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I'm struggling on these 2 problems:

  • How many 16 bit strings are there containing six 0's and ten 1's with no consecutive 0's?

  • How many 8 bit strings with exactly 2 1's are there such that the 1's are not adjacent?

For the first one, I am guessing they must alternate between 0 and 1 but I have no clue how to set this up. How can I use combinations to find this?

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Don't space out; select the spaces!   Put 0,0,0,0,0, and 0 into _1_1_1_1_1_1_1_1_1_1_

You have ten 1 and eleven spaces to place the six 0 such that only one 0 goes in any space. How many ways can you select six of these eleven spaces?

Second verse, same as the first. (Only the numbers are changed.)

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  • $\begingroup$ So, for the first one it is C(16,10) * C(16,6) ? Is that correct? $\endgroup$ – user3015986 Sep 17 '15 at 3:54
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    $\begingroup$ No ... What are you even counting there? The count of ways to select 6 of 11 items is $^{11}\mathrm C_6$ $\endgroup$ – Graham Kemp Sep 17 '15 at 4:05
  • $\begingroup$ Sorry I didn't quite think it through. So would have been C(11,6) * C(16,10) I am guessing. $\endgroup$ – user3015986 Sep 17 '15 at 4:45
  • $\begingroup$ @user3015986 Stop guessing. Think about what you are counting. You are given 10 ones and 6 zeroes. You don't need to select them! You have them. You just need to count ways to arrange them. So we place the ones down in a row. There's just one way to do that; they're all ones. Then we take the zeroes and place them in the spaces before, after, or between the ones so that there's at most 1 zero in each space. There are 11 such spaces and 6 zeros. So we only need to count the ways to select 6 of the 11 spaces. Just that. $\endgroup$ – Graham Kemp Sep 17 '15 at 5:10
  • $\begingroup$ I had the similar idea which my book showed, but I think I got it mixed up with something else or didn't understood the question completely. But thanks so much for your help. Now I get it. $\endgroup$ – user3015986 Sep 17 '15 at 5:12

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