Combinations and how they differ from permutations, deciding whether to multiply or add combinations, using subtraction from a total for at-least problems, building and reading Pascal's triangle, and using the binomial theorem to expand a power or to find a single coefficient.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 10 — Counting Methods and Probability
Use Combinations and the Binomial Theorem
Objectives
Five outcomes. Remove the ordering, and permutations become combinations.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.2 Use Combinations and the Binomial Theorem §10.2, pp. 690-695 — the lesson these objectives are drawn from
Warm-up
Lesson 10.1 counted orderings: 5P3 is 60 ways to arrange 3 objects chosen from 5.
Discussion prompt
From the letters A, B, C, D, E, how many three-letter arrangements are there? And how many three-letter GROUPS, if ABC and ACB count as the same group?
Hint: How many arrangements does each group produce?
Answer:
\[ \;_5P_3 = 60 \text{ arrangements} \]
Each group of three letters can be arranged in 3 factorial, or 6, ways, so the 60 arrangements fall into groups of 6.
\[ \frac{60}{6} = 10 \text{ groups} \]
That division is the whole of this lesson's first idea. A combination is a permutation with the ordering divided out.
Concept
A combination is a selection of r objects from n where order does not matter. Its count is the permutation count divided by r factorial, since each selected group can be ordered in r factorial ways that all count as one.
combination — A selection of r objects from a group of n objects in which the order is not important. The count is n factorial divided by the product of n minus r factorial and r factorial.
\[ \;_nC_r = \frac{n!}{(n-r)!\,r!} \]
Deciding between the two is a single question: would rearranging the chosen objects give a different outcome? Medals yes, a card hand no.
Figure (svg): Two columns comparing permutations with combinations
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.2 Use Combinations and the Binomial Theorem §10.2, pp. 690-690
Section
Section 1
Concept
The number of ways to choose r objects from n distinct objects, without regard to order, is n factorial divided by the product of n minus r factorial and r factorial. The extra factorial removes the orderings of the chosen group.
\[ \;_nC_r = \frac{\;_nP_r}{r!} \]
A card hand, a committee and a pizza's toppings are combinations. Medals, seating and passwords are permutations.
Figure (svg): The combination formula beside the permutation formula, with the extra division explained
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.2 Use Combinations and the Binomial Theorem §10.2, pp. 690-690 — Combinations of n Objects Taken r at a Time
Picture it
The two formulas side by side.
Figure (svg): The combination formula beside the permutation formula, with the extra division explained
The combination count is always the smaller, by exactly a factor of r factorial. For r equal to 5 that factor is 120.
Worked example
Example 1, both parts.
\[ \text{From a } 52 \text{-card deck, count } 5 \text{-card hands, and hands with all } 5 \text{ cards one colour.} \]
Any five cards
Why: Fifty-two factorial over 47 factorial times 5 factorial.
\[ 52 C 5 \]
Compute
Why: Fifty-two times 51 times 50 times 49 times 48, over 120.
\[ 2, 598, 960 \]
One colour: choose the colour
Why: Two colours, choose 1.
\[ 2 C 1 = 2 \]
One colour: choose the cards, then multiply
Why: Five of the 26 cards in that colour.
\[ 2 \cdot 65, 780 = 131, 560 \]
Figure (svg): Two card-hand counts, one for any five cards and one for five of a single colour
\[ 2{,}598{,}960; \qquad 131{,}560 \]
Verify: check the second against the first
Why: One hundred thirty-one thousand five hundred sixty is about 5 percent of 2,598,960, which is plausible: about half the cards are one colour, and getting five in a row from that half is roughly one half to the fifth, or 3 percent, doubled for the two colours. Rough estimates like this catch an answer that is out by a factor of ten.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.2 Use Combinations and the Binomial Theorem §10.2, pp. 690-690
Sorting
Would rearranging give a different outcome?
Sort into buckets
Sort each situation.
The test is always the same sentence: does swapping two of the chosen objects give a different answer? If not, divide by r factorial.
