Counting Without Listing

Mathematics · Probability · ages 15-16

Name ______________________ Date ____________

A stall offers 2 sizes, 7 flavours and 5 toppings. How many different orders are possible?
Answer: ______________
Adding the number of options at each stage gives the number of possible outcomes.
Circle one: True False
You are picking an outfit: 4 shirts, 3 pairs of pants, and 2 pairs of shoes. How many complete outfits can you make?
Answer: ______________
A set menu offers 4 starters, 6 mains and 3 puddings. How many different three-course meals are possible?
Answer: ______________
5 racers finish a race, and gold, silver, and bronze medals go to three different racers (nobody can win two medals). How many ways can the medals be given out?
- 15
- 60
- 125
- 10
A padlock code is four digits, each from 0 to 9, and digits may repeat. How many codes are there?
Answer: ______________
Mia counts the ways to give three different prizes to three of 10 people 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
- sshe has forgotten that the prizes are different from one another in value
A drinks machine offers 6 drinks and 3 cup sizes, but the largest cup only works with 2 of the drinks. What is the count?
- 18, since 6 × 3 counts each of the drinks against each of the sizes available
- 9, adding the six drinks to the three cup sizes offered
- 12, since two of the three sizes work with all six drinks
- 14, since two sizes take all six drinks and the largest takes only two
A count of outcomes can be used as the denominator of a probability only when the outcomes are equally likely.
Circle one: True False
A code is two different letters from a choice of 4, then two digits from 0 to 9 that may repeat. 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

Answer key

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

Counting Without Listing W1-mt_74gSbYIG8p-s1

70 · Multiply the stages: 2 × 7 × 5 = 70 different orders.
False · Adding counts the options themselves, not the ways of combining them. With 3 breads and 5 fillings, adding gives 8, which is neither a number of sandwiches nor anything else useful.
24 · Multiply the option count at each of the three stages.
72 · Multiply the options at each stage: 4 × 6 × 3 = 72 different meals.
60 · Once a racer wins a medal, they're out of the running for the next one, so the choices shrink at each stage.
10000 · Each of the four positions has 10 options and repeats are allowed, so 10 × 10 × 10 × 10 = 10,000 codes.
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.
14, since two sizes take all six drinks and the largest takes only two · The stages are not free choices, so 6 × 3 overcounts. Two sizes give 6 each, which is 12, and the largest gives 2, so the real total is 14.
True · Dividing by the total shares the chance out evenly between the outcomes. If some are likelier than others, that sharing is wrong and the fraction means nothing.
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.