How many ways can 8 teachers be distributed among $4 $ schools? There are several ways that the teachers can be divided amongst $4$ schools, namely here are the possible choices I came up with:
$1) 1 1 1 5$
$2) 1 1 2 4$
$3) 1 1 3 3$
$4) 1 2 2 3$
$5) 2 2 2 2$
now given the fact that say $2213$ is the same as $1 2 2 3$ it was omitted. With out repeats I believe these 5 are the only possibilities. 
1) 
${8 \choose 5} \times {3 \choose 1} \times {2 \choose 1} \times {1 \choose 1}$:
$\frac{8!}{5!3!} \times \frac{3!}{1!2!} \times \frac{2!}{1!1!} \times 1$
Which comes out to 
$56 \times 3 \times 2 \times 1 = 336$
2) 
${8 \choose 4} \times {4 \choose 2} \times {2 \choose 1} \times {1 \choose 1}$:
$\frac{8!}{4!4!} \times \frac{4!}{2!2!} \times \frac{2!}{1!1!} \times 1$
Which comes out to
$70 \times 6 \times 2 \times 1= 840$
3) 
${8 \choose 3} \times {5 \choose 3} \times {2 \choose 1} \times {1 \choose 1}$
$\frac{8!}{3!5!} \times \frac{5!}{3!2!} \times \frac{2!}{1!1!} \times 1$
Which comes out to 
$56 \times 10 \times \times 2 = 1,120$
4) 
${8 \choose 3} \times {5 \choose 2} \times {3 \choose 2} \times {1 \choose 1}$
$\frac{8!}{3!5!} \times \frac{5!}{2!3!} \times \frac{3!}{2!1!} \times \frac{1!}{1!0!}$
Which comes out to:
If 8 new teachers are to be divided amongst 4 new schools how many divisions are possible?
$56 \times 10 \times 3 \times 1= 1,680$ 
5) 
${8 \choose 2} \times {6 \choose 2} \times {4 \choose 2} \times {2 \choose 2}$
$\frac{8!}{2!6!} \times \frac{6!}{2!4!} \times \frac{4!}{2!2!} \times \frac{2!}{2!0!}$
Which comes out to:
$28 \times 15 \times 6 \times 1 = 2,520$
What am I missing?

The total number of combinations for $\color\red{111}\color\green{5}$:
$$\frac{(1+1+1+5)!}{1!\times1!\times1!\times5!}\times\frac{(\color\red3+\color\green1)!}{\color\red3!\times\color\green1!}=1344$$

The total number of combinations for $\color\red{11}\color\green{2}\color\orange{4}$:
$$\frac{(1+1+2+4)!}{1!\times1!\times2!\times4!}\times\frac{(\color\red2+\color\green1+\color\orange1)!}{\color\red2!\times\color\green1!\times\color\orange1}=10080$$

The total number of combinations for $\color\red{11}\color\green{33}$:
$$\frac{(1+1+3+3)!}{1!\times1!\times3!\times3!}\times\frac{(\color\red2+\color\green2)!}{\color\red2!\times\color\green2!}=6720$$

The total number of combinations for $\color\red{1}\color\green{22}\color\orange{3}$:
$$\frac{(1+2+2+3)!}{1!\times2!\times2!\times3!}\times\frac{(\color\red1+\color\green2+\color\orange1)!}{\color\red1!\times\color\green2!\times\color\orange1}=20160$$

The total number of combinations for $\color\red{2222}$:
$$\frac{(2+2+2+2)!}{2!\times2!\times2!\times2!}\times\frac{\color\red4!}{\color\red4!}=2520$$
