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Year 11 Maths

Combinations

Understand how to count selections where order does not matter, using the nCr formula and its connection to Pascal's triangle.

What Are Combinations?

A combination is a selection of items from a larger set where the order does not matter. For example, choosing 3 students from a class of 10 to form a committee is a combination problem because it does not matter in which order the students are picked.

This is different from a permutation, where order matters (e.g., choosing a president, secretary, and treasurer from the same group of 10 would be a permutation).

The nCr Formula

Number of combinations of r items from n:

nCr = n!/r!(n - r)!

n = total number of items to choose from

r = number of items being chosen

Permutations vs Combinations

Permutations (order matters)

Selecting A then B is different from B then A.

Formula: nPr = n! / (n-r)!

Combinations (order does not matter)

Selecting A then B is the same as B then A.

Formula: nCr = n! / [r!(n-r)!]

Pascal's Triangle Connection

The values of nCr appear as entries in Pascal's triangle. Each row corresponds to a value of n, and each position within the row corresponds to a value of r.

Pascal's Triangle (rows 0 to 5)

1
11
121
1331
14641
15101051

Row 4, position 2 gives 4C2 = 6. Row 5, position 2 gives 5C2 = 10.

Key property: nCr = nCn-r. This symmetry is visible in Pascal's triangle, where each row is a palindrome.

Applying the Formula

To decide whether a problem requires combinations, ask: "Does the order of selection matter?" If the answer is no, use combinations.

Common Combination Scenarios

  • Choosing a committee or team from a group
  • Selecting toppings for a pizza
  • Picking lottery numbers (without order)
  • Choosing cards from a deck (e.g., a poker hand)

Key Vocabulary

Combination

A selection of items where order does not matter. Denoted nCr or C(n, r).

Factorial (n!)

The product of all positive integers from 1 to n. For example, 5! = 5 x 4 x 3 x 2 x 1 = 120.

Pascal's Triangle

A triangular array where each entry equals the sum of the two entries directly above it. Row n contains the values nC0 through nCn.

Permutation

An arrangement of items where order matters. Denoted nPr.

Worked Examples

1

Calculate 6C2.

Step 1: Write out the formula: 6C2 = 6! / (2! x 4!)

Step 2: Expand: = (6 x 5 x 4!) / (2 x 1 x 4!)

Step 3: Cancel 4!: = (6 x 5) / 2 = 30/2

Answer: 6C2 = 15

2

A team of 3 is to be chosen from 8 athletes. How many different teams are possible?

Step 1: Order does not matter (it's a team), so use combinations. n = 8, r = 3.

Step 2: 8C3 = 8! / (3! x 5!) = (8 x 7 x 6) / (3 x 2 x 1)

Step 3: = 336/6

Answer: There are 56 different teams possible.

3

Show that 5C2 = 5C3.

Step 1: 5C2 = 5! / (2! x 3!) = (5 x 4) / (2 x 1) = 10

Step 2: 5C3 = 5! / (3! x 2!) = (5 x 4) / (2 x 1) = 10

Answer: Both equal 10, confirming the symmetry property nCr = nCn-r.

Knowledge Check

Select the correct answer for each question. Click "Check Answer" to see if you are right.

Question 1

What is the value of 7C3?

Question 2

A pizza shop offers 10 toppings. How many ways can you choose 4 toppings?

Question 3

Which row of Pascal's triangle contains the value 6C2 = 15?

Question 4

If nC2 = 28, what is the value of n?

Question 5

A committee of 5 is to be chosen from 12 people. How many different committees are possible?

Key Concepts Summary

Year 11: Functions Year 11: Binomial Theorem