Placing dolls on shelves I have this problem for my school project.
I have eleven different dolls, and have to put them on four shelves.
Each shelve can hold all of the dolls. It matters which doll is on which shelf, and also the order of the dolls matters. Ex. (ABCD | EFGH | I | JK) is not the same as (ABCD | EFGH | I | KJ ).
I have done some research, but everyone in their problems seems to not care about the order, or has more things of a kind.
Question 
a) How could I compute number of possibilities to arrange the dolls, if no constraints are presented?
b) Verify my thought process for "on each shelf must be at least two dolls"
ad b)
Using function V(k, n) = n!/(n-k!) I'd choose two dolls for each shelf
V(2, 11) * V(2, 9) * V(2, 7) * V(2, 5) and then use the solution of the first problem, only this time for three dolls only. Is this a valid solution or am I an idiot?
Thanks
 A: You can imagine the whole proces like this:


*

*You can first order the dolls in some order. You have $11!$ possibilities.

*Then you divide them in this order between the shelves. This means that you have to take $x_1$ dolls from the left and put them on the first shelf, $x_2$ from the remaining to put on the second shelf, etc. 


The number of possibilities how to divide them between shelves is the number solutions of 
$$x_1+x_2+x_3+x_4=11$$ 
such that each $x_i$ is integer and $x_i\ge 0$. 
Number of such solutions is
$$\binom{11+3}3.$$
More generally for $x_1+\dots+x_k=n$ you get $\binom{n+k-1}{k-1}$. See Wikipedia article on stars and bars or the links to other questions on this site posted in comments.
The total number of solutions is
$$\binom{14}3\cdot11!=\frac{14!}{11!}$$

For the second part you can notice that now we have similar problem, but we want integer solutions
$$x_1+x_2+x_3+x_4=11$$
such that $x_i\ge2$.
However, if we denote $y_i=x_i-2$ for $i=1,2,3,4$ we get new equation
$$y_1+y_2+y_3+y_4=3$$
with restrictions $y_i\ge0$. 
So this is again the problem of the same type as above and we have
$$\binom63$$
solutions.
If we add $11!$ possible orderings, we get
$$\binom63\cdot11!=20\cdot11!$$ possibilities.

If I understood correctly what you mean by your solution, then you get the same result. You wrote:

V(2, 11) * V(2, 9) * V(2, 7) * V(2, 5) and then use the solution of the first problem, only this time for three dolls only. Is this a valid solution

$$V(11,2)\cdot V(9,2)\cdot V(7,2)\cdot V(5,2) =11\cdot10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4 = \frac{11!}{3!}.$$
You want to multiply it by solution for a) with three dolls which is
$$\binom 63\cdot 3!$$
By multiplying the partial results together you get
$$\binom 63\cdot 3!\cdot\frac{11!}{3!} = \binom63\cdot11!$$
which is the same result as above. 
