<s>[INST] Distribute $13$ identical balls in $6$ cells. Find the number of distributions such that at least $10$ balls will be in the first 3 cells together Let $13$ identical balls be distributed in $6$ cells. Find the number of distributions in which there are at least $10$ balls in the first $3$ cells together.
My attempt:
First I'll divide into cases: $A_{0},A_{1},A_{2},A_{3}$
$\sum_{i=0}^{3}A_{i}=$ all distributions possible.
This is a question where the order matters and reparations
is allowed. Given $n$ balls and $k$ cells, there are: $\binom{n+k-1}{n}$
ways to distribute $n$ balls to $k$ cells.
$A_{0}$ - there are $13$ balls in the first $3$ cells, and $0$ balls in the $3$ other cells thus $k_{1}=3,n_{1}=13\Longrightarrow$$\binom{n+k-1}{n}=\binom{15}{13}=105$ ways to distribute $13$ balls between $3$ cells. The distribution of balls in the first $3$ cells is disjointed to the distribution of balls in the other $3$ cells, therefore we'd apply the multiplication principle:
number of ways to distribute $0$ balls in $3$ cells: 
$\binom{2}{0}=1\Longrightarrow A_{0}=\binom{15}{13}=105$
$A_{1}$ - there are $12$ balls in the first $3$ cells,
and $1$ ball in the $3$ other cells:
$k_{1}=3,n_{1}=12,k_{2}=3,n_{2}=1 \Longrightarrow\binom{n_{1}+k_{1}-1}{n_{1}}\cdot\binom{n_{2}+k_{2}-1}{n_{2}}=\binom{15}{12}\cdot\binom{3}{1}=455\cdot3=1365$
$A_{2}$- there are $11$ balls in the first $3$ cells,
and $2$ ball in the $3$ other cells:
$k_{1}=3,n_{1}=11,k_{2}=3,n_{2}=2$}$\Longrightarrow\binom{n_{1}+k_{1}-1}{n_{1}}\cdot\binom{n_{2}+k_{2}-1}{n_{2}}=\binom{15}{11}\cdot\binom{4}{2}=1365\cdot6=8190$}
$A_{3}$- there are $10$ balls in the first $3$ cells,
and $3$ ball in the $3$ other cells:
$k_{1}=3,n_{1}=11,k_{2}=3,n_{2}=3 \Longrightarrow\binom{n_{1}+k_{1}-1}{n_{1}}\cdot\binom{n_{2}+k_{2}-1}{n_{2}}=\binom{15}{10}\cdot\binom{5}{3}=3003\cdot10=30030$}
$\sum_{i=0}^{3}A_{i}=105+1365+8190+30030=39690$

I'm not sure that my answer is correct, it seems to me that my result is to big.

 [/INST] You're very close, but have made a small mistake in calculating ${n_1 + k_1 - 1 \choose n_1}$: the top argument does not always equal $15$. You should find:
$$A_0 = {15 \choose 2} {2 \choose 2} = 105$$
$$A_1 = {14 \choose 2} {3 \choose 2} = 273$$
$$A_2 = {13 \choose 2} {4 \choose 2} = 468$$
$$A_3 = {12 \choose 2} {5 \choose 2} = 660$$
$$A_0 + A_1 + A_2 + A_3 = 105 + 273 + 468 + 660 = 1506$$
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