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d) C(10,3) + C(10,4) + ... + C10,10) | d) C(10,3) + C(10,4) + ... + C10,10) | ||
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| + | ---- | ||
| + | this is also how i did it, but is there no other shorter way if the bit string was length 1000? | ||
Revision as of 10:13, 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)
this is also how i did it, but is there no other shorter way if the bit string was length 1000?
