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+ | How many bitstrings of length 10 have exactly 4 zeros? | ||
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We know that every bitstring has exactly 4 zeros, meaning that it also has exactly 6 ones. So, I think that the solution to this problem would be to find the number of ways that 4 zeros can be placed into 10 spaces and then 6 ones placed into the remaining 6 spaces. | We know that every bitstring has exactly 4 zeros, meaning that it also has exactly 6 ones. So, I think that the solution to this problem would be to find the number of ways that 4 zeros can be placed into 10 spaces and then 6 ones placed into the remaining 6 spaces. | ||
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C(10,4)*C(6,6) = (10!/(4!*6!))*(6!/(0!*6!)) = (7*8*9*10)/(1*2*3*4) * 1 = 5040/24 = 210 --[[User:Msstaffo|Msstaffo]] 21:52, 9 March 2009 (UTC) | C(10,4)*C(6,6) = (10!/(4!*6!))*(6!/(0!*6!)) = (7*8*9*10)/(1*2*3*4) * 1 = 5040/24 = 210 --[[User:Msstaffo|Msstaffo]] 21:52, 9 March 2009 (UTC) | ||
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Latest revision as of 17:35, 22 October 2010
Midterm practice question, MA453, Spring 2009, Prof. Walther
How many bitstrings of length 10 have exactly 4 zeros?
We know that every bitstring has exactly 4 zeros, meaning that it also has exactly 6 ones. So, I think that the solution to this problem would be to find the number of ways that 4 zeros can be placed into 10 spaces and then 6 ones placed into the remaining 6 spaces.
So, placing 4 indistinguishable objects (zeros) into 10 spaces:
C(10,4)
And, placing 6 indistinguishable objects (ones) into the remaining 6 spaces:
C(6,6)
Gives us
C(10,4)*C(6,6) = (10!/(4!*6!))*(6!/(0!*6!)) = (7*8*9*10)/(1*2*3*4) * 1 = 5040/24 = 210 --Msstaffo 21:52, 9 March 2009 (UTC)