Discrete Math
How many bit strings contain exactly six 1's and ten 0's if every 1 is immediately followed by a 0?
Last edited on
treat that as 10
so you have 6 times the number 10
101010101010
which is also 6 zeros.
so now you may add 4 zeros anywhere. this is what you must count. some places are the same if you try to do it in code, eg adding 0 to the least and next to least digit of the above is the same result.
anyway, now you iterate (or find a math way to do it):
101010101010 0000
1010101010010 000
1010101001010 000
...
the problem probably wants leading zeros to be included in the result.
so you have 6 times the number 10
101010101010
which is also 6 zeros.
so now you may add 4 zeros anywhere. this is what you must count. some places are the same if you try to do it in code, eg adding 0 to the least and next to least digit of the above is the same result.
anyway, now you iterate (or find a math way to do it):
101010101010 0000
1010101010010 000
1010101001010 000
...
the problem probably wants leading zeros to be included in the result.
Last edited on
yes. 10 things (6 10's and 4 zeros), choose 4 (the zeros), the 'combination' computation.
Sigh. But trying to get him there... did not succeed.
Sigh. But trying to get him there... did not succeed.
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