Real World Applications

Wondering why you need to know about binomials? Watch these videos to understand how the Binomial Theorem is used in the real world.

Some uses of binomial distributions in the real world include

Market research interviews: Modeling the probability of people agreeing to an interview in a busy city center.
Multiple-choice exams: Estimating the number of correct answers a student can guess randomly and the probability of passing the exam.
Probability of bananas being brown.

Connections

Binomial is not just an "stand-alone" topic. It can link with many different mathematical topics. Investigate some of these connections below.

Using Partial Fractions with Binomial Expansion.

Partial Fractions simplify the problem, making the terms more manageable

To do this,

Express the fraction with constants (A, B) and linear factors in denominators.
Determine constants by multiplying both sides and selecting x values.
Substitute constants back into the partial fractions expression.

Proof of the Binomial Theorem using Calculus

Calculus can be applied to solve the Binomial Theorem, even though it involves continuous numbers while the theorem deals with discrete numbers.
Differentiating a function multiple times can help identify patterns in coefficients, which can be useful in solving problems like the Binomial Theorem.
Setting x to 0 and analysing coefficients can help connect the derivatives to the binomial coefficients.
Using derivatives and the connection between them and binomial coefficients, one can derive the Binomial Theorem formula: nCr = n! / (r! * (n-r)!).
This approach demonstrates the versatility of calculus in solving problems from different mathematical areas and the importance of identifying patterns through differentiation.

University Level Mathematics

Feeling up for a challenge? See how the information you have learned so far can be used as a basis to explore more difficult mathematical concepts.

Modular Arithmetic

Pascal's triangle contains some interseting patterns, specifically when you consider the location of multiples. Investigate some of these patterns by clicking on the options below.

Highlight Multiples of

We are able to locate mulitiples in the triangle using the modular arithmetic.

a ≡ b (mod n) represents congruence in modular arithmetic, where a and b have the same remainder when divided by n.
Congruence can be transformed into an equation by expressing a as a = k * n + b, where k is an unknown integer.
The relationship between a and b can also be expressed as a - b = k * n, meaning a minus b is a multiple of n.
The concept of congruence extends to negative numbers, and sometimes it is necessary to add or subtract multiples of n to demonstrate congruence.
Understanding congruence and its different interpretations is essential for solving equations, doing proofs, and working with modular arithmetic.

Permutations and Combinations

Permutations are arrangements in which the order of items matters, while combinations involve selecting items without considering the order.
To calculate permutations, use the formula nPr = n! / (n-r)!, where n is the total number of items, and r is the number of items selected.
To calculate combinations, use the formula nCr = n! / [(n-r)! * r!], where n is the total number of items, and r is the number of items selected.
When dealing with repeating items, divide the total permutations by the factorials of the repetitions of each repeating item.

Combinatorics is a branch of mathematics concerned with the study of finite structures. Enumerative combinatorics is an area of combinatorics that deals with counting the number of elements in a finite set. Let's investigate some of the most common combinatorial problems.

Answer: \(n_1 \cdot n_2 \cdot ... \cdot n_k\)

There are \(n_1\) ways of choosing the first entry of a \(k\)-tuple in \( S_1 \times S_2 \times \dots \times S_k \). Then, for each choice of the first entry, there are \(n_2\) possible ways of choosing the second entry. Then, for each choice of the first and second entries, there are \(n_3\) possible ways of choosing the third entry. Continuing in this way, we arrive at the answer.

Answer: \( \frac{n!}{(n-k)!} \)

There are \(n\) possible ways of choosing the first entry of a \(k\)-tuples in the product of \(k\) copies of \(S\). But, for each choice of the first entry, there are only \((n-1)\) possible ways of choosing the second entry (as entries cannot be repeated). Then, for each of the first and second entries, there are \((n-2)\) possible ways of choosing the third entry. Continuing in this way, we arrive at the answer.

Answer: \( \binom{n}{k} \). That's right, it's the binomial coefficient! There are \(n!/(n-k)! \) possible ways of choosing an ordered \(k\)-tuple without repeated entries from a set with \(n\) elements. On the other hand, given a set of \(k\) elements, there are \(k!\) possible ways of arranging them to form an ordered \(k\)-tuple. Therefore, the total number of subsets of \(S\) with \(k\) elements is
\[ \frac{n!}{k!(n-k)!} = \binom{n}{k}. \]

These are generic questions. In practice, we may need to combine some of these to calculate our answer. Let's try a real life example.

Four different dice are rolled.

a) In how many outcomes will at least one five appear?

Answer =

b) In how many outcomes will the highest die be a five?