 # 6.1 Counting

### Counting range sizes

Most people think of counting as saying “1, 2, 3, …” perhaps while pointing to items being counted, but counting also involves approaches that use math to avoid having to explicitly count every item. One such approach counts the items in a range by applying simple math to the range’s low and high values.

PARTICIPATION ACTIVITY

6.1.1: A range’s items can be counted using simple math: High – Low + 1.Start2x speed

1

2

3

4

5

1

2

3

4

5

22 23 24 25 26 27 28 29

6

7

8

29

22

+ 1

= 7 + 1 = 8

20

34

Theater seats

34 – 20 + 1 = 14 + 1 = 15 seats keyboard_arrow_upCaptions

1. Counting is usually thought of as counting up “1 2 3 …” perhaps while pointing to items. How many balls? “1 2 3 4 5”.
2. But some counting benefits from a little math. For a range 1-5, most people know the range has 5 numbers. But how about 22-29? Sure, a person could count up.
3. A faster approach is: 29 – 22 + 1.
4. A row of seats in a theater are numbered 20 to 34. Is that enough for a group of 15 friends? 34 – 20 + 1 = 14 + 1 = 15. Perfect!

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6.1.2: Counting ranges.

Count the number of values in the range.

1)

1 to 10

` `

10

10 – 1 + 1 = 9 + 1 = 10.
Most people don’t need to use the math when the range starts with 1, realizing that the number of values equals the high value (because counting from 1 to 10 means 10 values exist).

2)

10 to 15

` `

3)

25 to 75

` `

51

75 – 25 + 1 = 51.
Some people wonder why the result isn’t just 50. While 50 is the difference, so 75 – 25 = 50, and 25 + 50 = 75, the question instead asks how many values are in the range. Basically, the + 1 is needed to account for the first value in the range, 25.

4)

In Europe, a building’s floors start at 0 (rather than 1 as in the U.S.). If a building’s top floor is numbered 8, how many floors does the building have?

` `

9

8 – 0 + 1 = 9.
One can confirm via counting these values: 0 1 2 3 4 5 6 7 8. Counting up for each value yields 1 2 3 4 5 6 7 8 9. The range 1 to 8 has 8 values, but when starting with 0, that 0 must be accounted for.

5)

12 people want to sit one row of the center section of a theater. That section has seats numbered 15 to 25. How many seats are in a row of that section?

` `

11

25 – 15 + 1 = 10 + 1 = 11.
Too bad, the row is short 1 seat.

6)

A hotel wants two-digit numbers for all rooms, so 10 to 99. How many possible rooms can the hotel have?

` `

90

99 – 10 + 1 = 90.Feedback?

Counting items by counting up by 2s or 3s

People often have to count items, like chairs in a row, people in a restaurant, or players on a soccer field. Counting up by 1s works, but may be slow. People have strong visual skills, so a faster way to count is to visually group items by 2s and then counting up by 2s: 2 4 6 8 … Even faster is counting by 3s: 3 6 9 …

A soccer referee must count players on the field before each half. How many total players are in the photo below? (Ignore the ref). First try moving across the scene and counting the players by 1s: 1 2 3 4… Then repeat but grouping the players into 3s and counting up by 3 for each group: 3 6 9 done. An example grouping is shown further below, starting from the bottom left and moving clockwise around the scene. ### Counting arrangements

People commonly count possible arrangements of something involving choices, like the possible arrangements of a letter and a digit (A9, B5, X0, …). Listing out each arrangement and counting is sometimes possible, but often the number is too large. A bit of math can help: The total arrangements can be found by multiplying the number of possible items for each position. For a letter (26 possibilities) and a digit (10 possibilities), the total possible arrangements is 26 × 10 = 260.

PARTICIPATION ACTIVITY

6.1.3: Counting arrangements is done by multiplying each position’s number of possible items.Start2x speed

A

B

1

2

3

= 12 arrangements

×

2 keyboard_arrow_upCaptions

1. A store owner wants to divide the store into regions like A1 and B3. Using AB and 1 2 3, how many 2-letter arrangements are possible?
2. The store owner starts with A and 1, then A 2, then A 3.
3. Next the owner does the same for B: B 1, B 2, B 3.
4. The first position had 2 choices, A and B. For each of those 2 choices, the second position had 3 choices: 1, 2, and 3. So total arrangements is 2 × 3 = 6.
5. What about 3 positions, with A B, then 1 2 3, then L R? Each of the first two position’s 6 arrangements has 2 choices, L or R.
6. A and B each has 3 choices, 1 2 3. Each of those has 2 choices, L R. Total arrangements is 2 × 3 × 2 = 12.

