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Counting Principles, Permutations, and Combinations: 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 configurations

A camera uses exactly one lens. Each of two wide lenses has 3 allowed settings, and a third lens has 5 allowed settings. How many lens-and-setting configurations are possible?
1. Hint 1
  Configurations using different lenses form non-overlapping cases.
2. Hint 2
  Count the configurations for the two wide lenses together, then add those for the third lens.
Answer

11 configurations.

Full solution

The wide lenses contribute $2\cdot3=6$ configurations.
The third lens contributes $5$ more.
These cases share no lens, so
$6+5=11.$
A uniform product of three lenses and one setting count would not work because the setting counts differ.

Answer

11 configurations.

Key idea

Add separate case counts when different first choices leave different numbers of options.
Problem 1: Mark complete
Problem 2 Two list lengths

For an integer $n\ge4$ , simplify $\frac{P(n,4)}{P(n,3)}$ to an expression with no factorials.
1. Hint 1
  Each permutation count has one factor per filled slot.
2. Hint 2
  Expand both shrinking products and cancel their shared positive factors.
Answer

$n-3$ .

Full solution

The numerator has factors $n,(n-1),(n-2),(n-3)$ and the denominator has the first three.
All shared factors are positive because $n\ge4$ , so
$\frac{P(n,4)}{P(n,3)}=n-3.$
This is also the number of unused objects available after three slots have been filled.

Answer

$n-3$ .

Key idea

Extending a no-repeat ordered list by one slot multiplies its count by the number of objects still available.
Problem 2: Mark complete
Problem 3 Lists and selections

For an integer $k\ge2$ , a program lists every ordered selection of $k$ distinct objects from a finite collection of at least $k$ objects, exactly once. There are $24$ times as many lists as unordered selections. Find $k$ .
1. Hint 1
  Compare the lists that belong to one unordered selection.
2. Hint 2
  Express that multiplicity as a factorial and compare it with 24.
Answer

$k=4$ .

Full solution

Each unordered selection of $k$ distinct objects appears once in each of its $k!$ orders, so the ratio of lists to selections is $k!$ .
Since $4!=24$ , and $3!=6$ and $5!=120$ miss it, $k=4$ .

Answer

$k=4$ .

Key idea

When each selected set of k distinct objects appears exactly once in every ordering, dividing the list count by k factorial counts the sets.
Problem 3: Mark complete
Problem 4 Matching ends

A six-character string contains exactly two X's, two Y's and two Z's. Its first and last characters must match. How many strings are possible?
1. Hint 1
  The matching ends use both copies of one symbol.
2. Hint 2
  Choose that symbol, then arrange the four remaining characters.
Answer

18 strings.

Full solution

The first and last characters are the two copies of one symbol, so there are $3$ choices for the ends.
The four middle places then hold two copies each of the other two symbols, in
$\frac{4!}{2!2!}=6$
distinguishable orders.
By the multiplication principle there are $3\cdot6=18$ strings.

Answer

18 strings.

Key idea

Fix required positions before dividing out the reorderings of identical objects.
Problem 4: Mark complete
Problem 5 Optional features

A device has four independent on-or-off settings labeled A, B, C, and D. It is usable if at least one of A or B is on. How many usable settings patterns are there?
1. Hint 1
  Separate all patterns into usable ones and ones where both required alternatives are off.
2. Hint 2
  Count every four-setting pattern, then subtract the patterns with A and B both off.
Answer

12 patterns.

Full solution

Every setting has two choices, so there are $2^4=16$ patterns.
An unusable pattern has A and B fixed off, while C and D remain free, giving $2^2=4$ patterns.
Hence
$16-4=12.$

Answer

12 patterns.

Key idea

Count at least one of several required alternatives by subtracting the case where all those alternatives are absent.
Problem 5: Mark complete
Problem 6 A short program

A program plays four different recordings in order. There are 3 jazz recordings and 4 classical recordings available. The opening recording must be jazz, the closing recording classical, and the two middle recordings may be from either type. How many programs are possible?
1. Hint 1
  Assign the restricted positions before the unrestricted ones.
2. Hint 2
  After choosing the opening and closing recordings, five recordings remain for two ordered middle positions.
Answer

240 programs.

