Combinatorics Problem with symbols and spaces Here is my problem that I have to solve:
An agent will send a secret code made up of 12 different symbols across a secure
wire. In addition to the 12 symbols, the agent will also send a total of 45 blank spaces between the symbols, with at least three spaces between each pair of
consecutive symbols. How many ways are there for the agent to send such a
message?
Just the wording of this problem has me thoroughly confused and lost.
Now, I figure that since the agent will put at least 3 spaces between a pair of consecutive symbols, then at least 36 spaces would be used. But that's all I have figured out, if I even did that correctly.
Could someone help me solve this problem?
Thank you for any help.
 A: You have $12$ symbols, so there are $11$ gaps between them.  Each gap has to have at least $3$ spaces, so we have accounted for $33$ spaces.  Your question is then the number of ways to distribute the remaining $12$ spaces between the gaps.  This is the number of weak compositions of $12$ into $11$ parts, which you can solve by a stars and bars argument.  You also need to assess how many choices there are for the message.
A: Make $12$ boxes, $11$ of a symbol followed by $3$ blanks, $\fbox{S---}$, and a last just $\fbox{S}$
There are $12$ blanks left. Exclude the first and the last symbols to get a total of $22$ entities.
Place the boxes in this string of $22$ in $\binom{22}{12}$ ways.
If the symbols of the secret code are in a fixed order, we are done, else multiply by $12!$.
A: You'll use up $33$ spaces fulfilling the "at least 3 spaces between a pair of consecutive symbols" condition.  We'll add these in last.  This leaves us with $12$ spaces to place.  One approach is to place these spaces down first and then add the unique symbols in one-by-one:
$$\_B\_B\_B\_B\_B\_B\_B\_B\_B\_B\_B\_B\_$$
Here the $B$'s represent the spaces and the $\_$'s are gaps where we can place the unique symbols.  Now since the blanks should fall "between" the symbols so we should select a first and last symbol and place them at the ends.  These are our "bookends" between which everything else will fall.  There are $12 \cdot 11$ ways to select the first and last symbol leaving us with the following string:
$$S_{first}\_B\_B\_B\_B\_B\_B\_B\_B\_B\_B\_B\_B\_S_{last}$$
There are $10$ remaining unique symbols.  Let's number them $1-10$.  As we can see from the number of gaps there are $13$ places we can place $S_1$.  Once we place $S_1$ our string will look something like this:
$$S_{first}\_B\_B\_B\_B\_B\_B\_B\_B\_B\_B\_S_1\_B\_B\_S_{last}$$
Notice that the number of available gaps increased by one (since the next symbol can fall on either side of the last one).  Thus there are $14$ ways to place $S_2$.  This pattern continues all the way up to $S_{10}$ yielding $13 \cdot 14 \cdots 22 = \frac{22!}{12!}$.  The last step is to add the $33$ spaces in but there is only $1$ way to do this.  In all there are
$$12 \cdot 11 \cdot \frac{22!}{12!} \cdot 1 = \frac{22!}{10!}$$
different possible strings.
A: First of all, if the agent used all 12 symbols, wouldn't the minimum amount of spaces be 33 instead of 36? Even so, this shouldn't be relevant, since the total is 45 spaces anyways, so they'd all end up being used. Also, I'm a little confused about what the word "consecutive" is supposed to mean in this problem, but I'm going to guess and say that it means "adjacent".
Say that the symbols are the letter A through L and they're distributed over the 45 spaces. I'll assign a symbol to the space just to make things easier.... Instead of a space, I'll make the space be the letter M. Also, I'm assuming that all spaces are the same. Then, find the number of ways to organize 12 letters and 45 Ms, which ends up being $$57!/45!$$ After that, subtract the number of organizations in which there are less than 3 spaces between the symbols, which would take place in a line less than 45 character-counts long (minimum number of spaces + twelve letters), and the number of ways to organize those twelve letters and 33 spaces would be $$(45!/33!) - 1$$ so the answer would end up being $$57!/45! - 45!/33! + 1$$
