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This page sets out the method for the product rule for counting in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the The product rule for counting topic page.
When you see a question about codes, menus, outfits or arrangements, draw one slot for each decision and write the number of choices under it.
When a digit is fixed, or two items cannot be together, fill the restricted slot first.
If the choices are independent, multiply the counts. Do not add them.
Combining 4 choices and 5 choices gives 20 total combinations.
A captain and a vice-captain from 8 people is 8×7, because the second person is different.
Choosing 2 people when order does not matter uses the ordered count and then halves it.
The second choice is not automatically equal to the first when a person cannot be picked twice. A rule can also remove a choice at one stage only. If a PIN cannot start with 0, the first digit has 9 choices and the later digits can still be 0.
Say whether order matters before you decide to divide.
Suppose a café offers 4 starters and 6 main courses, and a meal is one of each. Each starter pairs with all 6 mains, so there are 4 groups of 6. That is 4 × 6 = 24 meals. Adding 4 + 6 = 10 counts the dishes on the menu. It does not count the meals a customer can build. The product is the count of pairs.
A third stage extends the same idea. 3 desserts give 24 × 3 = 72 three-course meals. You do not add the 3 at the end. Every meal already counted can take any of the desserts, so the total is 3 times as large as 24. If a stage is compulsory, it belongs in the product. If it is optional, that is a different question and you have to count the 'skip it' choice as well.
The order of multiplication does not change the total. 4 × 6 × 3 and 3 × 6 × 4 are both 72. The order in which you fill the stages can matter for how you count the restrictions. It does not matter for the final product once those stage-counts are correct.
Method: 1. List the choices at the first stage. 2. Multiply by the choices at the next stage. 3. The product is the number of different combinations.
Five athletes are choosing a gold, a silver and a bronze, with no one winning two medals. The gold has 5 choices. The silver has 4 left. The bronze has 3 left. The number of podiums is 5 × 4 × 3 = 60. If the same person could win more than one medal, each place would have 5 choices and the total would be 5^3 = 125. The question has to say whether repeats are allowed.
The drop is one per stage only if exactly one item is used up at each stage. After two medals have been given, three athletes remain, not two. A slip is to write 5 × 4 × 2 by reducing too fast. Say the count out loud: 'five, then four, then three'. The last factor is the number still available, which is 5 - 2 = 3.
This is not the same as choosing an unordered group of three. 60 counts ordered podiums: gold to A and silver to B is different from the swap. If the question says 'how many ways can three people be chosen' with no jobs attached, order does not matter and the product rule on its own overcounts. For codes, medals and passwords, order matters, so the product is the right count.
Method: 1. The first place can be any of the people or items. 2. Each later place has one fewer choice, because a choice cannot be used twice. 3. Multiply the shrinking list of choices.
A three-digit code uses digits from 0 to 9 with no repeats, and the first digit cannot be 0. Deal with the restricted stage while the choice is still open. The first digit has 9 options: 1 to 9. It does not have 10, and it does not have 8. Zero is the only digit removed at this stage.
The second digit may be 0. It may not copy the first digit. That leaves 9 digits: the full set of 10, minus the one already used. The third digit misses two used digits, so it has 8 choices. The product is 9 × 9 × 8 = 648. Notice the second factor is 9, not 8. Students who write 9 × 8 × 7 have banned zero from the second digit as well.
If zero were allowed in front, the count would be 10 × 9 × 8 = 720. Those extra 72 codes are the ones that start with 0, and a three-digit code usually rejects them. Read the question for both rules: no repeats, and no leading zero. Each rule changes a different factor. Apply the tighter rule on the stage it belongs to.
Method: 1. Deal with the restricted first digit before the others. 2. A first digit from 1 to 9 has 9 choices, not 10. 3. Then multiply by the choices left for the later digits.
A code can take letters from one pool and digits from another. Two different letters chosen from A, B, C, D and E, followed by two different digits from 0 to 9, uses four stages. Letters: 5 then 4. Digits: 10 then 9, and 0 is allowed because these are not a place-value number. The total is 5 × 4 × 10 × 9.
Work it in two blocks if that is clearer. Letter pairs: 5 × 4 = 20. Digit pairs: 10 × 9 = 90. Then 20 × 90 = 1800. Every letter pair joins every digit pair, which is why the blocks multiply. Adding 20 + 90 = 110 counts the pairs as if you chose letters or digits, not letters and then digits.