Worked example
Guided Practice 1 to 4.
\[ \text{Find } \;_8C_3, \; \;_{10}C_6, \; \;_7C_2, \; \;_{14}C_5. \]
First: 8 factorial over 5 factorial times 3 factorial
Why: Three hundred thirty-six over 6.
\[ 56 \]
Second: 10 factorial over 4 factorial times 6 factorial
Why: Five thousand forty over 24.
\[ 210 \]
Third: 7 times 6 over 2
Why: Forty-two over 2.
\[ 21 \]
Fourth: 14 times 13 times 12 times 11 times 10 over 120
Why: Two hundred forty thousand two hundred forty over 120.
\[ 2002 \]
Figure (svg): The solution to Worked example four combinations shown as a ladder of expressions, one row per algebraic move
\[ 56, \; 210, \; 21, \; 2002 \]
Verify: check the second by symmetry
Why: Ten choose 6 should equal 10 choose 4, since choosing which 6 to take is the same as choosing which 4 to leave. Ten choose 4 is 210 as well. That symmetry, n choose r equals n choose n minus r, holds always and often makes the arithmetic shorter.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.2 Use Combinations and the Binomial Theorem §10.2, pp. 691-691
Trap
\[ \;_6C_2 = \frac{6!}{(6-2)!} = \frac{720}{24} = 30 \]
Use the permutation formula
Why: The extra factorial in the denominator is omitted.
\[ 30 \quad \text{(wrong)} \]
Thirty is 6P2, the number of ORDERED pairs. Each unordered pair was counted twice, once in each order.
\[ \;_6C_2 = \frac{6!}{4!\,2!} = \frac{720}{48} = 15 \]
Divide by r factorial as well
Why: Each chosen group of r has r factorial orderings, all counting as one.
\[ \frac{30}{2!} = 15 \quad \checkmark \]
This is lesson exercise 11's printed error. The combination count is always the smaller of the two, which is a quick sanity check.
Fill the middle
The relationship between the two counts.
Fill in the blanks
\;_5C_3 = \frac6___ = \frac______} = 10
Why: Three factorial is 6, and 60 over 6 is 10. Every group of three can be ordered in six ways, all of which count as the same combination.
Matching
Divide by both factorials.
Match the pairs
Why: Each is a descending product of r factors divided by r factorial: 8 times 7 times 6 over 6, and so on. Computing it that way avoids two very large factorials.
Prediction
Commit before reasoning.
Predict first
For the same n and r with r greater than 1, which is larger, nPr or nCr?
Correct: nPr, by a factor of r factorial.
\[ \;_nP_r = r! \cdot \;_nC_r \]
Why: The combination formula is the permutation formula with an extra r factorial in the denominator, so the permutation count is r factorial times larger. For r equal to 5 that is a factor of 120, and for r equal to 10 it is over three and a half million. The two agree only when r is 0 or 1, since 0 factorial and 1 factorial are both 1.
Section
Section 2
Concept
When a selection must satisfy two conditions at once, count each and multiply. When a selection may satisfy one condition or another, count each and add.
\[ \;_{18}C_2 \cdot \;_{10}C_1 = 1530 \]
This is Lesson 10.1's rule applied to combinations rather than to single choices, and it is the same distinction that will govern probabilities in Lesson 10.4.
Figure (svg): Two card-hand counts, one for any five cards and one for five of a single colour
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.2 Use Combinations and the Binomial Theorem §10.2, pp. 691-691 — Decide to multiply or add combinations
Picture it
Example 1b: two stages multiplied together.
Figure (svg): Two card-hand counts, one for any five cards and one for five of a single colour
Choosing which colour is one combination and choosing the five cards is another, and the two happen together, so they multiply.
