Number and Algebra

Furthermore

Official Guidance, clarification and syllabus links:

Not required:

Permutations where some objects are identical.

Circular arrangements.

Formula Booklet: AHL 1.10

Combinations	\( ^nC_r = \dfrac{n!}{r!(n-r)!} \)
Permutations	\( ^nP_r = \dfrac{n!}{(n-r)!} \)

The product principle states that if there are \(n\) different ways of performing an action, and for each of these there are \(p\) different ways of performing a second independent action, then there are \(np\) different ways of performing the two operations in succession.

The sum principle is a counting rule that tells us how many ways we can perform an action. If we have \(n\) different ways of performing an action OR \(p\) different ways of performing a different, mutually exclusive action, then the sum principle tells us that there are a total of \(n+p\) different ways we can perform the action.

For example:

How many ways are there of getting from A to D?

How many ways are there of getting from C to D?

How many ways are there of getting from B to D?

There are \(2 \times 3 = 6\) ways of getting from A to D (product principle).

There are \(3 \times 3 = 9\) ways of getting from C to D (product principle).

There are \(6 + 9 = 15\) ways of getting from B to D (sum principle).

Counting principles, including permutations and combinations, are fundamental concepts in combinatorics that are used to count the number of ways to choose or arrange objects in a set. Permutations and combinations are different ways of counting and are often used in probability problems.

Permutations:

$$\large{_{n}P_{r} = \frac{n!}{(n-r)!}}$$

Where n is the number of objects in a set and r is the number of objects chosen. This formula calculates the number of ordered arrangements of r objects chosen from a set of n objects.

For example, suppose we have a set of 5 different letters: A, B, C, D, E. How many ways are there to arrange 3 of these letters?

We can use the permutation formula:

$$\large{_{5}P_{3} = \frac{5!}{(5-3)!} = \frac{5\times4\times3\times2\times1}{2\times1} = 60}$$

Therefore, there are 60 different ways to arrange 3 letters from a set of 5 letters.

Combinations:

$$\large{{n \choose r} = \frac{n!}{r!(n-r)!}}$$

Where n is the number of objects in a set and r is the number of objects chosen. This formula calculates the number of ways to choose r objects from a set of n objects, without regard to order.

For example, suppose we have a set of 5 different letters: A, B, C, D, E. How many ways are there to choose 3 of these letters?

We can use the combination formula:

$$\large{{5 \choose 3} = \frac{5!}{3!(5-3)!} = \frac{5\times4\times3\times2\times1}{3\times2\times1\times2\times1} = 10}$$

Therefore, there are 10 different ways to choose 3 letters from a set of 5 letters.

The Binomial Theorem is a formula that allows us to expand expressions of the form $$(a+b)^n$$ where n is a positive integer. However, it can be extended to allow for fractional and negative indices, which can be useful in certain situations.

The Binomial Theorem for fractional and negative indices is given by:

\( (1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots \)

This expansion is valid when \( |x| < 1 \) and \( n \) is any real number, including fractions and negatives.

Tip worth remembering

To find the binomial expansion of \( (2+x)^{ -\frac12} \) reduce the first term to one by taking out a factor of 2. It then becomes \( 2^{- \frac12} (1+ \frac{x}{2})^{-\frac12} \)

Then use the formula:

This is the formula to remember as it makes the working as easy as possible.

$$ 2^{-\frac12} \left(1+ \frac{x}{2} \right)^{-\frac12} = \dfrac{1}{\sqrt{2}} \left(1 - \dfrac{x}{4} + \dfrac{3x^2}{32} - \dfrac{5x^3}{128} + \cdots \right) $$