Counting Outcomes with the Product Rule · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.
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Counting without listing

Mathematics · Probability · ages 15-16
Name ______________________   Date ____________

  1. A stall offers 2 sizes, 7 flavours and 5 toppings. Choose one of each, and every combination is available. How many different orders are possible?

    Answer: ______________

  2. For any staged choice, adding the numbers of options always gives the total number of complete outcomes.

    Circle one:   True   False

  3. A badge has one of 2 shapes. You may add one of 3 border styles or choose no border. Every shape allows every border choice. How many badge designs are possible?

    Answer: ______________

  4. A set menu offers 4 starters, 6 mains and 3 puddings. Choose one of each. Every combination is available. How many different three-course meals are possible?

    Answer: ______________

  5. A code has two different letters from A, B, C, then a digit. If the first letter is A, the digit is 0 or 1. If the first letter is B or C, the digit is 0, 1, 2 or 3. How many codes are allowed?

    • 24 codes
    • 20 codes
    • 12 codes
    • 16 codes

  6. A padlock code has four digit positions, each using 0 to 9. Digits may repeat, and 0 may be first. How many codes are there?

    Answer: ______________

  7. Mia counts the ways to give three different prizes to three different people from a group of 10 as 10 × 10 × 10 = 1000. Why is that too big?

    • she has multiplied when she should have added the three stages together instead
    • she should have used 8 × 3, since there are three separate prizes to be given out in total
    • her count includes one person winning two or three prizes, which is not allowed here
    • she has treated the three prizes as identical, even though each prize is different

  8. A machine offers 6 drinks and 3 cup sizes. Two sizes work with all 6 drinks; the largest works with only 2. How many drink-and-size orders are available?

    • 18 orders: treat every size as allowing all six drinks.
    • 9 orders: add the drink count and the cup-size count.
    • 12 orders: count only the two unrestricted cup sizes.
    • 14 orders: add twelve orders and two orders.

  9. The rule probability = wanted outcomes / total outcomes applies generally when all the counted outcomes are equally likely.

    Circle one:   True   False

  10. A code uses two different letters from a choice of 4, then two digits from 0 to 9. Digits may repeat, and 0 is allowed in either digit position. How many codes are there?

    • 1200, since the letters shrink from 4 to 3 and both digits stay free
    • 1600, since all four of the stages keep their full number of options available
    • 120, since 4 × 3 × 10 counts the letters and only one of the digits
    • 400, since 4 × 10 × 10 leaves the second letter out of the count
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Answer key

For grown-ups. Fold this page away before handing over the rest.

Counting without listing W1-mt_74gSbYIG8p-s1

  1. 70 · Multiply the stages: 2 × 7 × 5 = 70 different orders.
  2. False · Adding counts individual options, not generally complete outcomes. With 2 hats and 3 scarves, every pairing allowed, adding gives 5 options but multiplying gives 6 outfits.
  3. 8 · The border stage has 3 styles plus no border, giving 4 choices. Multiply 2 × 4 = 8 designs.
  4. 72 · Multiply the options at each stage: 4 × 6 × 3 = 72 different meals.
  5. 20 codes · Starting with A gives 1 × 2 × 2 = 4 codes. Starting with B or C gives 2 × 2 × 4 = 16. Add 4 + 16 = 20.
  6. 10000 · Each of the four positions has 10 options and repeats are allowed, so 10 × 10 × 10 × 10 = 10,000 codes.
  7. her count includes one person winning two or three prizes, which is not allowed here · Leaving 10 at every stage lets the same person be picked again. Once someone has won, only 9 are left, then 8, giving 10 × 9 × 8 = 720.
  8. 14 orders: add twelve orders and two orders. · The cup sizes have different numbers of drink choices. Two sizes give 2 × 6 = 12 orders, and the largest gives 2. Add the separate cases: 12 + 2 = 14.
  9. True · The rule assigns the same share of probability to each counted outcome. Equal likelihood makes those shares valid.
  10. 1200, since the letters shrink from 4 to 3 and both digits stay free · The letters cannot repeat, so that pair gives 4 × 3 = 12. The digits are free, so they give 10 × 10 = 100. Altogether 12 × 100 = 1200 codes.
Worksheet · LightMySky