1. permutations and combination

how many different signals can be formed with 3 dots(.),2 dashes(-),4 stars(*)?

2. Originally Posted by grgrsanjay
how many different signals can be formed with 3 dots(.),2 dashes(-),4 stars(*)?

$PR_{\;9}^{\;3,\;2,\;4}=\dfrac{9!}{3!\;2!\;4!}=\ldo ts$

Fernando Revilla

3. it isint given that all the nine have to be used

it has certainly more no of ways then what you provided

4. If you are not required to use all the symbols then the problem gets more complicated. I can think of a couple of ways to approach it. One way is to break the problem into cases depending on the number of signals used, say $0 \leq r \leq 9$. Then for a given r, find all the solutions in non-negative integers to

$x_1 + x_2 + x_3 = r$
where
$0 \leq x_1 \leq 3$
$0 \leq x_2 \leq 2$
$0 \leq x_3 \leq 4$

Given one solution to the equation, the number of signals that can be sent with that number of symbols is
$\binom{r}{x_1 \; x_2 \; x_3} = \frac{r!}{x_1 ! x_2 ! x_3 !}$
(a multinomial coefficient). All this looks pretty tedious, although I guess it can be done.

I'm too lazy to do it that way, so I would prefer a generating function approach. Let $a_r$ be the number of signals that can be sent with r of the symbols and define

$f(x) = \sum_r \frac{1}{r!} a_r x^r$

(an exponential generating function). Then it's clear (once you know how to do it) that

$f(x) = (1 + x + \frac{1}{2!} x^2 +\frac{1}{3!} x^3) (1 + x + \frac{1}{2!} x^2) (1 + x + \frac{1}{2!} x^2 +\frac{1}{3!} x^3 + \frac{1}{4!} x^4)$

Expand f (here it helps to have access to a computer algebra system, or you can use Wolfram Alpha), write the result in exponential generating form as in the definition of f (notice the factors 1/r!), and you have the coefficients $a_r$. Your final answer is then

$\sum_{r=0}^9 a_r$

The case r=0 corresponds to a "signal" of no symbols, so you need to decide whether that is acceptable.