Product Rule for Counting

For each choice from Block A, think about how many choices there are from Block B, then from Block C.

1. One subject from each block, so multiply the numbers of choices (don't add).
2. 5\times4\times6=120
3. Answer: 120

Count the number of choices for each of the five positions, then multiply (don't add).

1. There are 26 choices for each letter and 10 choices for each digit.
2. Repeats are allowed, so the choices don't go down.
3. Multiply: 26\times26\times10\times10\times10
4. =676\,000

Count the choices for the first song, then the second, and so on: each song played leaves one fewer.

1. First song: 7 choices; second: 6; third: 5; and so on down to 1
2. 7\times6\times5\times4\times3\times2\times1=5040
3. Answer: 5040

Think about how many students could win each prize, remembering that a student who has won one prize can still win another.

1. Each prize can go to any of the 25 students, because a student can win more than one prize.
2. So there are 25 choices for each of the 3 prizes.
3. 25\times25\times25=25^3=15\,625
4. 25\times24\times23 would be right only if a student could win at most one prize.
5. Answer: 25^3

A flag on the pole can't be used again, so each position has one fewer choice than the one above it.

1. Top flag: 7 choices
2. Middle flag: 6 choices left
3. Bottom flag: 5 choices left
4. 7\times6\times5=210
5. Answer: 210

Fill in the first and last digits first: which digits are square numbers, and how must a multiple of 5 end?

1. First digit: a square number, so 1, 4 or 9: 3 choices (0 can't start a 4-digit integer)
2. Last digit: 0 or 5 for a multiple of 5: 2 choices
3. Second and third digits: any digit 0 to 9, so 10 choices each
4. 3\times10\times10\times2=600
5. Answer: 600

First list the different digits that appear in the number, then deal with the first digit before the others.

1. The different digits are 0, 2, 5, 6 and 9: five digits
2. First digit: not 0, so 4 choices
3. Second digit: any of the 4 digits not yet used (0 is allowed now)
4. Third digit: 3 choices left
5. 4\times4\times3=48
6. Answer: 48

Split into two cases, electric and not electric, count each with the product rule, then add.

1. Petrol or diesel: 2\times5\times2=20
2. Electric: 1\times3\times2=6
3. Add the two cases: 20+6=26
4. Answer: 26

Fill the front place first: who is allowed to stand there?

1. Front place: anyone except Alex, so 4 choices
2. Second place: 4 people left (Alex is now allowed)
3. Then 3, 2 and 1 choices for the other places
4. 4\times4\times3\times2\times1=96
5. Answer: 96

Fill the most restricted places first: the last digit must be odd, and the first digit cannot be 0

1. Last digit (odd): 1, 3, 5, 7 or 9, so 5 choices.
2. First digit: not 0 and not the last digit, so 10-2=8 choices.
3. Middle digit: any digit except the two already used, so 8 choices.
4. 5\times8\times8=320

Work out the number of PINs altogether and the number with four different digits.

1. All PINs: 10\times10\times10\times10=10\,000
2. PINs with four different digits: 10\times9\times8\times7=5040
3. Every other PIN has at least one repeated digit.
4. 10\,000-5040=4960
5. Answer: 4960

Fill the first and last places before the middle four, remembering that a consonant used at the start can't be used again at the end.

1. The consonants are M, R, K and T (A and E are vowels)
2. First letter: 4 choices
3. Last letter: 3 consonants left
4. Middle four places: the 4 letters left in any order, 4\times3\times2\times1=24
5. 4\times3\times24=288
6. Answer: 288

Each match has a home team and an away team: count the choices for each, in terms of n, and set the product equal to 182.

1. Home team: n choices
2. Away team: any other team, so n-1 choices
3. n(n-1)=182
4. n^2-n-182=0
5. (n-14)(n+13)=0
6. n is positive, so n=14
7. Answer: n=14

Split into two cases: the last digit is 0, or the last digit is 2, 4, 6 or 8, because the first digit can't be 0.

1. Case 1, last digit 0: first digit 9 choices, then 8, then 7
2. 9\times8\times7=504
3. Case 2, last digit 2, 4, 6 or 8: 4 choices
4. First digit: not 0 and not the last digit, so 8 choices; then 8 and 7
5. 4\times8\times8\times7=1792
6. 504+1792=2296
7. Answer: 2296

Split by the first digit: if it is 1 or 2 the other digits are free, but if it is 3 the second digit is restricted too.

