Tree diagrams and the fundamental counting principle, counting arrangements with and without repeated characters, factorials and permutations of n objects, permutations of n objects taken r at a time, and permutations of objects that include repeats.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 10 — Counting Methods and Probability
Apply the Counting Principle and Permutations
Objectives
Five outcomes. One principle, and everything else in the lesson is that principle organised.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.1 Apply the Counting Principle and Permutations §10.1, pp. 682-687 — the lesson these objectives are drawn from
Warm-up
You can list outcomes and count them, and you have used exponents to record repeated multiplication.
Discussion prompt
A shop sells 3 kinds of board and 2 kinds of boot. List every possible pairing, then count them. Is there a way to know the count without listing?
Hint: How many boot choices go with each board?
Answer:
Each of the 3 boards pairs with each of the 2 boots, so there are 3 groups of 2.
\[ 3 \cdot 2 = 6 \]
That multiplication is the fundamental counting principle, and it is the only idea in this lesson. Everything else is that principle applied to slightly different questions.
Concept
If one event can happen in m ways and another in n ways, both together can happen in m times n ways. The principle extends to any number of stages, and every counting formula in the lesson comes from it.
permutation — An ordering of objects. The number of permutations of n distinct objects is n factorial, and the number of ways to order r of them chosen from n is n factorial over the quantity n minus r, factorial.
\[ m \cdot n \cdot p \cdot \;\dots \]
The one thing to decide before multiplying is whether an object may be used more than once, since that changes every count after the first stage.
Figure (svg): Two columns comparing counting with repetition against counting without it
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.1 Apply the Counting Principle and Permutations §10.1, pp. 682-685
Section
Section 1
Concept
A tree diagram draws one branch for each complete set of choices. Counting the branches gives the answer, and multiplying the number of choices at each stage gives the same answer without the drawing.
\[ 3 \cdot 2 = 6; \qquad 12 \cdot 55 \cdot 11 = 7260 \]
The principle scales where the diagram does not. Seven thousand branches cannot be drawn, but three numbers can be multiplied.
Figure (svg): The fundamental counting principle stated for two events and for more
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.1 Apply the Counting Principle and Permutations §10.1, pp. 682-682 — Fundamental Counting Principle
Picture it
Example 1: three boards, each with two boot options.
Figure (svg): A tree diagram with three board choices branching into two boot choices each
Six branches, and 3 times 2 is 6. The tree explains why the multiplication is right; the multiplication is what you actually use.
Worked example
Example 1 and Guided Practice 1.
\[ \text{Count the choices from } 3 \text{ boards and } 2 \text{ boots, then from } 3 \text{ bicycles and } 3 \text{ wheel sizes.} \]
Draw the first stage
Why: One branch for each board.
\[ 3\text{ branches} \]
Branch each of those
Why: Each board splits into 2 boot choices.
\[ 3 \times 2 \]
Count the ends
Why: Six complete paths through the tree.
\[ 6\text{ choices} \]
Repeat for the bicycles
Why: Three types, each with 3 wheel sizes.
\[ 3 \times 3 = 9 \]
Figure (svg): A tree diagram with three board choices branching into two boot choices each
\[ 3\cdot 2 = 6; \qquad 3\cdot 3 = 9 \]
Verify: check by listing the bicycle options
Why: Mountain with 20, 22 and 24 inch wheels; racing with the same three; BMX with the same three. That is 9, matching the multiplication. Listing is a good check while the numbers are small, and it is exactly what the multiplication replaces once they are not.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.1 Apply the Counting Principle and Permutations §10.1, pp. 682-683
Fill the middle
Example 2.
Fill in the blanks
12 \cdot 55 \cdot 11 = 7260
Why: Twelve times 55 is 660, and 660 times 11 is 7260. Each stage multiplies, however many stages there are.
Worked example
Example 2.
\[ \text{Count the ways to frame a picture with } 12 \text{ styles, } 55 \text{ colours and } 11 \text{ mat shades.} \]
Identify the stages
Why: Style, then colour, then mat.
