$$\DeclareMathOperator{cosec}{cosec}$$

# 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.

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:

$$(a+bx)^{n} = a^{n} \sum_{k=0}^{\infty} {n \choose k} \left(\frac{bx}{a}\right)^k$$

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

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

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