How many bit strings of length 12 contain?

Question

a. exactly three 1s?
b. at most three 1s?
at least three 1s?
d. an equal number of 0s and 1s?

Answer 1

this is a combination and permutations question. but i cant understand the question.

Answer 2

a) you have 12C3 ways to pick the location of the 1s, and the rest must be 0s, so 12x11x10/3x2x1 = 220 ways.

b) at most 3 ones: add 12C2 = 12x11/2x1 = 66 and 12C1 = 12 and 12C0 = 1: 220 + 66 + 12 + 1 = 299 ways.

c) this is just the total number of possible 12-bit strings minus the number of strings with 2, 1, or 0 1s: 2^12 - (66 + 12 + 1) = 4096 - 79 = 4017

d) to have an equal number of 1s and 0s, there must be 6 1s and 6 0s. There are 12C6 = 12x11x10x9x8x7/6x5x4x3x2x1 = 924 ways to choose the locations of the 1s...and the rest have to be 0s, so 924 is the answer.

I think.

Answer 3

I wont answer your question. But I WILL answer a more general form of it.

'How many bit strings of length n contain?
a. Exactly k 1's or k 0's
b. At most k 1's.
c. At least k 1's
d. An equal number of 0s and 1s'

Answer :
a. n C k = n! / ((n-k)! k!)
b. Sum_{from i = 0}^{to i = k} n C i
c. At least k1's is 2^n - the answer from b + n C k.
d. This is impossible for n odd but for n even its equal to n C n/2


For your problem just n = 12, and k = 3 will do I believe.
Although, if you define a bit string as starting with a leading 1 then I suggest you take n = 11 and k = 2, for parts a-c at least.

Answer 4

a) We need to choose 3 out of 12 positions in 12C3 = 220 ways.
b) We need to choose 0, 1, 2, or 3, in 12C0 + 12C1 + 12C2 + 12C3 = 299 ways.
c) We can have everything *except* choosing 0, 1, or 2 ones. There are 2^12 bit strings in total, so 2^12 - 12C0 - 12C1 - 12C2 = 4017 ways.
d) We need 6 ones, so 12C6 = 924 ways.