Tag Archives: Year 11 Specialist Mathematics

May 28, 2025 · 1:33 pm

A Counting Question

How many three digit numbers can you form from the digits 1, 2, 3, 4 and 5 if

(a) the digits must occur in increasing order?

(b) adjacent digits differ by 2?

Cambridge Year 11 Specialist Mathematics Skill Sheet 1A

(a) There are $5\times 4\times 3=60$ permutations of three digits from the five digits, but how many of those are in the right order?

Each set of 3 digits has 6 arrangements ( $3\times 2\times 1=6$ ).

For example, if the set is ${1, 2, 3}$ , then the possible arrangements are:

123, 132, 213, 231, 312, and 321.

Only one of those arrangements is in numerical order.

Hence what we want is $\begin{pmatrix}5\\3\end{pmatrix}=10$

(b) I think this one is more about creating a list. I shall start with 1

135, 531, 131, 242, 353, 424, 535, 315

There are 8 possibilities.

Using a Vector Method to Find an Angle Bisector

Points $A$ and $B$ are defined by the position vectors $\mathbf{a}=3\mathbf{i}+4\mathbf{j}$ and $\mathbf{b}=12\mathbf{i}+5\mathbf{j}$ .

Find a vector that bisects $\angle{AOB}$ .

If we think about how we add vectors using the parallelogram rule

Finding the resultant vector using the parallelogram rule

we can take advantage of the geometric properties of parallelograms (or of a rhombus).

If $\mathbf{a}$ and $\mathbf{b}$ are unit vectors, then the parallelogram is a rhombus, and the diagonal (i.e the resultant) bisects the angle.

We need to find the sum of the unit vectors.

$|\mathbf{a}|=\sqrt{3^2+4^2}=5$

$\therefore \hat{\mathbf{a}}=\frac{3}{5}\mathbf{i}+\frac{4}{5}\mathbf{j}$

$|\mathbf{b}|=\sqrt{12^2+5^2}=13$

$\therefore \hat{\mathbf{b}}=\frac{12}{13}\mathbf{i}+\frac{5}{13}\mathbf{j}$

The vector that bisects $\angle{AOB}$ is

$\frac{3}{5}\mathbf{i}+\frac{4}{5}\mathbf{j}+\frac{12}{13}\mathbf{i}+\frac{5}{13}\mathbf{j}=\frac{99}{65}\mathbf{i}+\frac{64}{65}\mathbf{j}$

Counting Techniques

Four teachers decide to swap desks at work. How many ways can this be done if no teacher sits at their previous desk?
Mathematics Specialist Units 1&2 Cambridge

I like this question as it seems easy until you start thinking about it. I think the best approach is a tree diagram.

If we think of the four teachers as A, B, C and D. Then A can no longer sit in A, so the options are B, C and D for the first desk.

For the second desk, If B is in the first desk, then A, C or D could be in the second. If C is in the first desk, then A or D could be in the second (B can’t be in the same desk). If D is in the first desk, then A or C can be in the second desk.

Tag Archives: Year 11 Specialist Mathematics

A Counting Question

Using a Vector Method to Find an Angle Bisector

Counting Techniques

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