In how many ways a binary code consisting of 12 binary digits can be created, so that 10 or more digits are 1s?
Question
In how many ways a binary code consisting of 12 binary digits can be created, so that 10 or more digits are 1s?
Solution 1
This is a combination problem. The formula for combinations is C(n, r) = n! / [r!(n-r)!], where n is the total number of items, r is the number of items to choose, and "!" denotes factorial.
We want to find the number of ways to choose 10, 11, or 12 digits to be 1s out of 12 digits. So, we need to calculate C(12, 10) + C(12, 11) + C(12, 12).
- C(12, 10) = 12! / [10!(12-10)!] = 66
- C(12, 11) = 12! / [11!(12-11)!] = 12
- C(12, 12) = 12! / [12!(12-12)!] = 1
So, there are 66 + 12 + 1 = 79 ways to create a binary code consisting of 12 binary digits so that 10 or more digits are 1s.
Solution 2
This is a combination problem. We want to find the number of ways to choose 10, 11, or 12 digits to be 1s out of 12 total digits.
The formula for combinations is C(n, r) = n! / [r!(n-r)!], where n is the
Similar Questions
How many unique values are possible using a single binary digit?
ow many binary strings of length 8 are there that contain two or fewer 1s?
How many four-digit numbers which are divisible by 12 can be formed using the digits 0, 2, 4, 5 and 6 such that no digit is used more than once and 0 is not used in the leftmost position?
How many bit strings of length 10 either begin with three 0s or end with two 0s
Q1. As discussed in the first assignment, it can be helpful to study the probability oferrors and therefore it is useful to know the cardinalities of sets of words and validcodewords — this is where combinatorics comes in.(a) In Question 3 of Assignment 1, it was stated that the set of all bit strings oflength four has 16 elements and the set of all bit strings of length 12 has4096. Justify these cardinalities.2 marks(b) Suppose that a code is made up six digits from the range 0–9 (inclusive)and satisfies the rules:• The first three digits must be odd (and non-zero).• No number is repeated in the first five digits.• The sixth digit is the remainder of the sum of the first five digits dividedby 10.i. How many valid codewords are there?4 marksii. What is the probability that a random selected six digit number is a validcodeword?2 marks(c) One common mistake humans make when reading out numbers is transpo-sitions. Transposition is the special case of swapping letters of a word givenby swapping adjacent letters. (In Assignment 1 we mentioned that ISBN-10could detect such errors.)How many distinguishable rearrangements (not just transpositions) of theISBN-10 code0-205-08005-7are there?2 marksTotal: 10 marksPage 3 of 8
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.