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6.1.4: Counting arrangements.

1)

A person will label chairs using one letter each. Possible letters are just A, B, C, D. How many unique labels are possible? (Hint: Just count those letters).

` `

4

Counting here is just normal counting: 1 2 3 4 (A is 1, B is 2, C is 3, and D is 4). 4 letters are possible.

2)

A person will label chairs using an arrangement of two letters. The first letter can be A, B, C, or D. The second can be L or R. How many unique labels are possible?

` `

8

The first position has 4 possible items: A, B, C, or D. For each of those, the second position has 2 possible items: L or R. So the total possible arrangements is 4 × 2 = 8. Those arrangements are AL, AR, BL, BR, CL, CR, DL, and DR.

3)

A sports team’s jerseys have two-digit numbers. The first digit can be 1, 2, 3, 4, or 5. The second can be 0, 1, 2, 3, 4, 5. How many possible 2-digit numbers are possible?

` `

30

The first digit has 5 possible items, and the second has 6. The total arrangements is 5 × 6 = 30. Those arrangements are 10, 11, 12, 13, 14, 15, 20, 21, 22, 23, 24, 25, …, 50, 51, 52, 53, 54, 55.

4)

A sports team’s jerseys have two-digit numbers. The first digit can be 1-5. The second can be 0-5. How many possible 2-digit numbers are possible?

` `

30

This question is the same as the previous, but the ranges are not listed out explicitly. A person may recognize that 1-5 has 5 items (the “standard” range which is easy to count) and that 0-5 has 6 items (because of the 0 being one more item).

5)

A sports team’s jerseys have two-digit numbers. The first digit can be 1-9. The second can be 0-9. How many possible 2-digit numbers are possible?

` `

90

1-9 has 9 items. 0-9 has 10 items. 9 × 10 = 90.

6)

People will be assigned a 2-digit ID consisting of a letter A-Z and a number 0-9, as in G5. How many unique IDs are possible? (Note: A-Z is 26 letters).

` `

260

A-Z has 26 possible items, and 0-9 has 10. Possible arrangements are thus 26 × 10 = 260. Note that a 2-digit number would have only had 10 × 10 = 100 arrangements. Using a letter increases the possibilities.

7)

How many 3-letter .com web domain names exist, like abc.com or cnn.com? (Case doesn’t matter; ABC.com is the same as abc.com)

` `

17576

a-z has 26 possible items. A 3-letter domain name has 3 such positions, each with 26 possible items, so 26 × 26 × 26 = 17,576 possibilities. Don’t expect to be able to register one for your next website; all 17,576 3-letter domain names have been taken, in fact all were taken by the year 2000. Buying one today could cost tens of thousands of dollars for seemingly meaningless 3-letter arrangements, and millions of dollars for arrangements that spell a word.

8)

This door lock has 5 buttons. Unlocking the door requires pressing the correct 4-button sequence. How many possible combinations exist? Assume that the same button can be pressed multiple times, and ignore the lock button. ` `

625

4 buttons must be pressed (like 4 positions). For each press, 5 possible items exist. So the possible combinations are 5 × 5 × 5 × 5 = 625.Feedback?

How long would guessing the combination take?

(True story): The keypad door lock shown above has the option of pressing a 4 or 6 button sequence to unlock the door. Frank’s wife felt pressing just 4 buttons would be too easy to guess, so the keypad was setup for pressing a 6 button sequence. Frank felt like requiring 6 buttons was excessive and grumbled slightly each time he had to press 6 buttons to get into the house (especially when mistyping). Finally, he decided to do a little calculation to convince his wife to reduce to 4 buttons.

