Line 16: Line 16:
  
 
Total = 386
 
Total = 386
 +
 +
Another way to solve this is by noting there are just as many strings with more 0s as 1s than more 1s than 0s. So take the number of all strings, subtract out the ones with 5 0s and 5 1s, and then divide by 2.

Revision as of 17:34, 14 September 2008

How many strings of length 10 have.

a) exactly three 0s?

C(10,3) = 120


b) more 0s than 1s?

10 0s : C(10,10) = 1 9 0s : C(10,9) = 10 8 0s : C(10,8) = 45 7 0s : C(10,7) = 120 6 0s : C(10,6) = 210

Total = 386

Another way to solve this is by noting there are just as many strings with more 0s as 1s than more 1s than 0s. So take the number of all strings, subtract out the ones with 5 0s and 5 1s, and then divide by 2.

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Basic linear algebra uncovers and clarifies very important geometry and algebra.

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