Multiply the first two
Why: Twelve times 55.
\[ 660 \]
Multiply by the third
Why: Six hundred sixty times 11.
\[ 7260 \]
State the answer
Why: Every combination of the three.
\[ 7260\text{ ways} \]
Figure (svg): The solution to Worked example three stages shown as a ladder of expressions, one row per algebraic move
\[ 12 \cdot 55 \cdot 11 = 7260 \]
Verify: sanity-check the size
Why: Each of the 12 styles comes in 55 colours, which is already 660 frames, and each of those pairs with 11 mats. A tree would need 7260 end branches, which is why nobody draws one. The principle handles any number of stages with the same single multiplication.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.1 Apply the Counting Principle and Permutations §10.1, pp. 683-683
Trap
\[ 3 \text{ boards}, \; 2 \text{ boots} \]
Add the two counts
Why: The totals are combined as though the choices were alternatives.
\[ 3+2 = 5 \quad \text{(wrong)} \]
Listing gives six pairings, not five. Adding would be right for choosing ONE item from two separate lists, which is a different question.
\[ 3 \cdot 2 = 6 \]
Multiply, because each board pairs with every boot
Why: Both a board AND boots are being chosen, not one or the other.
\[ \text{6 branches on the tree} \quad \checkmark \]
The word AND signals multiplication and the word OR signals addition. That distinction runs through the whole of Chapter 10.
Sorting
Ask whether the word is AND or OR.
Sort into buckets
Sort each situation.
Reading the question for AND or OR before touching any arithmetic settles which operation is needed, and it will do so again for probabilities in Lesson 10.4.
Matching
Multiply the stages.
Match the pairs
Why: Every row is the same operation applied to a different number of stages. Notice how quickly the totals grow: adding one stage with 11 options multiplies the answer elevenfold.
Prediction
Commit before reasoning.
Predict first
Why does a tree with 3 first-stage branches and 2 second-stage branches each have exactly 6 endings?
Correct: Because each of the 3 branches splits into the same 2, giving 3 groups of 2.
\[ 2+2+2 = 3 \cdot 2 = 6 \]
Why: Multiplication is repeated addition of equal groups, and that is exactly the tree's structure: the same number of second-stage options hangs off every first-stage branch. If the boot choice depended on the board — say carving boards came only in soft boots — the groups would be unequal and the principle would not apply directly. Independence of the stages is the hidden condition.
Section
Section 2
Concept
If an object may be reused, every stage keeps its full number of choices. If not, each stage after the first has one fewer, since the objects already used are gone.
\[ 26^3 \text{ against } 26 \cdot 25 \cdot 24 \]
Without repetition always gives the smaller count, and the gap widens quickly as the number of stages grows.
Figure (svg): License plate counts computed with and without repeated characters
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.1 Apply the Counting Principle and Permutations §10.1, pp. 683-683 — Use the counting principle with repetition
Picture it
Example 3, both parts.
Figure (svg): License plate counts computed with and without repeated characters
Forty-five million with repetition and thirty-two million without. Same slots, different rules, and the second is about 70 percent of the first.
Worked example
Example 3.
\[ \text{Count plates of the form letter, digit, digit, letter, letter, letter, with and without repetition.} \]
With repetition: count each slot
Why: Twenty-six for each letter, 10 for each digit.
\[ 26 \cdot 10 \cdot 10 \cdot 26 \cdot 26 \cdot 26 \]
With repetition: multiply
Why: Twenty-six to the fourth, times 100.
\[ 45, 697, 600 \]
Without repetition: letters run down
Why: Twenty-six, then 25, 24 and 23.
\[ 26 \cdot 25 \cdot 24 \cdot 23 \]
Without repetition: digits too
Why: Ten then 9, and multiply everything.
\[ 32, 292, 000 \]
Figure (svg): License plate counts computed with and without repeated characters
\[ 45{,}697{,}600; \qquad 32{,}292{,}000 \]
Verify: compare the two
Why: The second is about 71 percent of the first, and that is entirely due to the four letters: 26 times 25 times 24 times 23 is 358,800 against 26 to the fourth, which is 456,976. The digits contribute a factor of 90 against 100. Every restricted slot shrinks the total a little more.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.1 Apply the Counting Principle and Permutations §10.1, pp. 683-683
Fill the middle
Example 3b.
