Solving 6 1s and 8 0s Bit Strings

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SUMMARY

The problem involves calculating the number of distinct bit strings that can be formed using six 1s and eight 0s, totaling 14 bits. The correct approach is to use the combination formula C(n, r) = n!/(r!(n-r)!), specifically C(14, 6) to determine the arrangements of the six 1s among the 14 slots. The remaining eight slots will automatically be filled with 0s, making the calculation of C(8, 8) unnecessary, as it equals 1 and does not affect the total count.

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Homework Statement


How many different bit strings can be formed using six 1s and eight 0s?



Homework Equations



C(n,r) = n!/(r!*(n-r)!)



The Attempt at a Solution



since there's six 1s and eight 0s there are 14 slots. So I'm guessing for the possible six 1s it would be C(14,6) and the remaining eight 0s would have to be C(8,8) since there are only 8 slots left after six 1s are chosen so it would be C(14,6)*C(8,8)?

The last part C(8,8) seems kind of werid so I'm not very convinced this is correct. Any help is apperciated thanks.
 
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Yeah, that's right. The factor of C(8,8), which equals 1, is correct, but usually you just gloss over that since you know the rest of the slots have to be filled with what's left.
 

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