Worked example
Example 2a and Guided Practice 5.
\[ \text{From } 18 \text{ comedies, } 10 \text{ histories and } 10 \text{ tragedies, count sets of } 2 \text{ comedies and } 1 \text{ tragedy; then } 3 \text{ tragedies and } 2 \text{ histories.} \]
First: choose the comedies
Why: Two of 18.
\[ 18 C 2 = 153 \]
First: choose the tragedy and multiply
Why: One of 10, then multiply.
\[ 153 \cdot 10 = 1530 \]
Second: choose the tragedies
Why: Three of 10.
\[ 10 C 3 = 120 \]
Second: choose the histories and multiply
Why: Two of 10, then multiply.
\[ 120 \cdot 45 = 5400 \]
Figure (svg): The solution to Worked example multiply two combinations shown as a ladder of expressions, one row per algebraic move
\[ 1530; \qquad 5400 \]
Verify: check why multiplication is right
Why: Every one of the 153 comedy pairs can be joined with any of the 10 tragedies, giving 153 groups of 10. The two choices are made together and neither restricts the other, which is exactly the condition the counting principle needs.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.2 Use Combinations and the Binomial Theorem §10.2, pp. 691-691
Sorting
Read for AND against OR.
Sort into buckets
Sort each counting task.
The fourth is worth noticing: five hearts and five diamonds cannot both happen in a five-card hand, so the cases really are exclusive and adding is right.
Worked example
Example 2b.
\[ \text{From } 38 \text{ plays, count the sets of AT MOST } 3 \text{ plays.} \]
List the cases
Why: Zero, 1, 2 or 3 plays.
Count each
Why: One, 38, 703 and 8436.
Add them
Why: The cases are alternatives, not simultaneous.
\[ 9178 \]
Note the zero case
Why: Reading no plays at all is one of the possibilities.
\[ 38 C 0 = 1 \]
Figure (svg): An at-least count computed by subtracting the unwanted cases from the total
\[ 1+38+703+8436 = 9178 \]
Verify: check the zero case is included
Why: The book prints a caution about exactly this: at most 3 includes reading none, and 38 choose 0 is 1. Leaving it out gives 9177, wrong by one — a small error, but the same omission in a probability calculation later would be a real one.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.2 Use Combinations and the Binomial Theorem §10.2, pp. 691-691
Error analysis
A student counts sets of exactly 2 comedies and 1 tragedy.
Annotate
On: \( \;_{18}C_2+\;_{10}C_1 = 153+10 = 163 \)
Adding would answer a different question: how many sets consist of either two comedies or one tragedy, which is not what was asked.
Fill the middle
Example 2a.
Fill in the blanks
\;_10C_2 \cdot \;____C_1 = 153 \cdot ___ = 1530
Why: Ten choose 1 is 10, so each of the 153 comedy pairs joins with any of 10 tragedies. Choosing one object from n is always n ways.
Matching
Decide the operation first.
Match the pairs
Why: Three of the four multiply and one adds. The one that adds is the one whose cases are alternatives — you read either none, or one, or two, or three plays, never several of those at once.
Prediction
Commit before reasoning.
Predict first
Counting sets of at most 3 plays, why is 38 choose 0 included?
Correct: Because reading no plays satisfies at most 3, and there is exactly one way to choose nothing.
\[ \;_nC_0 = \frac{n!}{n!\,0!} = 1 \]
Why: At most 3 means 3 or fewer, and zero is fewer than 3. There is exactly one way to select nothing — take no plays — which is why n choose 0 is 1 for every n rather than 0. The empty selection is genuinely one of the possibilities, and omitting it is the error the book warns about beside this example.
Section
Section 3
Concept
For an at-least problem with many cases, count every possibility and subtract the few unwanted ones. Each object is either taken or not, so n objects give 2 to the n total selections.
\[ 2^{12}-(\;_{12}C_0+\;_{12}C_1+\;_{12}C_2) = 4017 \]
The total of 2 to the n is itself the sum of every combination from 0 to n, so this is not a new idea but a shortcut through a long sum.
Figure (svg): An at-least count computed by subtracting the unwanted cases from the total
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.2 Use Combinations and the Binomial Theorem §10.2, pp. 691-691 — Subtracting possibilities
Picture it
Example 3: attending at least 3 of 12 games.
Figure (svg): An at-least count computed by subtracting the unwanted cases from the total
Adding directly needs ten combinations; subtracting needs three and one power of 2. The answer is 4017 either way.
