Be systematic. Take each filling in turn and pair it with every type of bread before moving on to the next one.

Chicken and white, chicken and brown

Cheese and white, cheese and brown

Falafel and white, falafel and brown

Working through the filling choices in order like this is what makes sure nothing is missed and nothing is written twice. There are 6 combinations altogether.

Each one of the 6 combinations from part (a) can be joined with any one of the 6 sauces, so every combination from part (a) splits into 6.

\(6 \times 6 = 36\)

Listing all 36 would take a while, which is exactly why multiplying the number of choices is worth knowing.

Each option has three of the four fillings, so each option leaves out exactly one of them. Leave out each one in turn.

Leave out egg (e): \(t c h\)

Leave out tuna (t): \(e c h\)

Leave out cheese (c): \(e t h\)

Leave out ham (h): \(e t c\)

The option shown leaves out tuna (t). Changing the order of the letters in an option does not make a new option, so there are four options in all.

Be systematic: fix the hundreds digit, list every number that starts with it, then move on to the next digit.

Starting with 5: \(567\) and \(576\)

Starting with 6: \(657\) and \(675\)

Starting with 7: \(756\) and \(765\)

Any of the 3 cards can go first, either of the remaining 2 can go second, and the last card is then forced, so there are \(3 \times 2 \times 1 = 6\) numbers:

\(567,\ 576,\ 657,\ 675,\ 756,\ 765\)

Repeats are allowed, so nothing is used up: each of the three columns can be filled by any one of the 3 cards, independently of the others.

\(3 \times 3 \times 3 = 27\)

Take the four in the order they are given, and pair each one only with the people listed after it. That way no pair is ever written twice.

Isla: Isla and Jack, Isla and Kaya, Isla and Liam

Jack: Jack and Kaya, Jack and Liam

Kaya: Kaya and Liam

Liam has already been paired with all three of the others, so there is nothing new to add.

That gives 6 pairs.

There are 7 choices for head student. Once that person is fixed, 6 people are left to choose from for deputy head student.

\(4 \times 3 = 42\)

That is exactly twice the 6 pairs in part (a), because a pair does not say who is which, while the two roles do: Isla as head student with Jack as deputy head student is a different outcome from Jack as head student with Isla as deputy head student.

There are \(4\) cards to choose for the tens digit, and then \(4 - 1\) cards left for the units digit.

\(4 \times 3 = 12\)

A number is even when its last digit is even, and only one of the four cards shows an even digit: \(6\).

So every even number she can make ends in \(6\), and any of the other \(3\) cards can go in front of it.

\(36,\ 56,\ 96\)

Put the largest digit in the tens column and the next largest in the units column.

\(9\) then \(6\), giving \(96\).

Omar has already chosen Art, which is in block 3, so no other subject from block 3 can be chosen. The subjects ruled out cannot be chosen either. That leaves the allowed subjects in each of the other blocks: Block 1: Music Block 2: English, Physics and Spanish Block 4: Chemistry The other two subjects must come from two different blocks. Go through the blocks in order: block 1 with block 2, then block 1 with block 4, then block 2 with block 4. Within each pair of blocks, take each allowed subject of the first block with each allowed subject of the second. A group is never made from two subjects in the same block, and no pair of blocks is used twice, so nothing is repeated or missed.

Subject 1	Subject 2	Subject 3
Art	Music	English
Art	Music	Physics
Art	Music	Spanish
Art	Music	Chemistry
Art	English	Chemistry
Art	Physics	Chemistry
Art	Spanish	Chemistry

The number of groups from blocks 1 and 2 is \(3\), from blocks 1 and 4 is \(1\) and from blocks 2 and 4 is \(3\). \(3 + 1 + 3 = 7\), so there are seven groups.

Start with the totals that leave the fewest choices.

The smallest total is \(10\). The three smallest numbers are \(2\), \(2 + 1\) and \(2 + 2\), which add to \(9\). The total is 1 more, so the only way is to replace the largest of those three by the next number up.

