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Re: 2 - MATH SCHOOL TONIGHT

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vlekje513 is not online. vlekje513
Joined: 28 Dec 2010
Total Posts: 9057
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 is not online. vlekje513
Joined: 28 Dec 2010
Total Posts: 9057
08 Nov 2016 11:58 AM
bump
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vlekje513 is not online. vlekje513
Joined: 28 Dec 2010
Total Posts: 9057
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 is not online. vlekje513
Joined: 28 Dec 2010
Total Posts: 9057
08 Nov 2016 12:29 PM
24 views, no answer, is it that hard? :(
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vlekje513 is not online. vlekje513
Joined: 28 Dec 2010
Total Posts: 9057
08 Nov 2016 12:38 PM
I'm dead.
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litalela is not online. litalela
Joined: 30 Mar 2010
Total Posts: 6267
08 Nov 2016 12:43 PM
thats just permutations

use a calculator


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vlekje513 is not online. vlekje513
Joined: 28 Dec 2010
Total Posts: 9057
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 is not online. litalela
Joined: 30 Mar 2010
Total Posts: 6267
08 Nov 2016 12:57 PM
do you need to write a program or just find the answer


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litalela is not online. litalela
Joined: 30 Mar 2010
Total Posts: 6267
08 Nov 2016 12:57 PM
#################################################################################################################################
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litalela is not online. litalela
Joined: 30 Mar 2010
Total Posts: 6267
08 Nov 2016 12:58 PM
goo(dot)gl/dtEIAQ

that should help, formula for permutations


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vlekje513 is not online. vlekje513
Joined: 28 Dec 2010
Total Posts: 9057
08 Nov 2016 12:58 PM
Just find the answer, with calculation of course.
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vlekje513 is not online. vlekje513
Joined: 28 Dec 2010
Total Posts: 9057
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 is not online. litalela
Joined: 30 Mar 2010
Total Posts: 6267
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 is not online. Casualist
Joined: 26 Jun 2014
Total Posts: 4443
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 is not online. Casualist
Joined: 26 Jun 2014
Total Posts: 4443
08 Nov 2016 01:06 PM
Ignore my my C, I jumped the gun and posted something ridiculously wrong
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vlekje513 is not online. vlekje513
Joined: 28 Dec 2010
Total Posts: 9057
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 is not online. litalela
Joined: 30 Mar 2010
Total Posts: 6267
08 Nov 2016 01:31 PM
silly rabbit

you dont know math


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vlekje513 is not online. vlekje513
Joined: 28 Dec 2010
Total Posts: 9057
08 Nov 2016 01:34 PM
am i the silly rabbit
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vlekje513 is not online. vlekje513
Joined: 28 Dec 2010
Total Posts: 9057
08 Nov 2016 01:47 PM
paid now https://forum.roblox.com/Forum/ShowPost.aspx?PostID=201195666#201195735
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VilgO is not online. VilgO
Joined: 15 Feb 2011
Total Posts: 518
08 Nov 2016 02:05 PM
Have you tried googling combinatorics and number of combinations?
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Isosta is not online. Isosta
Joined: 10 May 2015
Total Posts: 14729
08 Nov 2016 02:06 PM
what are you trying to do....


$.get('gud')
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vlekje513 is not online. vlekje513
Joined: 28 Dec 2010
Total Posts: 9057
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 is not online. MiniNob
Joined: 14 May 2013
Total Posts: 822
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 is not online. vlekje513
Joined: 28 Dec 2010
Total Posts: 9057
08 Nov 2016 02:21 PM
NumLetters P 3???
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MiniNob is not online. MiniNob
Joined: 14 May 2013
Total Posts: 822
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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