Deal with the choices one at a time.

There are 4 ways to choose the starter. Whichever starter is chosen, there are 6 ways to choose the main course, so there are

\(4 \times 6 = 24\)

ways to choose a starter and a main course together. Each of those 24 pairs can be finished with any one of the 5 desserts:

\(24 \times 5 = 120\)

So there are 120 different meals. When choices are made one after another and every choice is free, you multiply.

The starter and the main course are unchanged: \(4 \times 6 = 24\) ways.

The dessert now has 6 outcomes, not 5 — the 5 desserts on the menu, and "no dessert", which is a choice like any other.

\(24 \times 6 = 144\)

So there are 144 different meals — exactly 24 more than in part (a), which are the 24 starter-and-main meals with no dessert.

Take the positions one at a time. Each of the 3 letter positions can be filled in 26 ways, and repeats are allowed, so nothing is used up:

\(26^{ 3 } = 17,576\)

Each of the 2 digit positions can be filled in 4 ways:

\(4^{ 2 } = 16\)

Every set of letters can go with every set of digits, so multiply:

\(17,576 \times 16 = 281,216\)

The letters are unaffected, so that part is still \(26^{ 3 } = 17,576\).

Of the digits 1 to 4, exactly 2 are odd. Each of the 2 digit positions now has 2 choices instead of 4:

\(2^{ 2 } = 4\)

\(17,576 \times 4 = 70,304\)

A route is one road for each of the three stages, and the three choices are independent — any road out of Ashford can be followed by any road out of Barton.

\(2 \times 4 \times 3 = 24\)

The outward journey can be made in 24 ways, as in part (a).

Now fix one outward journey. Coming back, one road out of Denby has already been used, so there are \(3 - 1 = 2\) left; then \(4 - 1 = 3\); then \(2 - 1 = 1\).

\(2 \times 3 \times 1 = 6\) return routes for each outward route.

Every one of the 24 outward routes can be paired with every one of those 6 return routes:

\(24 \times 6 = 144\)

One starter, one main course and one dessert are chosen, so the number of ways is

\(3 \times 8 \times (\text{number of desserts})\)

The starters and main courses give \(3 \times 8 = 24\) ways, so the claim needs

\(24 \times (\text{number of desserts}) = 280\)

Divide the claim by 24 to find the number of desserts:

\(280 \div 24 = 11\frac{2}{3}\)

The number of desserts must be a whole number. Here the answer is not a whole number, so there cannot be that many desserts.

So the answer is No.

Idris chooses one combination only, so the combinations are alternatives. For each one, the choices are made together, so multiply the stock numbers within it.

an iced bun and a muffin: \(7 \times 4 = 28\) ways.

a flapjack and a muffin: \(11 \times 4 = 44\) ways.

an iced bun, a flapjack and a muffin: \(7 \times 11 \times 4 = 308\) ways.

No way of choosing belongs to two of these combinations, so add the three totals:

\(28 + 44 + 308 = 380\)

So there are 380 different ways to choose.

Choose the president first: any of the 15 members, so 15 ways.

That member has now been used up, so the secretary can be any of the 14 members left.

Two members have been used up, so the social secretary can be any of the 13 left.

\(15 \times 14 \times 13 = 2,730\)

The numbers go down by one each time because each post takes a person out of the pool.

Nobody is used up now, so every post has all 15 members to choose from:

\(15 \times 15 \times 15 = 15^3 = 3,375\)

That is 645 more than in part (a) — those are the ways in which somebody holds two of the posts, or all three.

This is the difference the topic is about: "may be repeated" gives a power, "may not be repeated" gives a product that falls by one each step.

There are 6 cards to choose from for the hundreds digit.

That card has been used, so 5 remain for the tens digit, and then 4 remain for the units digit.

\(6 \times 5 \times 4 = 120\)

Writing all 120 of them out would be slow and easy to get wrong, which is why the counting is done by multiplying rather than by listing.

A number is odd exactly when its units digit is odd, so choose the units digit first — the position with the restriction always goes first.

3 of the cards are odd: \(3,\ 5,\ 7\). So there are 3 choices for the units digit.

The other 3 cards, \(2,\ 6,\ 8\), cannot go in the units place.

That leaves 5 cards for the hundreds digit and then 4 for the tens digit:

\(3 \times 5 \times 4 = 60\)

Check it against part (a): the other 3 cards give \(3 \times 5 \times 4 = 60\) even numbers, and \(60 + 60 = 120\).

Count the choices for each digit, then multiply, because every choice for one digit can go with every choice for the others.

Method A:

1st digit, between 2 and 8 inclusive: 7 digits, 2 to 8. Number of choices: \(7\)

2nd digit, odd: 5 digits: 1, 3, 5, 7, 9. Number of choices: \(5\)

3rd digit, between 2 and 8 inclusive: 7 digits, 2 to 8. Number of choices: \(7\)

4th digit, less than 6: 6 digits, 0 to 5. Number of choices: \(6\)

\(7 \times 5 \times 7 \times 6 = 1470\)

Method B:

1st digit, less than 8: 8 digits, 0 to 7. Number of choices: \(8\)

2nd digit, greater than 3: 6 digits, 4 to 9. Number of choices: \(6\)

3rd digit, any digit: 10 digits, 0 to 9. Number of choices: \(10\)

4th digit, less than 3: 3 digits: 0, 1, 2. Number of choices: \(3\)

\(8 \times 6 \times 10 \times 3 = 1440\)

Comparing the totals, \(1470 > 1440\), so Method A gives the greater number of codes.

Count how many digits from 1 to 9 are allowed in each box.

Box 1, a cube number: 1 and 8, so 2 choices.

Box 2, a multiple of 3: 3, 6 and 9, so 3 choices.

Box 3, less than 6: 1, 2, 3, 4 and 5, so 5 choices.

Box 4, less than 4: 1, 2 and 3, so 3 choices.

Repeats are allowed, so each box is chosen independently. Multiply the numbers of choices:

\(2 \times 3 \times 5 \times 3 = 90\)

There are 90 possible passcodes and only one is correct, so the probability of a correct guess is

\(\frac{1}{ 90 }\)

Deal with the choices one at a time.

There are 6 ways to choose the starter. Whichever starter is chosen, there are 5 ways to choose the main course, so there are

\(6 \times 5 = 30\)

ways to choose a starter and a main course together. Each of those 30 pairs can be finished with any one of the 3 desserts:

\(30 \times 3 = 90\)

So there are 90 different meals. When choices are made one after another and every choice is free, you multiply.

The starter and the main course are unchanged: \(6 \times 5 = 30\) ways.

The dessert now has 4 outcomes, not 3 — the 3 desserts on the menu, and "no dessert", which is a choice like any other.

\(30 \times 4 = 120\)

So there are 120 different meals — exactly 30 more than in part (a), which are the 30 starter-and-main meals with no dessert.