Fill in the blanks
26 \cdot 10 \cdot 9 \cdot 25 \cdot 24 \cdot 23 = 32___292___000
Why: Three letters have already been used, so 23 remain for the fourth. Each slot loses exactly one option per earlier slot of the same type.
Worked example
Guided Practice 2.
\[ \text{Repeat for three letters followed by four digits.} \]
With repetition: letters
Why: Twenty-six cubed.
\[ 17, 576 \]
With repetition: digits
Why: Ten to the fourth.
\[ 10, 000 \]
With repetition: multiply
Why: Seventeen thousand five hundred seventy-six times ten thousand.
\[ 175, 760, 000 \]
Without repetition
Why: Twenty-six times 25 times 24, then 10 times 9 times 8 times 7.
\[ 78, 624, 000 \]
Figure (svg): The solution to Worked example a different plate format shown as a ladder of expressions, one row per algebraic move
\[ 175{,}760{,}000; \qquad 78{,}624{,}000 \]
Verify: compare the gap with the earlier format
Why: Here the restricted count is about 45 percent of the unrestricted one, against 71 percent before. Four digit slots without repetition costs a factor of 5040 over 10,000, which is a much heavier reduction than two slots costing 90 over 100. More slots means a bigger relative penalty.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.1 Apply the Counting Principle and Permutations §10.1, pp. 683-683
Error analysis
A student counts three-letter codes with no repeated letters.
Annotate
On: \( 26 \cdot 26 \cdot 26 = 17{,}576 \)
The reduction is small for three slots and large for many. Reading the question for the words repeated or distinct decides which count is wanted.
Sorting
Look for the words repeated, distinct or different.
Sort into buckets
Sort each situation.
The no-repetition situations are exactly the ones the permutation formula was built for, which is why it appears in the next idea rather than here.
Comparison
Fill the blanks. Same slots, different rules.
Comparison matrix
| Format | With repetition | Without |
|---|---|---|
| 3 letters | 26^3 = 17,576 | 26 * 25 * 24 = 15,600 |
| 4 digits | 10^4 = 10,000 | 10 * 9 * 8 * 7 = 5040 |
| Texas plate | 45,697,600 | 32,292,000 |
| New York plate | 175,760,000 | 78,624,000 |
The digit rows show the effect most clearly: three-letter codes lose only 11 percent, while four-digit codes lose almost half.
Prediction
Commit before reasoning.
Predict first
Forbidding repeats costs more, proportionally, in four digit slots or in four letter slots. Which?
Correct: The digits, since 10 options shrink to 7 while 26 shrink only to 23.
\[ \frac{10\cdot 9\cdot 8\cdot 7}{10^4} = 0.504; \qquad \frac{26\cdot 25\cdot 24\cdot 23}{26^4} \approx 0.785 \]
Why: Four digits fall from 10,000 to 5040, keeping about half; four letters fall from 456,976 to 358,800, keeping about 79 percent. Removing one option from a small pool matters far more than removing one from a large pool — the same reason that repeats are almost irrelevant when choosing from thousands of items and decisive when choosing from a handful.
Section
Section 3
Concept
An ordering of n objects is a permutation. Counting them by the fundamental principle gives n times n minus 1 times n minus 2 and so on down to 1, which is written n factorial.
\[ n! = n(n-1)(n-2)\cdots 3 \cdot 2 \cdot 1 \]
Zero factorial is defined to be 1, a convention that makes the formulas of the next idea work at their boundaries.
