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Basic Counting Principles: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Camera controls

The grid shows the allowed camera settings. The camera currently uses Portrait mode with the flash On. How many new settings change both the mode and the flash choice?

The allowed camera settings, with the mode down the side and the flash choice across the top.

Text description of this figure

A grid with three equal rows and two equal columns. The rows are labeled Landscape, Portrait and Macro from top to bottom, under the heading Mode. The columns are labeled On and Off from left to right, under the heading Flash. Every cell holds one identical dot, and no cell is highlighted or numbered.
1. Hint 1
  Each cell combines a mode with a flash choice.
2. Hint 2
  Keep the rows and columns whose labels differ from the current choices, then count the cells where they meet.
Answer

$2$ settings.

Full solution

The new mode can be Landscape or Macro, giving two choices.
The new flash choice must be Off, giving one choice.
The number of new settings is
$2\times1=2.$

The two cells represent Landscape with Off and Macro with Off.
Each changes both controls, and no other cell does.

Answer

$2$ settings.

Key idea

A grid can count outcomes that meet restrictions by using the rows and columns still allowed.
Problem 1: Mark complete
Problem 2 Multiples of $8$ or $11$

How many whole numbers from $1$ to $60$ , including both, are a multiple of $8$ or a multiple of $11$ ?
1. Hint 1
  Each counted number is a single choice from the multiples of $8$ or the multiples of $11$ , so ask whether the two groups share any member.
2. Hint 2
  Count the multiples of $8$ up to $60$ , and separately count the multiples of $11$ up to $60$ .
3. Hint 3
  A number in both groups would be a common multiple of $8$ and $11$ , so compare their least common multiple with $60$ .
Answer

$12$ numbers.

Full solution

The multiples of $8$ from $1$ to $60$ are $8$ , $16$ , $24$ , $32$ , $40$ , $48$ and $56$ , which is seven numbers.
The multiples of $11$ from $1$ to $60$ are $11$ , $22$ , $33$ , $44$ and $55$ , which is five numbers.

A number in both groups would be a common multiple of $8$ and $11$ .
Their least common multiple is $88$ , which is greater than $60$ , so no number from $1$ to $60$ is in both groups.
The lists agree: none of $11$ , $22$ , $33$ , $44$ and $55$ is a multiple of $8$ .

Each counted number is one number from one of two separate groups, so the group sizes add.
$7+5=12.$

Multiplying,
$7\times5=35,$
would count pairs made of one multiple of $8$ and one multiple of $11$ , but each outcome here is a single number.

Answer

$12$ numbers.

Key idea

Decide between adding and multiplying from what one outcome contains: one item from groups that share no member adds, and one item from each group multiplies.
Problem 2: Mark complete
Problem 3 The walking tour

The tree shows every allowed walking tour. A tour starts at Gate, passes through one middle location, and ends at one room. How many different complete tours are shown?

The tree of allowed walking tours, starting at Gate.

Text description of this figure

A tree drawn from left to right. A starting dot labeled Gate has one branch to a middle dot labeled Courtyard and one branch to a middle dot labeled Hall. From Courtyard, one branch goes to a dot labeled Library and one to a dot labeled Studio. From Hall, one branch goes to a dot labeled Gallery, one to a dot labeled Kitchen, and one to a dot labeled Loft. There are no arrows, distances or numbers.
1. Hint 1
  Follow a whole path from Gate to an ending room to identify one outcome.
2. Hint 2
  Count the endings reached through each middle location separately, then combine those separate groups of complete tours.
Answer

$5$ tours.

Full solution

Through Courtyard there are two tours, ending at Library or Studio.
Through Hall there are three, ending at Gallery, Kitchen, or Loft.
The groups are separate, so
$2+3=5.$

The two middle locations have different numbers of choices afterward.
Counting their endings separately accounts for every tour once.

Answer

$5$ tours.

Key idea

When branches offer different numbers of later choices, count each branch separately and add its outcomes.
Problem 3: Mark complete
Problem 4 A sticker strip

A sticker strip is a row of squares, and each square shows one of five shapes: a star, a heart, a moon, a leaf, or a drop. Every strip has the same number of squares, and every such row of shapes is allowed, including repeated shapes. There are exactly $625$ possible strips. How many squares does each strip have?
1. Hint 1
  Adding a square gives five versions of every shorter strip.
2. Hint 2
  Compare successive powers of $5$ with the total number of strips.
Answer

$4$ squares.

Full solution

Each square offers five choices, so a strip with $n$ squares has $5^n$ possible patterns.
The counts for one through four squares are
$\begin{aligned}5^1&=5\\5^2&=25\\5^3&=125\\5^4&=625\end{aligned}$

The count reaches $625$ at four squares.
Each additional square multiplies the count by five, so no other positive whole number of squares gives that total.

Answer

$4$ squares.

Key idea

For repeated choices, the exponent tells how many stages are needed to produce the stated total.
Problem 4: Mark complete
Problem 5 Installing the updates

A technician must install four different updates, each exactly once. Any order is allowed. The technician also chooses one of two starting times, morning or evening, with every update order available at either time. How many complete plans are possible?
1. Hint 1
  A complete plan contains an update order and a starting time.
2. Hint 2
  Count the update orders one position at a time, removing each used update from the choices.
3. Hint 3
  Each update order can be paired with either starting time.
Answer

$48$ plans.

Full solution

The four update positions have $4$ , $3$ , $2$ , and $1$ choices as updates are used.
The number of orders is
$4\times3\times2\times1=24.$

Each order works with either of the two starting times.
$24\times2=48.$

As a check, morning plans and evening plans are separate groups of $24$ , giving $24+24=48$ .