\(2 + 3 + 5 = 10\)

The largest total is \(36\). The three largest numbers are \(13\), \(13 - 1\) and \(13 - 2\), which add to \(36\). That is exactly the total, so no other three numbers can make it.

\(11 + 12 + 13 = 36\)

Cross those six numbers off. The numbers left are \(4,\ 6,\ 7,\ 8,\ 9,\ 10\), and the two totals still to make are \(18\) and \(26\).

Choose three of the numbers left to make \(18\), and check that the other three make \(26\):

\(4 + 6 + 8 = 18\)

\(7 + 9 + 10 = 26\)

Every number from 2 to 13 has now been used once, so the four calculations are

\(11 + 12 + 13 = 36\); \(7 + 9 + 10 = 26\); \(2 + 3 + 5 = 10\); \(4 + 6 + 8 = 18\)

Each side of four numbers must add up to 27. So the blanks on a side must add up to 27 minus the given numbers on that side. A corner belongs to two sides, so a blank corner has to suit both of them.

On the top side, \(A + B = 15\). Of the numbers still unused, only 3 + 12, 4 + 11 make 15. On the right side, \(B + C = 13\). Of the numbers still unused, only 1 + 12 make 13. On the bottom side, \(F + E = 15\). Of the numbers still unused, only 3 + 12, 4 + 11 make 15. On the left side, \(E + D = 17\). Of the numbers still unused, only 6 + 11 make 17.

Putting what the sides allow together, B can only be 12, so \(B = 12\).

The top side has only one blank, A. \(A = 27 - (2 + 10 + 12) = 3\). The right side has only one blank, C. \(C = 27 - (12 + 9 + 5) = 1\).

Putting what the sides allow together, E can only be 11, so \(E = 11\).

The bottom side has only one blank, F. \(F = 27 - (5 + 7 + 11) = 4\). The left side has only one blank, D. \(D = 27 - (11 + 8 + 2) = 6\).

Check each side: top: \(2 + 3 + 10 + 12 = 27\); right: \(12 + 1 + 9 + 5 = 27\); bottom: \(5 + 7 + 4 + 11 = 27\); left: \(11 + 6 + 8 + 2 = 27\).

Pair each flavour with itself and with the flavours listed after it, so that no option is written twice.

c: cc, cs, cb, cp

s: ss, sb, sp

b: bb, bp

p: pp

That gives 10 options. The four with the same flavour twice (cc, ss, bb and pp) are part of the list.

The whole circle is \(360^\circ\) and stands for 576 scoops.

\(360 \div 576 = 0.625\)

So one scoop is \(0.625^\circ\).

Each scoop is \(0.625^\circ\), so multiply each number of scoops by \(0.625\).

Strawberry: \(112 \times 0.625 = 70^\circ\)

Banana: \(160 \times 0.625 = 100^\circ\)

Pistachio: \(224 \times 0.625 = 140^\circ\)

The sector for chocolate is \(80 \times 0.625 = 50^\circ\), and the four angles add up to \(50 + 70 + 100 + 140 = 360^\circ\).

Rule each radius by counting the marks, which are \(10^\circ\) apart, and write each name in its sector.

Start with the "can" clues, because each one can be made in only one way.

22p with three coins: 2p + 10p + 10p = 22p

30p with four coins: 5p + 5p + 10p + 10p = 30p

All of these coins are among the six. The two groups use three and four coins, which is more than six, so they share some. Take each value as many times as it is needed in either group, and count a shared coin once:

2p, 5p, 5p, 10p, 10p

That is five coins, so there is one more coin to find. It is 2p, 5p or 10p. Test each against the "cannot" clues.

A sixth coin of 2p would mean Anya could make 2p + 2p + 5p = 9p using three coins, but that is a total Anya cannot make. So the sixth coin is not 2p.

A sixth coin of 10p would mean Anya could make 2p + 10p + 10p + 10p = 32p using four coins, but that is a total Anya cannot make. So the sixth coin is not 10p.

The sixth coin is 5p, so the six coins are

2p, 5p, 5p, 5p, 10p, 10p