Figure (svg): The permutation formula derived by cancelling the unwanted tail of a factorial
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.1 Apply the Counting Principle and Permutations §10.1, pp. 684-684 — Permutations and factorials
Picture it
Example 4: ten bobsledding teams.
Figure (svg): The permutation formula derived by cancelling the unwanted tail of a factorial
Ten factorial is 3,628,800 orderings of the whole field, and 10 times 9 times 8 is the 720 ways to fill three medal positions.
Worked example
Example 4, both parts.
\[ \text{With } 10 \text{ teams and no ties, count the full orderings and the medal orderings.} \]
Full ordering: count each place
Why: Ten for first, 9 for second, and so on to 1.
\[ 10! \]
Full ordering: compute
Why: The product of all ten factors.
\[ 3, 628, 800 \]
Medals: only three places
Why: Ten, then 9, then 8.
\[ 10 \cdot 9 \cdot 8 \]
Medals: compute
Why: Ninety times 8.
\[ 720 \]
Figure (svg): The permutation formula derived by cancelling the unwanted tail of a factorial
\[ 10! = 3{,}628{,}800; \qquad 720 \]
Verify: check the relationship between the two
Why: The medal count times 7 factorial equals 10 factorial, since the remaining seven teams can finish in any order. Seven factorial is 5040, and 720 times 5040 is 3,628,800 — confirming both answers at once.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.1 Apply the Counting Principle and Permutations §10.1, pp. 684-684
Fill the middle
Example 4.
Fill in the blanks
10! = 10 \cdot 9 \cdot 8 \cdots 2 \cdot 1 = 3628______800
Why: Ten factorial is 3,628,800. It is worth knowing the small factorials by heart: 5 factorial is 120, 6 is 720, 7 is 5040, 8 is 40,320.
Worked example
Guided Practice 3.
\[ \text{Repeat with } 12 \text{ teams.} \]
Full ordering
Why: Twelve factorial.
\[ 479, 001, 600 \]
Medals
Why: Twelve times 11 times 10.
\[ 1320 \]
Compare with ten teams
Why: The full count grew by a factor of 132.
Note the reason
Why: Adding two teams adds two more factors to the product.
Figure (svg): The solution to Worked example twelve teams instead shown as a ladder of expressions, one row per algebraic move
\[ 12! = 479{,}001{,}600; \qquad 1320 \]
Verify: compare the growth rates
Why: The medal count went from 720 to 1320, less than doubling, while the full ordering count went from 3.6 million to 479 million — a factor of 132, which is 12 times 11. Restricting to three places keeps the count manageable; ordering everything does not.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.1 Apply the Counting Principle and Permutations §10.1, pp. 684-684
Trap
\[ 10! \]
Multiply 10 by 1
Why: The exclamation mark is read as a formatting flourish.
\[ = 10 \quad \text{(wrong)} \]
Ten factorial is the product of every whole number from 10 down to 1, which is over three and a half million.
\[ 10! = 10 \cdot 9 \cdot 8 \cdots 2 \cdot 1 = 3{,}628{,}800 \]
Multiply the whole descending chain
Why: Each factor counts the choices remaining at one stage.
\[ 5! = 120, \; 6! = 720, \; 7! = 5040 \]
Factorials grow astonishingly fast: 13 factorial already exceeds six billion. Knowing the first few by heart saves time throughout the chapter.
Ranking
Smallest first.
Put in order
Why: Each step here multiplies by a good deal more than the last: from 6 to 8 factorial multiplies by 56, and from 8 to 10 by 90. That runaway growth is why counting problems reach astronomical numbers so quickly.
Sorting
All the objects, or only some?
Sort into buckets
Sort each situation.
The second kind is what the next idea's formula computes, and it is the more common question in practice.
Prediction
Commit before reasoning.
Predict first
Zero factorial is defined to be 1 rather than 0. Why is that useful?
