vlekje513
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| Joined: 28 Dec 2010 |
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| 08 Nov 2016 11:31 AM |
This is ungoogleable and I have no idea how to do this after a lot of time spent on this. This is my last resort. - We have asked on other forum's but no answer.
1 Take all letters of your name (including double letters). - For me this is Bbbeeoommrrrt. - For my friend who I am making this assignment with this is .
a How many different codes can you make with these letters? - I already have this answered. b How many different codes of 3 letters can you make?
c How many different codes of 4 letters can you make when atleast two of them are the same?
My answer to 1a: 13!/(3!*2!*2!*2!*3!) = 21.621.600 Basicly it's numberoflettersinname!/(double!*double!*double!*double!*double!)
pls halp
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vlekje513
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vlekje513
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| Joined: 28 Dec 2010 |
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| 08 Nov 2016 12:02 PM |
| My 19+ year old friends can't answer it either, pls help math wizards. |
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vlekje513
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| 08 Nov 2016 12:29 PM |
| 24 views, no answer, is it that hard? :( |
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vlekje513
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litalela
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| 08 Nov 2016 12:43 PM |
thats just permutations
use a calculator
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vlekje513
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| 08 Nov 2016 12:56 PM |
| Yeah it is, but I don't know how to make a formula out of 1b and 1c. |
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litalela
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| 08 Nov 2016 12:57 PM |
do you need to write a program or just find the answer
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litalela
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| 08 Nov 2016 12:57 PM |
| ################################################################################################################################# |
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litalela
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| 08 Nov 2016 12:58 PM |
goo(dot)gl/dtEIAQ
that should help, formula for permutations
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vlekje513
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| 08 Nov 2016 12:58 PM |
| Just find the answer, with calculation of course. |
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vlekje513
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| 08 Nov 2016 01:02 PM |
| That's how I calculated 1a, but 1b requires it to restrict to 3 letters, do you have any idea how to do that? |
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litalela
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| 08 Nov 2016 01:04 PM |
set the r value as the number of things you need to fill
so 3
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Casualist
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| 08 Nov 2016 01:05 PM |
b) N!/(N-3)!
c) N!/2
Where N is the number of unique characters in your name. |
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Casualist
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| 08 Nov 2016 01:06 PM |
| Ignore my my C, I jumped the gun and posted something ridiculously wrong |
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vlekje513
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| 08 Nov 2016 01:30 PM |
| b is incorrect Casualist, my friend has 13 letters in his name of which 9 are unique. So if everything was able to go 3 times it would be 13x13x13 which is 2197. But that formula said 15k+ |
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litalela
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| 08 Nov 2016 01:31 PM |
silly rabbit
you dont know math
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vlekje513
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vlekje513
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| Joined: 28 Dec 2010 |
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| 08 Nov 2016 01:47 PM |
| paid now https://forum.roblox.com/Forum/ShowPost.aspx?PostID=201195666#201195735 |
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VilgO
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| Joined: 15 Feb 2011 |
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| 08 Nov 2016 02:05 PM |
| Have you tried googling combinatorics and number of combinations? |
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Isosta
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| 08 Nov 2016 02:06 PM |
what are you trying to do....
$.get('gud') |
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vlekje513
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| 08 Nov 2016 02:14 PM |
VilgO, I have now and my mind is blown and confused.
Isosta, to get the answer of 1b and 1c. |
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MiniNob
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| 08 Nov 2016 02:19 PM |
| The answer to b is (NumLetters P 3)/RepeatedLetters1!/RepeatedLetters2!... for all of the letters that repeat |
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vlekje513
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MiniNob
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| 08 Nov 2016 02:24 PM |
this is all the combinations of 4 w/o repeated letters you can make with the name (NumLetters P 4)/RepeatedLetters1!/RepeatedLetters2!... (1)
and ask yourself How many an you make when NO LETTERS are the same
which is the same as asking how many combinations of 4 can you make with this: beomrt
which is answered by 5 P 4 (2)
To get the answer to the question you divide all of the combinations (1) by the ones that don't match (2). |
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