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October 21, 2014

October 21, 2014

Posted by **sweety** on Friday, September 7, 2007 at 12:59pm.

- math -
**bobpursley**, Friday, September 7, 2007 at 2:00pmIt appears to me that 1/2 of all possible strings start with 0, and 1/4 end with 00. So the issue then is how many bit strings of eight digits can be made.

check my thinking.

- math -
**Touseef Ahmed**, Sunday, September 9, 2007 at 2:56amIf the first bit (left most bit) is a 0, then it can be filled in only each of the bit string can be filled in one way and when bit strings of length eight end with the two bits 00. Each of the remaining five position is represent in the bit string can be filled in 2 ways (i.e., either by 0 or 1).

Hence, there are 1×2×2×2×2×2×1×1 = 25 = 32 bit strings of length eight either start with a 0 bit or end with the two bits 00.

- math -
**Aedamar**, Tuesday, October 7, 2008 at 1:16pmif it begins with 0 we have two to the power of 7 ways of choosing the rest, and if it ends in 00 then we have two to the power of 6 ways of choosing the rest. As the question uses the word OR we add these results to get our answer!!

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