Correct: So that the permutation formula still works when r equals n, giving n factorial over 0 factorial.
\[ \;_{n}P_{n} = \frac{n!}{0!} = n! \;\Longrightarrow\; 0! = 1 \]
Why: Ordering all n objects should give n factorial, and the formula n factorial over n minus r factorial gives n factorial over 0 factorial when r equals n. Defining 0 factorial as 1 makes the two answers agree. Defining it as 0 would make the expression undefined instead. There is also a counting reading: there is exactly one way to arrange nothing.
Section
Section 4
Concept
The number of ways to order r objects chosen from n distinct objects is n factorial divided by the quantity n minus r, factorial. The denominator cancels exactly the factors that fall past the r positions being filled.
\[ \;_{n}P_{r} = \frac{n!}{(n-r)!} \]
The formula computes nothing the counting principle could not, but it is a single calculator key rather than a chain of multiplications.
Figure (svg): The permutation formula derived by cancelling the unwanted tail of a factorial
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.1 Apply the Counting Principle and Permutations §10.1, pp. 685-685 — Permutations of n Objects Taken r at a Time
Picture it
Example 4b rewritten as a quotient of factorials.
Figure (svg): The permutation formula derived by cancelling the unwanted tail of a factorial
Ten times 9 times 8 becomes 10 factorial over 7 factorial, and 7 is 10 minus 3. That last observation is the whole formula.
Worked example
Example 5.
\[ \text{In how many orders can } 4 \text{ of } 12 \text{ songs be burned to a CD?} \]
Identify n and r
Why: Twelve songs available, 4 positions.
\[ n = 12, r = 4 \]
Write the formula
Why: Twelve factorial over 12 minus 4, factorial.
\[ 12! / 8! \]
Compute the two factorials
Why: Four hundred seventy-nine million and forty thousand three hundred twenty.
\[ 479, 001, \frac{600}{40}, 320 \]
Divide
Why: The quotient.
\[ 11, 880 \]
Figure (svg): The permutation formula derived by cancelling the unwanted tail of a factorial
\[ \;_{12}P_{4} = \frac{12!}{8!} = 11{,}880 \]
Verify: check by the counting principle
Why: Twelve choices for the first song, 11 for the second, 10 for the third and 9 for the fourth: 12 times 11 times 10 times 9 is 11,880. The factorial formula and the direct product agree, as they must — the denominator's whole job is to cancel the eight factors below 9.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.1 Apply the Counting Principle and Permutations §10.1, pp. 685-685
Fill the middle
Example 5.
Fill in the blanks
\;_8P____ = \frac______ = \frac______!}
Why: Twelve minus 4 is 8, so the denominator is 8 factorial. It cancels exactly the eight factors from 8 down to 1, leaving 12 times 11 times 10 times 9.
Worked example
Guided Practice 4 to 7.
\[ \text{Find } \;_{5}P_{3}, \; \;_{4}P_{1}, \; \;_{8}P_{5}, \; \;_{12}P_{7}. \]
First: 5 factorial over 2 factorial
Why: One hundred twenty over 2.
\[ 60 \]
Second: 4 factorial over 3 factorial
Why: Twenty-four over 6.
\[ 4 \]
Third: 8 factorial over 3 factorial
Why: Forty thousand three hundred twenty over 6.
\[ 6720 \]
Fourth: 12 factorial over 5 factorial
Why: Four hundred seventy-nine million over 120.
\[ 3, 991, 680 \]
Figure (svg): The solution to Worked example four permutations shown as a ladder of expressions, one row per algebraic move
\[ 60, \; 4, \; 6720, \; 3{,}991{,}680 \]
Verify: check the second by reasoning
Why: Choosing and ordering just 1 object from 4 is simply 4 ways, and the formula gives 24 over 6, which is 4. Any permutation with r equal to 1 must equal n, which is a quick check that the formula has been set up correctly.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.1 Apply the Counting Principle and Permutations §10.1, pp. 685-685
Error analysis
A student evaluates a permutation.
Annotate
On: \( \;_{12}P_{4} = 12!-8! = 479{,}001{,}600-40{,}320 \)
The counting principle settles it: 12 times 11 times 10 times 9 is 11,880, nowhere near half a billion. Checking against the direct product catches this in seconds.