How long might it take for someone to go through every combination? 4 presses is like 4 positions in the earlier animation above: __ __ __ __. Each position has 5 possible items: Button “1 2”, “3 4”, “5 6”, “7 8”, or “9 10”. So 5 × 5 × 5 × 5 = 625 possible combinations exist. Frank determined each 4-button guess takes about 8 seconds, so all 625 combinations could be guessed in 625 combinations × (8 seconds)/(1 combination) = 5000 seconds, which is 5000 seconds × (1 minute)/(60 seconds) = 83 minutes. And in reality, a person would hit the right combination sooner. On average that might be halfway through all combinations. So a person could guess the code in about 40 minutes and walk right into the house. 40 minutes is not very long at all.

Oops, she was right!

In contrast, a 6-digit code has 5 × 5 × 5 × 5 × 5 × 5 = 15,625 combinations, which would take 15,625 combinations × (8 sec)/(1 combination) × (1 min) / (60 sec) = 2083 minutes, or 34 hours (or 17 hours, which is half). Ah, that’s better. Frank no longer grumbles at having to press 6 buttons.

### Arrangements without repeats

Sometimes arrangements involve no repeats, like creating a 3-letter string using letters A-E but without any letter appearing more than once. In that case, the number of possible items decreases by 1 for each successive position. An arrangement of items in which the order matters is known as a permutation. Permutations are used to count arrangements where the number of original items equals the number of positions N. In these cases, the calculation is just N!, called N factorial. The number of permutations of 5 items is 5!, or 5 × 4 × 3 × 2 × 1 = 120.

PARTICIPATION ACTIVITY

6.1.5: Arrangements with and without repeats.Start2x speed

A

B

C

D

E

= 125 arrangements

AAA AAB AAC … EEC EED EEE

A

B

C

D

E

= 60 arrangements

D

D

C

D

C

A

A

B

C

3 × 2 × 1

= 6 arrangements

ABC ACB BAC BCA CAB CBA

“Permutations”

3! (3 factorial)

5

× 5

× 5

5

× 4

× 3 keyboard_arrow_upCaptions

1. If each position can be any letter A-E, then each position has 5 possibilities. The total possible arrangements of 3 letters is 5 × 5 × 5 = 125 arrangements.
2. If each letter can only appear once, the 1st position still has 5 possibilities, but the 2nd only has 4, and the 3rd only 3. If D appears first, the 2nd can only be A, B, C, or E.
3. The number of permutations of 3 items in 3 positions is 3 × 2 × 1 = 6 (or 3!, read as 3 factorial).

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6.1.6: Arrangements without repeats.

1)

How many 3-letter strings can be built from the letters W X Y Z if repeats are allowed?

` `

2)

How many 3-letter strings can be built from the letters W X Y Z if repeats are not allowed?

` `

24

For the first position, 4 choices exist. But once that position’s letter is chosen (suppose X), then only 3 choices remain for the second position (like W Y Z). And then once that position’s letter is chosen (suppose Z), then only 2 choices remain for the third position. So the total number of possibilities is 4 × 3 × 2 = 24.

3)

How many 4-letter strings can be built from the letters W X Y Z if repeats are not allowed? This question is the same as asking how many arrangements exist of the letters W X Y Z.

` `

24

The first position has 4 choices. The second has 3 choices (one is taken by the first position). The third position has 2 choices. The fourth position has just 1 letter to choose from. The total number of unique arrangements, known as permutations, is 4 × 3 × 2 × 1 = 24. That calculation is commonly written as 4!, and called 4 factorial.

4)

Jay is trying to fit 4 odd-sized suitcases into the trunk, left-to-right. Jay can’t figure out which ordering yields the best fit. If Jay decides to try all possible orderings, how many orderings must Jay try? .

` `

24

Because all four suitcases must be placed, the problem is a permutation problem. The calculation is 4!, so 4 × 3 × 2 × 1 = 24.

5)

A soccer coach can’t decide which arrangement of 3 defenders (Mia, Jay, Lee) works best. The positions are left, center, and right defender. If the coach decides to try every arrangement during a game to see which seems best, how many arrangements must the coach try?

` `

6

Because three players must be placed into three positions, the problem is a permutation problem. The calculation is 3!, so 3 × 2 × 1 = 6. Here are the permutations: Mia Jay Lee, Mia Lee Jay, Lee Mia Jay, Lee Jay Mia, Jay Mia Lee, Jay Lee Mia.Feedback?

CHALLENGE ACTIVITY

6.1.1: Counting.

433986.2690798.qx3zqy7Start

How many values are in the range? 0 to 50

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2

3

4

5CheckNext

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