Final answer:
To find the number of 14-bit strings with more 0s than 1s, one should calculate the combinations for having 1 to 6 ones and then sum these values, leading to a total of 6,475 such bit strings.
Step-by-step explanation:
The question relates to the combinatorics field of mathematics and asks how many 14-bit strings have more 0s than 1s. To solve this, we should consider that for a 14-bit string there are a total of 214 possible strings, but we only want the ones with more 0s than 1s. This means we need strings with 1 to 6 ones (as 7 ones would mean an equal number of 0s and 1s, which doesn't meet the condition).
Using combinatorics, for each number of 1s, the number of such bit strings can be calculated as "14 choose k" where k is the number of 1s, and we sum these calculations for k from 1 to 6. Hence, the answer is C(14,1) + C(14,2) + C(14,3) + C(14,4) + C(14,5) + C(14,6), where C(n, k) represents combinations of n items taken k at a time.
Now, let's calculate:
• C(14,1) = 14
• C(14,2) = 91
• C(14,3) = 364
• C(14,4) = 1,001
• C(14,5) = 2,002
• C(14,6) = 3,003
Add these up to get the total: 14 + 91 + 364 + 1,001 + 2,002 + 3,003 = 6,475. Thus, there are 6,475 bit strings of length 14 with more 0s than 1s.