Matching
Divide the factorials.
Match the pairs
Why: Each is a descending product with r factors: 5 times 4 times 3; 4 alone; 8 times 7 times 6 times 5 times 4; and 12 times 11 times 10 times 9. Reading the formula that way is faster than computing two large factorials and dividing.
Comparison
Fill the blanks. Same answer, different route.
Comparison matrix
| Question | Counting principle | Formula |
|---|---|---|
| 12P4 | 12 * 11 * 10 * 9 | 12!/8! |
| 10P3 | 10 * 9 * 8 | 10!/7! |
| Number of factors | r of them | the denominator cancels the rest |
| Better for mental work | the direct product | the formula, on a calculator |
The direct product is easier by hand and the formula is easier on a calculator with a permutation key, so both are worth being fluent in.
Prediction
Commit before reasoning.
Predict first
What is the value of n objects taken n at a time?
Correct: n factorial, since every object gets a position.
\[ \;_{n}P_{n} = \frac{n!}{0!} = n!; \qquad \;_{n}P_{0} = \frac{n!}{n!} = 1 \]
Why: Taking all n objects means ordering the whole set, which is n factorial by the previous idea. The formula agrees: n factorial over 0 factorial, and 0 factorial is 1. That consistency is exactly why 0 factorial is defined as it is. At the other extreme, n taken 0 at a time gives n factorial over n factorial, which is 1 — the one way to arrange nothing.
Section
Section 5
Concept
When some objects are identical, swapping them changes nothing you can see. Counting all n factorial orderings and then dividing by a factorial for each repeated object removes those duplicates.
\[ \frac{n!}{s_1!\cdot s_2!\cdots s_k!} \]
Each repeated object contributes its own factorial to the denominator, and objects appearing only once contribute 1 factorial, which changes nothing.
Figure (svg): Two words counted for distinguishable arrangements, with the repeated letters divided out
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.1 Apply the Counting Principle and Permutations §10.1, pp. 685-686 — Permutations with Repetition
Picture it
Example 6: MIAMI and TALLAHASSEE.
Figure (svg): Two words counted for distinguishable arrangements, with the repeated letters divided out
MIAMI has two repeated letters, so 120 is divided by 2 twice, giving 30. TALLAHASSEE has four, and 39,916,800 divided by 48 gives 831,600.
Worked example
Example 6.
\[ \text{Count the distinguishable permutations of the letters in MIAMI and in TALLAHASSEE.} \]
MIAMI: count the letters and repeats
Why: Five letters, with M twice and I twice.
\[ 5! / (2! \cdot 2!) \]
MIAMI: compute
Why: One hundred twenty over 4.
\[ 30 \]
TALLAHASSEE: count
Why: Eleven letters, A three times, L, S and E twice each.
\[ 11! / (3! \cdot 2! \cdot 2! \cdot 2!) \]
TALLAHASSEE: compute
Why: Thirty-nine million over 48.
\[ 831, 600 \]
Figure (svg): Two words counted for distinguishable arrangements, with the repeated letters divided out
\[ 30; \qquad 831{,}600 \]
Verify: check the idea on a tiny case
Why: The letters E, E and Y give 3 factorial, or 6, orderings if the two E's are labelled, but only three that look different: EEY, EYE and YEE. Dividing 6 by 2 factorial gives 3, matching the list. The same reasoning scales to any number of repeats.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.1 Apply the Counting Principle and Permutations §10.1, pp. 686-686
Fill the middle
Example 6a.
Fill in the blanks
\frac4___ = \frac______} = 30
Why: Two factorial times 2 factorial is 4, and 120 over 4 is 30. Each repeated letter contributes its own factorial to the denominator.