Worked example
Example 3.
\[ \text{From } 12 \text{ home games, count the ways to attend at least } 3. \]
Count all selections
Why: Each game is attended or not.
\[ 2 ^{12} = 4096 \]
List the unwanted cases
Why: Attending 0, 1 or 2 games.
\[ 12 C 0, 12 C 1, 12 C 2 \]
Compute them
Why: One, 12 and 66.
\[ \sum 79 \]
Subtract
Why: Four thousand ninety-six minus 79.
\[ 4017 \]
Figure (svg): An at-least count computed by subtracting the unwanted cases from the total
\[ 4096-79 = 4017 \]
Verify: explain why 2 to the 12 is the total
Why: For each of the 12 games there are two choices, attend or not, so the counting principle gives 2 multiplied by itself 12 times. That total is also the sum of 12 choose r for every r from 0 to 12, which is why subtracting three of those terms leaves the other ten.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.2 Use Combinations and the Binomial Theorem §10.2, pp. 691-691
Fill the middle
Example 3.
Fill in the blanks
\text4096 2^___ = ___
Why: Two to the twelfth is 4096, the number of subsets of a 12-element set. Every possible attendance pattern is one of them.
Worked example
Lesson exercise 18.
\[ \text{Count } 5 \text{-card hands containing at least } 1 \text{ spade.} \]
Count all hands
Why: Five cards from 52.
\[ 52 C 5 = 2, 598, 960 \]
Count the unwanted case
Why: No spades at all means 5 from the 39 non-spades.
\[ 39 C 5 = 575, 757 \]
Subtract
Why: Two million five hundred ninety-eight thousand nine hundred sixty minus 575,757.
\[ 2, 023, 203 \]
Note the saving
Why: Adding directly would need five separate cases.
Figure (svg): The solution to Worked example at least one spade shown as a ladder of expressions, one row per algebraic move
\[ 2{,}598{,}960-575{,}757 = 2{,}023{,}203 \]
Verify: sanity-check the proportion
Why: About 78 percent of all hands contain at least one spade, which is plausible: the chance of avoiding spades on five draws from a deck three quarters non-spade is roughly three quarters to the fifth, about 24 percent. At least one is the classic case where subtracting from the total is far shorter than adding the wanted cases.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.2 Use Combinations and the Binomial Theorem §10.2, pp. 694-694
Trap
\[ \text{at least } 3 \text{ of } 12 \]
Subtract the cases up to and including 3
Why: The boundary is included among the unwanted.
\[ 4096-(1+12+66+220) = 3797 \quad \text{(wrong)} \]
At least 3 INCLUDES attending exactly 3, so the 220 ways of doing that must stay in the count.
\[ 4096-(1+12+66) = 4017 \]
Subtract only the cases strictly below the boundary
Why: At least 3 excludes 0, 1 and 2 and nothing else.
\[ \text{exactly } 3 \text{ is wanted, so keep } \;_{12}C_3 \]
At least means the boundary counts; more than means it does not. Reading the phrase carefully is worth more here than any formula.
Sorting
Read the boundary carefully.
Sort into buckets
For at least 3 of 12 games, sort each case.
Only three of the thirteen cases are unwanted, which is precisely why subtracting is shorter than adding here.
Comparison
Fill the blanks. Count whichever run is shorter.
Comparison matrix
| Phrase | Direct sum | Better route |
|---|---|---|
| At most 3 of 38 | 4 terms | add them directly |
| At least 3 of 12 | 10 terms | subtract 3 terms from 2^12 |
| At least 1 spade | 5 terms | subtract the no-spade case |
| Exactly 2 comedies | 1 term | compute it directly |
The choice is purely practical: both routes give the same answer, and the shorter one is whichever side of the boundary has fewer cases.
Prediction
Commit before reasoning.
Predict first
Why does the sum of n choose r, over every r from 0 to n, equal 2 to the n?
