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Hi
I can't remeber how to do this. It has got something to do with factorials i think.
I want to know how many combinations there are for a 9 digit binary sequence where 5 of the bits must be 1 and 4 of the bits must be 0
e.g.
111110000
111101000
111100100
111100010
111100001
111010001
111001001
...
....
etc.
Can anybody tell me the answer and also possibly how the calculation is done?
Thanks?
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Hi mojorising;
Also the question is phrased incorrectly. In combinations order does not count so there is only one combination. You are asking how many permutations are there.
Welcome to the forum.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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actually no
4 in 9 and 5 in 9 are the same problem so the anser is
1 in 9 answer is 9
2 in 9 answer is 8 x 7
3 in 9 answer is 7 x 6 x 5
4 in 9 answer is 6 x 5 x 4 x 3 (same as 5 in 9)
(= 6!/2! ?)
= 360
?
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Hi;
Please see post # 2
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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OK, thanks Bobby.
I am trying to get me head around the calculation but in the meantime I will take your word for it!
Is there a link to somewhere that has an explanation of why that factorial ratio/product is the right answer?
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Hi;
but in the meantime I will take your word for it!
In math we do not have to take anyone's word for anything. Convince yourself of it, go here:
http://www.mathwarehouse.com/probabilit … -items.php
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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great thanks
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Hi;
You are welcome. After you watch the video come back if you have a question.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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hi mojorising
welcome to the forum!
Here's how I would explain it:
Pretend for a moment that the 9 objects are different. In the diagram below I've called them 1a 1b etc and I've shown two possible arrangements.
How many are there?
Well I've made it a permutation problem ... how many ways can you arrange 9 different objects ... 9!
But the objects are not all different. A zero looks much like another. so I've ended up counting the same arrangements over and over.
But how many times too many?
If you look at the two arrangements I've shown, you can see I got the second by just shuffling around some of the ones and, separately, some of the zeros.
I can shuffle the ones in 5! ways and I can shuffle the zeros in 4! ways.
So if I now drop the letters from the problem I need to divide 9! by those two factorials.
Thus 9! / (5!.4!)
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Thanks Bobby, video is good
Thanks Bob, that is good way of thinking about the problem
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Hi mojorising;
You are welcome.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi mojorising;
You are welcome. Combinatorics is fantastic!
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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