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Math Help - permutations and combination

  1. #1
    Member grgrsanjay's Avatar
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    permutations and combination

    how many different signals can be formed with 3 dots(.),2 dashes(-),4 stars(*)?
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    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by grgrsanjay View Post
    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
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  3. #3
    Member grgrsanjay's Avatar
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    it isint given that all the nine have to be used

    it has certainly more no of ways then what you provided
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    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.
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