Correct: Because each object is independently either taken or left, giving two choices n times.
\[ \;_nC_0+\;_nC_1+\dots+\;_nC_n = 2^n \]
Why: Counting subsets one way gives 2 to the n by the counting principle; counting them by size gives the sum of all the combinations. Both count the same thing, so they are equal. This identity is also visible in Pascal's triangle, where each row sums to a power of 2 — row 4 sums to 16, row 5 to 32.
Section
Section 4
Concept
Writing n choose r in a triangle, one row per value of n, gives Pascal's triangle. The first and last entries of every row are 1, and every other entry is the sum of the two nearest entries in the row above.
\[ \;_nC_r = \;_{n-1}C_{r-1}+\;_{n-1}C_r \]
Reading a value takes no arithmetic at all, provided the rows above have been written — which makes the triangle useful when several combinations from the same n are needed.
Figure (svg): Pascal's triangle to the sixth row, with each entry the sum of the two above it
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.2 Use Combinations and the Binomial Theorem §10.2, pp. 692-692 — Pascal's Triangle
Picture it
Rows 0 through 6, each built from the one above.
Figure (svg): Pascal's triangle to the sixth row, with each entry the sum of the two above it
Row 6 reads 1, 6, 15, 20, 15, 6, 1. The entry for 6 choose 2 is the third one, since the row starts at 6 choose 0.
Worked example
Example 4 and Guided Practice 6.
\[ \text{Use Pascal's triangle to find } \;_6C_2 \text{ and } \;_7C_2. \]
Write row 5
Why: One, 5, 10, 10, 5, 1.
Build row 6 by adding pairs
Why: One, then 1 plus 5, then 5 plus 10, and so on.
\[ 1, 6, 15, 20, 15, 6, 1 \]
Read the third entry
Why: The row starts at 6 choose 0, so 6 choose 2 is third.
\[ 15 \]
Build row 7 the same way
Why: One, 7, 21, 35, 35, 21, 7, 1.
\[ 7\text{ choose } 2\text{ is } 21 \]
Figure (svg): Pascal's triangle to the sixth row, with each entry the sum of the two above it
\[ \;_6C_2 = 15; \qquad \;_7C_2 = 21 \]
Verify: check against the formula
Why: Six times 5 over 2 is 15, and 7 times 6 over 2 is 21. Both match. The triangle is faster when several entries of one row are wanted and slower when a single entry from a large row is wanted, since every row above must be built first.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.2 Use Combinations and the Binomial Theorem §10.2, pp. 692-692
Fill the middle
Example 4.
Fill in the blanks
\text15 5: \; 1, 5, 10, 10, 5, 1 \;\Longrightarrow\; \text___ 6: \; 1, 6, ___, 20, 15, 6, 1
Why: Five plus 10 is 15, the third entry of row 6. Each entry is the sum of the two nearest above it, which is why the triangle can be extended with only addition.
Worked example
Reading structure out of the rows.
\[ \text{Check that row } 4 \text{ is symmetric and that its entries sum to } 2^4. \]
Write row 4
Why: One, 4, 6, 4, 1.
Check the symmetry
Why: Reading forward and backward gives the same list.
Explain the symmetry
Why: Choosing r to take is choosing n minus r to leave.
\[ n C r = n C(n - r) \]
Add the row
Why: One plus 4 plus 6 plus 4 plus 1.
\[ 16 = 2 ^{4} \]
Figure (svg): The solution to Worked example two properties of the triangle shown as a ladder of expressions, one row per algebraic move
\[ 1,4,6,4,1: \; \text{symmetric}, \; \text{sum } 16 \]
Verify: check both properties on row 6
Why: Row 6 is 1, 6, 15, 20, 15, 6, 1 — symmetric, and summing to 64, which is 2 to the sixth. Both properties hold for every row, and the sum property is the same identity that made the at-least shortcut work in the previous idea.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.2 Use Combinations and the Binomial Theorem §10.2, pp. 692-692
Error analysis
A student reads 6 choose 2 from row 6 of Pascal's triangle.
