The result for the given 14 length bit string is-
(a) The bit strings of length 14 which have exactly three 0s is 2184.
(b) The bit strings of length 14 which have same number of 0s as 1s is 17297280.
(c) The bit strings of length 14 which have at least three 1s is 4472832.
<h3>What is a bit strings?</h3>
A bit-string would be a binary digit sequence (bits). The size of the value is the amount of bits contained in the sequence.
A null string is a bit-string with no length.
The concept used here is permutation;
ⁿPₓ = n!/(n-x)!
Where, n is the total samples.
x is the selected samples.
(a) Because there are exactly three 0s, its other digits have always been one, hence the total of permutations is equal to;
n = 14 and x = 3 digits.
¹⁴P₃ = 14!/(14-3)!
¹⁴P₃ = 2184.
Thus, the bit strings of length 14 which have exactly three 0s is 2184.
(b) Because it is a bit string with 14 digits and it can only have digits 0 or 1, we must choose 7 0s and the remaining 7 1s, hence the number possible permutations is equal to;
n = 14 and x = 7 digits.
¹⁴P₇ = 14!/(14-7)!
¹⁴P₇ = 17297280.
Thus, the bit strings of length 14 which have same number of 0s as 1s is 17297280.
(c) The answer is identical to problem a, but the rest of a digits could be either 0 or 1, hence it must be doubled by 2¹¹ because there are 11 digits, each of which can be one of two options;
= 2184×2¹¹
= 4472832
Thus, he bit strings of length 14 which have at least three 1s is 4472832.
To know more about permutation, here
brainly.com/question/12468032
#SPJ4
The correct question is-
How many bit strings of length 14 have:
(a) Exactly three 0s?
(b) The same number of 0s as 1s?
(d) At least three 1s?
where s is the side.