I'm not sure I understand the problem, can someone clarify?


by mr_coffee
Tags: clarify
mr_coffee
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#1
Oct27-06, 07:30 PM
P: 1,629
Hello everyone. I'm not sure where to begin with this problem becuase I don't think I know what they are trying to tell me.

Heres what it says:

In morse code, Symobls are represented by variable-length sequences of dots and dashes. For example, A = .-, 1 = . - - - -, ? = . . - - . .
How many different symobls can be represented by sequences of seven or fewwer dots and dashes?

So are they saying, I have a total number of 7 either dots or dashes to work with and how many different permutations can i get with that? or how many different combinations can i get with that?

Order shouldn't matter, they didn't say it did, and also repeitions not allowed becuase you don't want 2, A's, ?'s, etc.

So that leaves me to think its combinations.

Would this be correct?

combinations formula is the following:
Choosing r items from n possibilities:
n!/[r! (n-r)!]

so your choose 2 items from 7 possibilities
r = 2
n = 7

7!/[2!5!] = 21 different symbols can be represeented by sequences of seven or fewer dots and dashes.
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mattmns
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#2
Oct27-06, 10:56 PM
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You are looking for sequences of Seven or fewer, so you are not done yet. I think you will have to compute 7 cases, but once you see the pattern the number for each case should be easy to find. Also, I would say you are looking at permutations, not combinations, since we can map different symbols to .- and -.

If this is the case, what would the number be?
mr_coffee
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#3
Oct28-06, 01:29 AM
P: 1,629
hm...if its permutations then it should be pretty straight forward using:
P(n,r) = n!/(n-r)!
That means order does matter, so
n = 7
r = 2
7!/(5)! = 42 different symbols which sounds more correct than 21.

matt grime
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#4
Oct28-06, 03:18 AM
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I'm not sure I understand the problem, can someone clarify?


42 symbols? That doesn't sound remotely correct. Why are you doing it with permutations or combinations? It is just asking how many strings of length 1,2,3,4,5,6, or 7 can be made up from 2 symbols, be they dots, dashes, or zeroes and ones. You are off by a factor of around 5.5.
mr_coffee
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#5
Oct29-06, 03:12 PM
P: 1,629
matt, i'm not sure i understand how your supose to figure this out. All the problems i've been doing invovle permutations or combinations thats why I thought thats how it would be solved.
WHen you say strings of length 1, 2, 3, 4, 5, 6, and 7, could i do it directly and say:
1 = . or -
2 = .- or .-
3 = .-. or -.. or ..- or --- or ...
4 = 4!
5 = 5!
6 = 6!
7 = 7!

but i'm getting huge numbers, i'm not sure i follow

or is it just

2^n? 2^7 = 128
matt grime
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#6
Oct29-06, 03:24 PM
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You've not counted strings of length 2 correctly.
mr_coffee
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#7
Oct30-06, 06:57 PM
P: 1,629
(2^7)+(2^6)+(2^5)+(2^4)+(2^3)+(2^2)+(2^1)= 2^8-2 = 254

hows that look?
mattmns
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#8
Oct30-06, 08:42 PM
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P: 1,119
That looks perfect.
mr_coffee
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#9
Oct30-06, 08:48 PM
P: 1,629
Thanks matt!


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