Fill a restricted stage first when the restriction is about position. For a four-digit PIN with no repeats and no leading zero, the factors are 9 × 9 × 8 × 7 = 4536. First digit: 9. Second: 9, including zero. Third: 8. Fourth: 7. The same 'say the remaining count' habit fixes the product. Then multiply straight through.
Method: 1. Count the letter choices and the digit choices separately. 2. Reduce the choices if repeats are not allowed inside a pool. 3. Multiply the letter product by the digit product.
The product rule says that if one choice can be made in m ways and a separate second choice in n ways, then the pair of choices can be made in m × n ways. Real questions dress this up as outfits, lunch deals, number plates, passwords or lockers. Before you multiply, list the separate choices in order and write the number of options beside each one. Phrases such as "one of each" or "a shirt, a pair of jeans and a jacket" tell you to multiply.
Two details change the numbers. First, can something be repeated? "Digits can be repeated" keeps the same number of options at every step, while "no digit is used more than once" removes one option each time. Second, does order matter? In a code or a podium, 4, 7, 2 is different from 7, 2, 4, so every position is its own choice. Write the number of options for each position in a row of boxes, then multiply along the row.
Method: 1. List each position or each separate choice in order. 2. Write the number of options for each one, using fewer options when repeats are banned. 3. Multiply the numbers together. 4. State the answer as a count, and check that it is a whole number.
Tip: Draw a row of empty boxes, one per choice, and write the number of options in each box.
A condition usually changes the number of options in just one position. A number is even when its last digit is 0, 2, 4, 6 or 8, so the last position has 5 options. It is odd when its last digit is 1, 3, 5, 7 or 9, which is also 5 options. It is a multiple of 5 when it ends in 0 or 5, so there are 2 options. A number with 4 digits cannot start with 0, so the first position has 9 options and the others have 10.
Always deal with the restricted position first, then fill in the others. For a 4-digit multiple of 5, the last digit has 2 options and the first digit has 9. The two middle digits are free, with 10 options each, so the count is 9 × 10 × 10 × 2 = 1800. If repeats are not allowed, the order of filling matters. Fill the restricted position first, so that you take the used digits away from the remaining positions correctly.
Method: 1. Find the positions that have a condition, such as the last digit or the first digit. 2. Count the options for the restricted positions first. 3. Count the options left for the free positions, taking away digits already used if repeats are banned. 4. Multiply all the counts together.
Some questions say "either this or that". A café lets a customer choose a soup and a main, or a main and a dessert. These are different cases. Inside each case you multiply, because the case needs one item from each list. Across the cases you add, because any one meal belongs to exactly one case. With 6, 5 and 4 options in three lists, the case "first list and second list" has 6 × 5 = 30 ways and the case "second list and third list" has 5 × 4 = 20 ways.
Show-that questions give you the total, so your working must reach exactly that number. Write each case on its own line with its product, then add the lines. A common mistake is to add the numbers of items first and then multiply, such as 6 + 5 + 4 = 15 followed by 15 × 15. Check that the cases do not overlap, and that you have not left out the case that uses all three lists.
Method: 1. List the separate cases the question allows. 2. For each case, multiply the numbers of options in that case. 3. Add the totals of the cases. 4. For show-that, finish with the exact total given in the question.
A question that asks for the number of codes with at least one repeated digit is easier by subtraction. Count all the possible codes, count the codes with no repeats, then subtract. For a 3-digit locker code the total is 10 × 10 × 10 = 1000 and the number with no repeats is 10 × 9 × 8 = 720. So 280 codes have at least one repeat. Counting the repeats directly would need many separate cases.
Comparison questions, such as "Which method gives the greater number of possible codes? You must show your working", need a count for each method and a clear conclusion. Write both counts, compare them in a sentence and name the winning method. Do not stop at the two numbers, because the final mark is for the statement. Use friendly arithmetic: 5 × 5 × 5 × 5 = 625 and 9 × 8 × 7 = 504 are quick by hand.
Method: 1. Decide whether it is easier to count the wanted outcomes or to count everything and subtract the unwanted ones. 2. Count the total, then count the unwanted outcomes. 3. Subtract to get the answer. 4. For comparisons, write both counts and a sentence naming which is greater.
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