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This is on discrete math

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How many bit strings of 15 have

a) exactly three 0s?

b) more 0s than 1s

Oct 21st, 2017

This deals with combinations...

a) If the bit string is of length 15 and you want exactly 3 zeroes then..

C(15,3) = 15!/[(15-3)!3!] = 15!/[12!3!] = 455 bit strings of 15 that have exactly three 0's

b) For more 0's than 1's you have to start at the majority of a set of 15...if 7.5 is half of 15 then anything greater than 7.5 would be the majority.

We stick with whole numbers so 8 would be the minimum number of bit strings that have more 0's than 1's. We don't stop here though. We have to account for 8,9,10,11,12...and 15 zeros!

So using similar methods from part a we have:

C(15,8) + C(15,9) + C(15,10) + C(15,11) + C(15,12) + C(15,13) + C(15,14) + C(15,15)...

15!/[(15-6)!8!] + 15!/[(15-9)!9!] +.....+ 15!/[(15-15)!15!] = 16,384 bit strings with more 0's than 1's.


Aug 14th, 2014

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