Product Rule for Counting

For each sandwich, there are 3 possible soups.

1. Each of the 4 sandwiches can be paired with any of the 3 soups.
2. There are 4\times3=12 different lunches.

Each notebook design can be paired with any of the pen colours.

1. For each of the 3 notebook designs, there are 5 pen colours.
2. The number of pairs is 3\times5=15.

Choose one card from each pile in turn.

1. Each of the 2 first cards can be paired with 4 second cards and 3 third cards.
2. There are 2\times4\times3=24 possible sets.

Divide the total number of meals by the number of main courses.

1. Each of the 6 main courses can be paired with every drink.
2. If there are d drinks, then 6d=30, so d=30\div6=5.

For each die result, the coin can land in either of two ways.

1. There are 6 possible die results and 2 possible coin results.
2. Together they give 6\times2=12 outcomes.

Once the entrance is chosen, one gate is unavailable as the exit.

1. There are 6 choices of entrance.
2. For each entrance there are 5 different exit gates, so there are 6\times5=30 pairs.

Work out how many posters each choice of colour produces.

1. For each colour there are 20\times2=40 design-and-size combinations.
2. The number of colours is 280\div40=7.

If there are d types of drink, how many ways are there to choose one sandwich and one drink?

1. By the product rule, if there are d types of drink, the number of ways is 14\times d=14d.
2. Mia's claim would need 14d=200, so d=200\div14=14.28\ldots
3. The number of types of drink must be a whole number, and 200 is not a multiple of 14 (14\times14=196 and 14\times15=210).
4. So Mia cannot be correct. Rounding to 14 types of drink gives 196 ways, not 200; and 186 comes from subtracting 14, which is wrong because the choices multiply.

There are the same number of choices at each of the three positions.

1. Each of the three positions has 4 choices because repetition is allowed.
2. The number of signals is 4^3=64.

Count the choices for each of the five positions, then multiply.

1. There are 5 positions. Each letter position has 26 choices and each digit position has 10 choices, because repeats are allowed.
2. By the product rule, multiply the numbers of choices: 26\times26\times10\times10\times10.
3. 26\times26=676 and 10\times10\times10=1000.
4. So there are 676\times1000=676\,000 different codes.

Count all shape-and-colour combinations, then remove the forbidden combination.

1. Without the restriction there would be 4\times3=12 badges.
2. Only the yellow star is excluded, so 12-1=11 badges are available.

Add the numbers of available sessions before counting the day choices.

1. There are 3+2=5 choices of session because the member chooses one from either group.
2. Each session can be paired with any of 4 days.
3. There are 5\times4=20 bookings.

Count each type of day out separately, then decide whether to add or multiply the two totals.

1. A museum day out needs a museum visit AND a lunch place, so multiply: 3\times4=12 ways.
2. An outdoor day out needs an outdoor visit AND a lunch place: 2\times5=10 ways.
3. A day out is one type OR the other, so add the two totals: 12+10=22.
4. Multiplying 12\times10 would count a museum day and an outdoor day together, which is not what is being chosen.

For each departure station, there are the same number of destination choices.

1. Each departure station gives n possible tickets.
2. For 3 departure stations, the total is 3\times n=3n.

The first-prize winner cannot also win second prize.

1. There are 8 choices for first prize and then 7 different competitors left for second prize.
2. There are 3 choices of judge, so the total is 8\times7\times3=168 outcomes.

Write the total as a product, then find the positive square root.

1. The number of icons is n\times n\times5=5n^2.
2. So 5n^2=320, giving n^2=64.
3. The count n must be positive, so n=8.

Let there be p flavours of popcorn and write the number of each type of deal in terms of p.

1. Let p be the number of flavours of popcorn. Within each type of deal, multiply the numbers of choices.
2. Drink and popcorn: 4p deals. Popcorn and sweet: 3p deals. Drink, popcorn and sweet: 4\times p\times3=12p deals.
3. A deal is one type or another, so add: 4p+3p+12p=19p.
4. So 19p=152, which gives p=152\div19=8 flavours of popcorn.

Count the possible last digits first, then the first and middle digits.

1. The last digit has 5 choices: 0, 2, 4, 6 or 8.
2. The first digit has 9 choices because it cannot be 0, and the middle digit has 10 choices.
3. Repetition is allowed, so the total is 9\times10\times5=450 codes.

Count the passwords of each pattern separately; remember a letter or digit that has been used cannot be used again.

1. There are 6 letters (A to F) and 9 digits. Count each pattern separately, then add.
2. Letter, digit, letter: 6 choices for the first letter, 9 for the digit, then 5 for the second letter (one letter is used). That is 6\times9\times5=270.
3. Digit, letter, digit: 9\times6\times8=432, since the second digit cannot repeat the first.
4. A password follows one pattern or the other, so add: 270+432=702 passwords.

Choose the last digit first, then count the choices left for each other position.

1. Fill the positions with restrictions first. The last digit must be odd: 1, 3, 5, 7 or 9, so it has 5 choices.
2. The first digit cannot be 0 and cannot be the digit already used at the end. The last digit is odd, so it is not 0; that leaves 10-2=8 choices.
3. Two different digits have now been used, so the second position has 10-2=8 choices and the third position has 7.
4. By the product rule, the total is 5\times8\times8\times7=2240 codes.
5. Starting with the first digit instead is harder: whether it is odd changes how many odd digits are left for the end.