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-Also, for the second question, with the forced zeros, you are correct. That's how I solved the problem. Just be sure to count everything twice by "inverting" your selection to find the string of five 1's. Once that is done, you need to take out the slight overcount of the overlay between the inversions. --mkburges | -Also, for the second question, with the forced zeros, you are correct. That's how I solved the problem. Just be sure to count everything twice by "inverting" your selection to find the string of five 1's. Once that is done, you need to take out the slight overcount of the overlay between the inversions. --mkburges | ||
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| + | -i had a totally different way of doing this problem but i think i ended up with the right answer--did anyone else get 222? | ||
Revision as of 17:24, 28 January 2009
do you think the question is asking you to find
5 consecutive 0's OR 5 consecutive 1's, NOT both
or
5 consecutive 0's AND/OR 5 consecutive 1's
? it says "either" but i'm not sure what it means.
I believe it will allow both, I.E 1111100000 and 0000011111 count. (make sure not to double count them!)
Speaking of double counting, I was trying to devise a way to help with double counting. Here it the method I'm trying use:
Start with 00000----- and figure out all the possibilities. Then move your 'forced zeros' over, so you're now working with -00000----. However, i realized there would be a lot of double counting (for example, 0000001111 would be counted in 1st position and 2nd position) So i was thinking I could force uniqueness by simply putting a 1 at the beginning of my 0s. So my second position would be 100000----, my third position would be -100000---, etc. I feel like this should be accurate, but am I undercounting now?
-I believe that the interpretation of this question is more up to how the student wants to do it. As long as you explain what you are doing between either way, you shouldn't be taken off any points. Worst case scenario is you'll have an overcount of 2 (1111100000 and 0000011111).
-Also, for the second question, with the forced zeros, you are correct. That's how I solved the problem. Just be sure to count everything twice by "inverting" your selection to find the string of five 1's. Once that is done, you need to take out the slight overcount of the overlay between the inversions. --mkburges
-i had a totally different way of doing this problem but i think i ended up with the right answer--did anyone else get 222?