1. First digit 1 or 2: 2\times9\times8\times7=1008
2. First digit 3: the second digit must be less than 5 and not 3, so 0, 1, 2 or 4: 4 choices
3. Then 8 and 7 choices: 1\times4\times8\times7=224
4. 1008+224=1232
5. Answer: 1232

Fill the restricted places first (the first digit, and the last digit for a multiple of 5), and split into cases by the last digit.

Part (a)

1. First digit: not 0, so 5 choices
2. Then 5, 4 and 3 choices for the other digits
3. 5\times5\times4\times3=300
4. Answer: 300

Part (b)

1. A multiple of 5 ends in 0 or 5
2. Ends in 0: 5\times4\times3=60
3. Ends in 5: first digit not 0 or 5, so 4 choices; then 4 and 3
4. 4\times4\times3=48
5. 60+48=108
6. Answer: 108

Part (c)

1. The first digit must be 5, 7 or 8
2. Ends in 0: first digit 5, 7 or 8, then 4 and 3 choices: 3\times4\times3=36
3. Ends in 5: first digit 7 or 8 (5 is used), then 4 and 3 choices: 2\times4\times3=24
4. 36+24=60
5. Answer: 60

The question doesn't say how many digits, so count the 1-digit, 2-digit, 3-digit, 4-digit and 5-digit integers separately and add.

Part (a)

1. Count the integers of each length and add.
2. 1 digit: 5
3. 2 digits: 5\times4=20
4. 3 digits: 5\times4\times3=60
5. 4 digits: 5\times4\times3\times2=120
6. 5 digits: 5\times4\times3\times2\times1=120
7. 5+20+60+120+120=325
8. Answer: 325

Part (b)

1. An even integer ends in 2 or 6, so there are 2 choices for the last digit
2. 1 digit: 2; 2 digits: 2\times4=8; 3 digits: 2\times4\times3=24
3. 4 digits: 2\times4\times3\times2=48; 5 digits: 2\times4\times3\times2\times1=48
4. 2+8+24+48+48=130
5. Answer: 130

Write the number of choices for the bottom, middle and top scoop in terms of n, multiply, and set this equal to 2730

Part (a)

1. Bottom scoop: n choices; middle: n-1; top: n-2
2. So n(n-1)(n-2)=2730
3. Three consecutive integers multiply to 2730, so they are close to \sqrt[3]{2730}\approx14.0
4. Try 13\times14\times15=2730 ✓
5. So n-2=13 and n=15

Part (b)

1. Bottom scoop: chocolate only, so 1 choice.
2. Middle scoop: not chocolate and not vanilla, so 15-2=13 choices.
3. Top scoop: 12 choices left.
4. 1\times13\times12=156

Write the number of girls in terms of the number of boys, b, and use the product rule to form an equation. Answer part (a) before attempting part (b).

Part (a)

1. Let the number of boys be b, so there are 15-b girls
2. Choose a boy (b ways) and a girl (15-b ways): b(15-b)=54
3. b^2-15b+54=0
4. (b-6)(b-9)=0, so b=6 or b=9
5. There are more girls than boys, so b=6
6. Answer: 6 boys

Part (b)

1. There are 15-6=9 girls
2. Choosing a first and a second girl: 9\times8=72
3. Each game is counted twice this way (A against B and B against A), so 72\div2=36 games
4. Total: 54+36=90
5. Answer: 90

Multiply the number of choices for each of the four characters, and remember n must be a whole number.

Part (a)

1. Letters: n\times n=n^2
2. Digits: 10\times9=90
3. Answer: 90n^2

Part (b)

1. 90n^2\geqslant50\,000
2. n^2\geqslant555.5\ldots
3. n\geqslant23.57\ldots
4. n is a whole number, so n=24
5. Check: 90\times23^2=47\,610 (too few) and 90\times24^2=51\,840
6. Answer: n=24

Part (c)

1. Letters: n choices, then n-1
2. Number of codes: 90n(n-1)
3. n=24: 90\times24\times23=49\,680, which is less than 50 000
4. n=25: 90\times25\times24=54\,000
5. Answer: n=25