Full solution

Choose the opening in $3$ ways and the closing in $4$ ways.
They come from different types, so they are distinct.
Five recordings remain for the first middle position and four for the second.
Thus
$3\cdot4\cdot5\cdot4=240.$
Every program determines exactly these four choices.

Answer

240 programs.

Key idea

Filling restricted positions first can leave a fixed count for the remaining ordered positions.
Problem 6: Mark complete
Problem 7 Photo selections

A display selects 2 different photos from 6. Each selected photo is then assigned either a thin border or a thick border. The photos have no display order. How many selections with border assignments are possible?
1. Hint 1
  First count the unordered pair, then count the border choices for that pair.
2. Hint 2
  Each particular selected photo has two border choices, even though the pair itself is unordered.
Answer

60 selections with border assignments.

Full solution

The two-photo set can be chosen in
$\binom62=15$
ways.
For each set, the two distinct photos have $2\cdot2=4$ border assignments.
Therefore
$15\cdot4=60.$
Swapping the photos in a list does not create another display, but changing a photo's border does.

Answer

60 selections with border assignments.

Key idea

An unordered selection can be followed by separate choices attached to its distinct members.
Problem 7: Mark complete
Problem 8 One card made distinct

Nine symbol cards carry R, R, R, R, A, B, C, D, E; cards carrying the same symbol are indistinguishable. One R card is replaced by an S card. A student claims this quadruples the number of distinguishable rows. Is the claim correct? Explain without evaluating $9!$ .
1. Hint 1
  Consider which positions become distinguishable after the replacement.
2. Hint 2
  Compare the repeated-symbol factorials before and after the replacement.
Answer

Yes; the number of rows becomes four times as large.

Full solution

Before the replacement there are $\frac{9!}{4!}$ distinguishable rows, and after it $\frac{9!}{3!}$ , since only three R cards remain alike.
The ratio is
$\frac{4!}{3!}=4.$
Equivalently, each original row has four R positions, and choosing which one becomes S gives four different new rows, each arising from only one original row.

Answer

Yes; the number of rows becomes four times as large.

Key idea

For unrestricted rows, replacing one of r identical symbols with a new symbol multiplies the count by r.
Problem 8: Mark complete
Problem 9 A middle label

From the labels 1 to 8, three different labels are selected, in no order. How many selections have middle label 4 or 5? Explain why the two cases can be counted separately and added.
1. Hint 1
  Once the middle label is fixed, the other two labels must lie on opposite sides of it.
2. Hint 2
  Count each case with the multiplication principle, then decide whether the two cases can overlap.
Answer

24 selections; each selection has exactly one middle label, so the two cases do not overlap.

Full solution

With middle label $4$ , one label comes from $1,2,3$ and the other from $5,6,7,8$ , in $3\cdot4=12$ ways.
With middle label $5$ , one comes from $1,2,3,4$ and the other from $6,7,8$ , in $4\cdot3=12$ ways.

A selection of three different labels has exactly one middle label, so no selection is counted in both cases, and together the cases hold every selection asked for.
The total is $12+12=24$ .

Answer

24 selections; each selection has exactly one middle label, so the two cases do not overlap.

Key idea

Cases that cannot overlap, and together cover every selection asked for, can be counted separately and added.
Problem 9: Mark complete
Problem 10 Allowed successors

A two-letter code starts with A, B, or C. After A, the next letter may be B or C. After B, it must be A. After C, it may be A or B. A student counts $3\cdot2=6$ codes. Is this correct? Give the correct count and justify it.
1. Hint 1
  Check whether each first letter allows the same number of second letters.
2. Hint 2
  Separate the codes according to their first letter.
Answer

No; 5 codes.

Full solution

The three first letters allow $2$ , $1$ , and $2$ completions.
Those counts are not all $2$ .
Adding the non-overlapping cases gives
$2+1+2=5.$
The codes are AB, AC, BA, CA, and CB, confirming the count.

Answer

No; 5 codes.

Key idea

When the number of completions varies with the first choice, sum the non-overlapping first-choice cases instead of assuming one fixed completion count.
Problem 10: Mark complete

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