Annotate
On: \( \text{row } 6: \; 1, 6, 15, 20, 15, 6, 1; \quad \;_6C_2 = 6 \)
Labelling the entries 6 choose 0 through 6 choose 6 underneath before reading removes the off-by-one entirely.
Matching
Each row is the combinations for that n.
Match the pairs
Why: Row n has n plus 1 entries, starting and ending at 1. The second entry is always n itself, since there are n ways to choose a single object.
Prediction
Commit before reasoning.
Predict first
Why is 6 choose 2 equal to 5 choose 1 plus 5 choose 2?
Correct: Because a group of 2 from 6 either contains a chosen object or does not, splitting the count in two.
\[ \;_6C_2 = \;_5C_1+\;_5C_2 = 5+10 = 15 \]
Why: Single out one of the six objects. Groups containing it need 1 more from the remaining 5, giving 5 choose 1; groups avoiding it need 2 from the remaining 5, giving 5 choose 2. Every group falls into exactly one case, so the counts add. That argument works for any n and r, which is why the triangle can be built entirely by addition.
Sorting
Properties of Pascal's triangle.
Sort into buckets
Sort each claim.
The four true properties are all readings of combination facts: n choose 0 is 1, the symmetry of choosing versus leaving, the subset count, and the fact that middling group sizes are the most numerous.
Section
Section 5
Concept
Expanding a binomial to the n gives terms of the form n choose r times a to the n minus r times b to the r, with r running from 0 to n. The coefficients are exactly row n of Pascal's triangle.
\[ (a+b)^n = \sum \;_nC_r\,a^{n-r}b^r \]
For a difference, write it as a sum with a negative second term. The negative then alternates the signs automatically.
Figure (svg): A binomial expansion with its coefficients taken from a row of Pascal's triangle
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.2 Use Combinations and the Binomial Theorem §10.2, pp. 693-693 — Binomial Theorem
Picture it
The expansion of a binomial to the fourth power.
Figure (svg): A binomial expansion with its coefficients taken from a row of Pascal's triangle
The exponents on the first letter count down from 4 and those on the second count up from 0, always adding to 4.
Worked example
Examples 5 and 6.
\[ \text{Expand } (x^2+y)^3 \text{ and } (a-2b)^4. \]
First: coefficients from row 3
Why: One, 3, 3, 1.
First: powers of x squared count down
Why: Six, then 4, then 2, then 0.
\[ x ^{6} + 3 x ^{4} y + 3 x ^{2} y ^{2} + y ^{3} \]
Second: write the difference as a sum
Why: A plus negative 2b, with coefficients 1, 4, 6, 4, 1.
\[ [a + (-2 b)] ^{4} \]
Second: expand the powers of negative 2b
Why: One, negative 2b, 4b squared, negative 8b cubed, 16b to the fourth.
Figure (svg): A binomial expansion with its coefficients taken from a row of Pascal's triangle
\[ x^6+3x^4y+3x^2y^2+y^3; \quad a^4-8a^3b+24a^2b^2-32ab^3+16b^4 \]
Verify: check the total degree of every term
Why: In the first expansion every term has degree 6 in x and y together when x squared counts as 2: x to the sixth, x to the fourth times y, and so on. In the second every term has degree 4. A term whose degrees do not add to n signals a slipped exponent, and the check takes seconds.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.2 Use Combinations and the Binomial Theorem §10.2, pp. 693-693
Fill the middle
Example 7.
Fill in the blanks
(3x+2)^6: \; \text___ x^4 \text___ 10-r = 4, \text___ r = ___
Why: Ten minus 4 is 6, so the wanted term is the one with r equal to 6. Solving for r first turns a full expansion into a single term's worth of work.