Worked example
Guided Practice 8 to 10.
\[ \text{Count the distinguishable permutations of MALL, KAYAK and CINCINNATI.} \]
MALL: four letters, L twice
Why: Twenty-four over 2.
\[ 12 \]
KAYAK: five letters, K twice, A twice
Why: One hundred twenty over 4.
\[ 30 \]
CINCINNATI: count the letters
Why: Ten letters, with C twice, I three times, N three times.
\[ 10! / (2! \cdot 3! \cdot 3!) \]
CINCINNATI: compute
Why: Three million six hundred twenty-eight thousand eight hundred over 72.
\[ 50, 400 \]
Figure (svg): The solution to Worked example three more words shown as a ladder of expressions, one row per algebraic move
\[ 12, \; 30, \; 50{,}400 \]
Verify: check the letter counts in the last word
Why: C, I, N, C, I, N, N, A, T, I: that is C twice, I three times, N three times, and A and T once each — ten letters in total, which matches. Miscounting the repeats is the main hazard here, so recounting the letters before dividing is time well spent.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.1 Apply the Counting Principle and Permutations §10.1, pp. 686-686
Trap
\[ \text{TALLAHASSEE: } A \text{ appears } 3 \text{ times} \]
Divide by 3
Why: The count of repeats is used directly.
\[ \frac{11!}{3 \cdot 2 \cdot 2 \cdot 2} = \frac{39{,}916{,}800}{24} \quad \text{(wrong)} \]
The three A's can be arranged among their own positions in 3 factorial, or 6, ways — not 3.
\[ \frac{11!}{3!\cdot 2!\cdot 2!\cdot 2!} = \frac{39{,}916{,}800}{48} = 831{,}600 \]
Divide by the factorial of each repeat count
Why: Each group of identical letters is internally rearrangeable in that many ways.
\[ 3! = 6, \text{ not } 3 \]
For letters appearing twice the two agree, since 2 factorial is 2 — which is exactly why the error hides until a letter appears three or more times.
Matching
Count the letters, then the repeats.
Match the pairs
Why: MIAMI and KAYAK give the same answer because both have five letters with two pairs of repeats, even though the letters themselves differ. Only the pattern of repetition matters, never which letters are involved.
Sorting
One for each repeated object.
Sort into buckets
Sort each word by how many factorials go underneath.
Letters appearing once contribute 1 factorial, which is 1, so they can be ignored entirely — but they still count toward the n on top.
Prediction
Commit before reasoning.
Predict first
Why does removing duplicate arrangements mean dividing rather than subtracting?
Correct: Because the arrangements come in equal-sized groups, and each group counts as one.
\[ 6 \text{ labelled orderings} \div 2 \text{ per group} = 3 \text{ distinguishable} \]
Why: With E, E and Y the six labelled orderings fall into three pairs, each pair looking identical, so the visible count is 6 divided by 2. Subtracting would assume a fixed number of duplicates rather than a fixed group size, and the group size is what stays constant. This is the same reason that counting objects arranged in equal rows uses division rather than subtraction.
Comparison
Fill the blanks. All four come from one principle.
Comparison matrix
| Question | Formula | Example |
|---|---|---|
| Choices at each stage, repeats allowed | multiply the counts | 26 * 10 * 10 * 26 * 26 * 26 |
| Order all n objects | n factorial | 10! = 3,628,800 |
| Order r of n objects | n!/(n - r)! | 12P4 = 11,880 |
| Order n objects with repeats | n! divided by a factorial per repeat | MIAMI: 5!/(2!2!) = 30 |
Every row is the fundamental counting principle with a different bookkeeping rule about what counts as the same outcome.
Pattern
Three questions decide which formula applies.
Order matters in everything counted here. Lesson 10.2 handles the questions where it does not.
OpenStax Algebra and Trigonometry 2e, §13.5 Counting Principles §13.5
Check
Repetition changes every stage after the first.
Check your understanding
How many plates of the form letter, digit, digit, letter, letter, letter allow repeats?
Answer: A
Why: Each letter has 26 choices and each digit 10, giving 26^4 times 100.
Check
Permutations. Divide, do not subtract.
