Discrete Maths , This is Hard, really!

• Jan 23rd 2010, 02:19 AM
primeimplicant
Discrete Maths , This is Hard, really!
There is a message that consists of 12 different symbols.This message is going to be sent through a transmition channel. The transmitter will additionally send 45 spaces between the symbols, with at least 3 spaces between each pair of symbols. In how many ways can the emitter send such a message?

My asnwer is: (using combinations with repetition)

$\frac{(11+12-1)!}{12!(11-1)!}. = 646,646
$

• Jan 23rd 2010, 02:34 AM
drumist
Since there are 3 spaces in between each symbol, 33 of the spaces are already accounted for. That leaves us with just 12 spaces we can place freely.

We have 11 spots to place additional spaces, so that gives us $\binom{12+11-1}{12} = \frac{22!}{12! \cdot 10!}$ possible combinations.

It's not clear to me whether the symbols are already in a predefined order, or if we can also move the 12 symbols in any order. If you can move them, then you should multiply the answer above by $12!$.
• Jan 23rd 2010, 03:40 AM
primeimplicant
Well, right, that was my original thought....apparently. But is it THAT easy? i am convinced that i am misssing something!!!