Answer

$48$ plans.

Key idea

Count an order with shrinking choices, then multiply by the choices of any further stage that is available with every order.
Problem 5: Mark complete
Problem 6 Four beats of sound

A sound sequence has four beats. Each beat uses tone A, B, or C, but a tone must differ from the tone immediately before it. A tone may return after an intervening beat. How many sequences are possible?
1. Hint 1
  The number of choices after the first beat matters, even when the particular available tones change.
2. Hint 2
  At each later beat, exclude just the tone used immediately before it.
3. Hint 3
  Multiply the counts for the four beats.
Answer

$24$ sequences.

Full solution

The first beat has three choices.
At each later beat, exactly one of the three tones is excluded, leaving two choices.
A tone used earlier than the preceding beat is available again.
$3\times2\times2\times2=24.$

This multiplication is valid because every possible earlier sequence leaves exactly two choices for its next beat.

Answer

$24$ sequences.

Key idea

When each stage offers the same number of choices after every earlier outcome, the counts multiply, even if the particular choices differ.
Problem 6: Mark complete
Problem 7 Tile samples

A tile catalog originally offers $6$ textures and $4$ colors, with every texture available in every color. It adds $2$ new textures and $1$ new color, again offering every combination. How many additional tile choices does the catalog now offer?
1. Hint 1
  Compare the total choices before the additions with the total afterward.
2. Hint 2
  Work out the new texture count and the new color count before multiplying.
3. Hint 3
  Subtract the old total from the new total.
Answer

$16$ additional choices.

Full solution

Originally each of the six textures pairs with four colors.
$6\times4=24.$

The new catalog has $6+2=8$ textures and $4+1=5$ colors.
$8\times5=40.$

The increase is
$40-24=16.$

Check by counting the additions separately: the new color gives six choices on the old textures, and the two new textures give $2\times5=10$ choices, for $6+10=16$ .

Answer

$16$ additional choices.

Key idea

When several stages gain choices, compare the full products before and after the change.
Problem 7: Mark complete
Problem 8 Volunteer lists

The chess list contains Ari, Bea, Cal, and Dev. The music list contains Cal, Dev, and Emi. One volunteer may be chosen from either list. Jules claims there are seven possible volunteers. Is the claim correct? Explain.
1. Hint 1
  A possible volunteer is a person, even when that name appears on both lists.
2. Hint 2
  Make one organized list in which each eligible name appears once.
Answer

No; there are $5$ possible volunteers.

Full solution

The eligible people are Ari, Bea, Cal, Dev, and Emi, so there are five possible volunteers.
Jules added the list sizes,
$4+3=7,$
but that counts Cal twice and Dev twice.

Counting each eligible person once gives $5$ , so the claim is false.

Answer

No; there are $5$ possible volunteers.

Key idea

Adding group sizes counts a single choice correctly when the groups share no members.
Problem 8: Mark complete
Problem 9 Recorder slots

A recorder stores four sounds in order. Each slot may contain a bell, a whistle, or a clap, and sounds may repeat. Nora groups the first two slots as one block and the last two as another. She writes $9^2$ for the number of possible records. Is her expression correct? Explain what its base and exponent count, and state how many records are possible.
1. Hint 1
  A group of two sounds needs its own count before it can serve as a single stage.
2. Hint 2
  Each sound in the first slot of a block can pair with any of the three sounds in its second slot.
3. Hint 3
  Choosing the first block leaves all possible choices available for the last block.
Answer

Yes; $81$ records. Base $9$ counts the ordered two-slot blocks, and exponent $2$ counts the two blocks in a record.

Full solution

Each block is an ordered pair of sounds.
There are three choices for its first sound and three for its second, including a repeat.
The block count is
$3\times3=9.$

Every first block can be followed by any of the nine last blocks.
Thus Nora's count is
$9\times9=9^2.$
The base counts the choices for a whole block, and the exponent counts the two blocks.

The total number of records is
$9^2=81.$
Counting the individual slots also gives
$3^4=81.$
Grouping the slots changes how the choices are organized, but every complete record is still counted once.

Answer

Yes; $81$ records. Base $9$ counts the ordered two-slot blocks, and exponent $2$ counts the two blocks in a record.

Key idea

A group of ordered choices can serve as one counting stage after its possible outcomes have been counted.
Problem 9: Mark complete
Problem 10 Status windows

Three labeled windows each show one of seven different status words. Under rule A, the three windows must show three different words. Under rule B, words may repeat. Mina claims rule B allows twice as many settings as rule A. Is her claim correct? Justify your answer.
1. Hint 1
  Work out a count for each rule before comparing them.
2. Hint 2
  Under one rule, a used word is unavailable for the next window; under the other, every word remains available.
3. Hint 3
  Compare the rule B count with twice the rule A count.
Answer

No; rule A allows $210$ settings and rule B allows $343$ , which is not twice $210$ ( $420$ ).

Full solution

For rule A, the window choices shrink as words are used.
$7\times6\times5=210.$

For rule B, all seven choices remain available in every window.
$7\times7\times7=343.$

Twice the rule A count is
$2\times210=420.$

Since $343$ is less than $420$ , rule B allows fewer than twice as many settings, so Mina's claim is false.

Answer

No; rule A allows $210$ settings and rule B allows $343$ , which is not twice $210$ ( $420$ ).

Key idea

Allowing repetition changes the choices at each later stage, so compare the complete products.
Problem 10: Mark complete

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