrace{\ \ \ \binom{n}{k}\ \ \ }^{\substack{\text{the number of}\\\text{ways to choose}\\\text{the $k$ people to}\\\text{have the $3$ or}\\\text{more balls}}}\overbrace{\binom{n-3k+n-1}{n-3k}}^{\substack{\text{use stars and bars}\\\text{to count the number}\\\text{of ways to distribute}\\\text{the remaining $n-3k$}\\\text{balls to $n$ people}}}\tag6
$$
The number of ways to be in none of the $S(j)$ is
$$
\begin{align}
\sum_k(-1)^{k-0}\binom{k}{0}N(k)
&=\sum_k(-1)^k\binom{n}{k}\binom{2n-3k-1}{n-3k}\tag7\\
&=\sum_{k=0}^{\lfloor n/3\rfloor}(-1)^k\binom{n}{k}\binom{2n-3k-1}{n-1}\tag8
\end{align}
$$
which matches $(5)$.
