Question

How many \(n\) -bit binary numbers have exactly \(r\) l's assuming \(n \geq r ?\) Prove by induction that for \(n \geq 1\), exactly half the \(n\) -bit binary strings have an even number of 1 's.

Short answer

Exactly \(\binom{n}{r}\) \(n\)-bit binary numbers have \(r\) ones; half have even 1s.

Step-by-step solution

Understanding the Problem

We begin by identifying how many ways we can create an \(n\)-bit binary number with exactly \(r\) ones. This is a combinatorial problem where we are choosing \(r\) positions out of \(n\) to place 1s. Mathematically, this is expressed using the binomial coefficient: \(\binom{n}{r}\).

Identify Base Case

For the induction proof that half the \(n\)-bit binary numbers have an even number of 1s, start with \(n = 1\). There are two \(1\)-bit numbers: 0 and 1. Exactly one of these has an even number (zero) of 1s.

Induction Hypothesis

Assume that for some \(k\), half of the \(k\)-bit binary numbers have an even number of 1s. This is our induction hypothesis: \( \sum_{r=0,2,4, ext{...}}^k \binom{k}{r} = \frac{1}{2} \cdot 2^k = 2^{k-1} \).

Induction Step

We need to prove that the induction hypothesis holds for \(k + 1\). Consider all \((k+1)\)-bit numbers: each \(k\)-bit number can form two \((k+1)\)-bit numbers by appending either a 0 or a 1 at the end. Appending a 0 keeps the parity of the number of 1s, while appending a 1 switches the parity.

Derive Conclusion by Induction

From the induction hypothesis, half of the \(k\)-bit numbers have an even number of 1s. Appending a 0 preserves the parity, and appending a 1 switches it. Thus, each \(k\)-bit sequence contributes equally to \((k+1)\)-bit sequences with even and odd number of 1s, keeping the total even. Therefore, half of the \((k+1)\)-bit numbers have an even number of 1s.

Key concepts

Combinatorial Problems

Combinatorics is the area of mathematics that deals with counting, arrangement, and combination of elements within a set. One common type of problem in combinatorics involves figuring out how many different ways we can choose items from a larger set, which is precisely what we explore when determining how to create an \(n\)-bit binary number with exactly \(r\) ones. In this context, the idea is to select \(r\) positions out of \(n\) to place the 1s, effectively choosing which of the \(n\) bit positions will be 1. This scenario is a classic combinatorial problem, often solved using a mathematical concept known as the binomial coefficient.

Understanding combinatorial problems empowers us to solve questions involving arrangements and selections in a systematic way. This is essential in various fields, including computer science, where binary numbers serve as the fundamental language of digital systems.

Binomial Coefficient

The binomial coefficient, often written as \(\binom{n}{r}\), is a mathematical term that counts the number of ways to choose \(r\) elements from an \(n\)-element set. It is expressed through the formula:

\[ \binom{n}{r} = \frac{n!}{r! (n-r)!} \]

where \(!\) denotes factorial, representing the product of all positive integers up to that number. The binomial coefficient shows up in the expansion of binomial expressions like \((x + y)^n\) and is crucial in solving our original exercise where we need to determine how many \(n\)-bit binary numbers contain exactly \(r\) ones.

In simple terms, \(\binom{n}{r}\) calculates the number of possible combinations, showing us how selection differs when order doesn't matter. This concept solves many practical problems across different fields, including probability and statistical physics.

Induction Proof

Induction is a powerful proof technique used in mathematics to prove statements for all natural numbers. The general procedure involves two main steps: proving a base case and proving an induction step. For our problem, we start with the base case of 1-bit binary numbers to show that half have an even number of 1s (i.e., zero 1s).

The induction hypothesis assumes that for \(k\)-bit numbers, half have an even number of 1s. This forms the basis for the next step, the induction step, which extends the proof to \((k+1)\)-bit numbers. By appending either '0' or '1' to \(k\)-bit sequences, we observe that appending a '0' keeps the parity of the number of 1s, while appending a '1' changes it.

This elegant process ensures that each progression from \(k\) to \(k+1\) preserves the statement's truth, thereby leveraging the beauty of induction to solve complex mathematical challenges effortlessly.

Parity of Numbers

The concept of parity refers to whether a number is even or odd. It is a fundamental property used in various areas of mathematics and computer science to simplify and solve problems. In binary numbers, parity plays a crucial role, especially when counting the number of 1s.

In our exercise, we are concerned with whether the number of 1s in a binary number is even or odd (its parity). By examining parity, we discovered that appending '0' to a binary string does not change its parity, whereas appending '1' does. This property allows us to systematically analyze binary strings of differing lengths and deduce their parity easily.

Understanding parity equips us with tools to solve diverse problems, from error detection in data transmission to designing efficient algorithms that run on binary-based systems.