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How many bit strings of length 12 contain at least three 1s?

A) 792
B) 924
C) 1024
D) 2048

1 Answer

2 votes

Final answer:

To find the number of 12-length bit strings with at least three 1s, calculate all possible strings
(2^12) and subtract the strings with fewer than three 1s. The correct answer is 4017, which does not match the provided options, indicating a potential error in the calculation or question options.

Step-by-step explanation:

To determine how many bit strings of length 12 contain at least three 1s, we can use combinatorics. First, calculate the total number of bit strings of length 12, which is
2^12. Then, subtract from this the number of bit strings that have fewer than three 1s (which would be strings with 0, 1, or 2 ones).

The number of ways to choose positions for the 1s in a bit string of length 12 is given by
C(n, k) = n! / (k! * (n - k)!), where n is the length of the bit string, and k is the number of 1s.

  • For 0 ones: C(12, 0) = 1 way
  • For 1 one: C(12, 1) = 12 ways
  • For 2 ones: C(12, 2) = 66 ways

Add up these possibilities: 1 + 12 + 66 = 79.

To find the number of bit strings with at least three 1s, subtract the number of strings with fewer than three 1s from the total number of strings:


2^12 - 79 = 4096 - 79 = 4017.

The correct answer is not listed among the options given by the student. Therefore, we need to double-check our calculations or the options provided.

User Zebraman
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