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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. | 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. | ||
| + | |||
| + | ---- | ||
| + | c) C(10,7) + C(10,8) + C(10,9) + C(10,10) = 176 | ||
| + | |||
| + | d) C(10,3) + C(10,4) + ... + C10,10) | ||
Revision as of 04:50, 17 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.
c) C(10,7) + C(10,8) + C(10,9) + C(10,10) = 176
d) C(10,3) + C(10,4) + ... + C10,10)