Check your understanding
Evaluate 12P4.
Answer: A
Why: It is 12 times 11 times 10 times 9, or equivalently 12! divided by 8!.
Check
Repeated letters. Factorials underneath.
Check your understanding
How many distinguishable arrangements do the letters of MIAMI have?
Answer: A
Why: There are 5! orderings, divided by 2! for the M's and 2! for the I's.
Real world
A password must be exactly 8 characters long, chosen from 26 lowercase letters, 26 uppercase, 10 digits and 10 symbols, with repeats allowed.
Discussion prompt
Count the possible passwords, then count them if only lowercase letters are allowed, and explain what the comparison means for security.
Hint: Each character is one stage with the full pool available.
Answer:
\[ \text{full pool: } 72^8 \approx 7.2 \times 10^{14} \]
\[ \text{lowercase only: } 26^8 \approx 2.1 \times 10^{11} \]
The full pool gives about 720 trillion passwords and lowercase alone about 209 billion — a factor of roughly 3400 between them, from the same eight characters.
A machine testing a billion passwords a second would exhaust the lowercase space in about three and a half minutes and the full space in about eight days. Neither is comfortable, which is why length matters more than variety: adding two characters to the lowercase password multiplies its count by 676, more than the entire gain from adding uppercase, digits and symbols. The counting principle is what makes that trade-off calculable, and it is the reason security advice shifted from complexity rules to longer passphrases.
Commit first
Answer, then rate your confidence honestly.
Predict first
With 3 boards and 2 boots available, how many equipment choices are there?
Correct: 6, by multiplying, since each board pairs with each boot.
\[ 3 \text{ boards} \times 2 \text{ boots} = 6 \text{ complete outfits} \]
Why: A tree makes it visible: each of the three boards branches into two boot options, giving three groups of two. Adding would answer a different question — how many single items are available to pick one of — and gives 5. The distinction is whether the choices are made together, signalled by AND, or as alternatives, signalled by OR. Getting this wrong at the first step makes every later formula in the chapter give the wrong answer, since they are all built on this multiplication.
Explain it
They can list outcomes but always want to add the counts.
Discussion prompt
In four sentences or fewer, explain why choices multiply rather than add.
Hint: Describe a menu.
Answer:
Imagine a menu with 3 main courses and 2 desserts. For every main course you pick, you still have both desserts to choose from.
So there are 3 groups of 2 meals, which is 6 — and multiplying is just a fast way of counting equal groups. Adding would answer a different question: how many single dishes are on the menu, which is 5.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For multiply or add, look for AND against OR in the wording. For repetition, look for the words distinct, different or repeated. For the permutation formula, write the descending product instead and count that it has r factors. For repeats, tally the letters in a column before dividing anything.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a counting page. Top left: draw the snowboard tree diagram in full, count the branches, and write the multiplication beside it. Top right: work the license plate problem both ways in parallel columns, and write one sentence on why the second is smaller. Middle: derive the permutation formula from 10 times 9 times 8, showing every line of the cancellation, and use it on two examples of your own. Bottom left: write the first eight factorials in a column and note how much each step multiplies by. Bottom right: count the distinguishable arrangements of three words with repeats, showing the letter tally for each, and write one sentence explaining why you divide by a factorial rather than by the repeat count.
If any of your denominators uses the number of repeats rather than its factorial, redo it: three identical letters contribute 6, not 3.
Recap
Five things, all built on one multiplication.
| If you see | Then |
|---|---|
| Choices made together, AND | Multiply the counts |
| Alternatives, OR | Add the counts |
| Repeats allowed | Every stage keeps its full count |
| Distinct objects, all used | n factorial |
| Distinct objects, r positions | n! over (n - r)! |
| Identical objects | Divide by one factorial per repeated group |
Lesson 10.2 asks the same questions with order removed, giving combinations and the binomial theorem.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.1 Apply the Counting Principle and Permutations §10.1, pp. 682-687 — everything on these slides traces back here
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