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Thanks Bobby, video is good
Thanks Bob, that is good way of thinking about the problem
great thanks
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?
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
?
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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