Worked example
Example 7 and Guided Practice 11 and 12.
\[ \text{Find the coefficient of } x^4 \text{ in } (3x+2)^{10}, \text{ of } x^5 \text{ in } (x-3)^7, \text{ and of } x^3 \text{ in } (2x+5)^8. \]
First: find which r gives x to the fourth
Why: Ten minus r equals 4.
\[ r = 6 \]
First: compute the term
Why: Two hundred ten times 81x to the fourth times 64.
\[ 1, 088, 640 \]
Second: 7 minus r equals 5
Why: R is 2, so 7 choose 2 times negative 3 squared.
\[ 21 \cdot 9 = 189 \]
Third: 8 minus r equals 3
Why: R is 5, so 8 choose 5 times 2 cubed times 5 to the fifth.
\[ 56 \cdot 8 \cdot 3125 = 1, 400, 000 \]
Figure (svg): The solution to Worked example find a single coefficient shown as a ladder of expressions, one row per algebraic move
\[ 1{,}088{,}640; \; 189; \; 1{,}400{,}000 \]
Verify: check the sign of the second
Why: The term is 7 choose 2 times x to the fifth times negative 3 squared, and a negative squared is positive — so the coefficient is positive 189. Had the exponent on the negative been odd, the coefficient would have been negative. Tracking the parity of r is what gets the signs right.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.2 Use Combinations and the Binomial Theorem §10.2, pp. 694-694
Trap
\[ (3x+2)^{10}, \; \text{coefficient of } x^4 \]
Use 10 choose 6 times 2 to the sixth
Why: The 3 in front of x is left behind.
\[ 210 \cdot 64 = 13{,}440 \quad \text{(wrong)} \]
The term is 3x raised to the fourth, which is 81 times x to the fourth — the 3 is raised too.
\[ \;_{10}C_6(3x)^4(2)^6 = 210 \cdot 81x^4 \cdot 64 \]
Raise the entire quantity, coefficient included
Why: The binomial's terms are 3x and 2, not x and 2.
\[ = 1{,}088{,}640x^4 \]
The same care applies to negative 2b in Example 6: the whole thing is raised, so 4 squared appears as well as the sign.
Ranking
Finding one coefficient in an expansion.
Put in order
Why: Step three is what makes this efficient: one value of r is found and only that term is computed, rather than all n plus 1 of them. For a tenth power that is a saving of ten terms out of eleven.
Comparison
Fill the blanks. One sign changes everything.
Comparison matrix
| Question | (a + b)^4 | (a - 2b)^4 |
|---|---|---|
| Coefficients from row 4 | 1, 4, 6, 4, 1 | 1, 4, 6, 4, 1 |
| Second term | 4a^3 b | -8a^3 b |
| Why it differs | the whole -2b is raised to the power | giving a factor of -2 |
| Signs | all positive | alternating |
The combinations are identical in both columns. Everything else comes from raising negative 2b rather than b, which contributes both a sign and a power of 2.
Prediction
Commit before reasoning.
Predict first
How many terms does the expansion of a binomial to the tenth power have before simplifying?
Correct: Eleven, since r runs from 0 to 10.
\[ r = 0, 1, \dots, n \;\Longrightarrow\; n+1 \text{ terms} \]
Why: The index r takes every whole value from 0 to n inclusive, giving n plus 1 terms — matching row n of Pascal's triangle, which has n plus 1 entries. The off-by-one is the same one that made 6 choose 2 the third entry rather than the second, and it comes from counting starting at zero rather than one.
Comparison
Fill the blanks. Order, then repetition, then structure.
Comparison matrix
| Question | Formula | Example |
|---|---|---|
| Order r of n, order matters | n!/(n - r)! | medals among teams |
| Choose r of n, order does not | n!/((n - r)! r!) | a 5-card hand |
| All subsets of n objects | 2^n | attend any set of 12 games |
| Coefficients of (a + b)^n | row n of Pascal's triangle | 1, 4, 6, 4, 1 for n = 4 |
The last two rows are the same numbers seen twice: the row of the triangle sums to the subset count, because both count all the ways to choose any number of objects.
Pattern
One question decides the formula, and one more decides the arrangement.
At least includes the boundary; more than does not. The same distinction applies to at most and fewer than.
OpenStax Algebra and Trigonometry 2e, §13.6 Binomial Theorem §13.6
Check
Combinations. Divide by both factorials.
Check your understanding
Evaluate 6C2.
Answer: A
Why: It is 6! divided by 4! times 2!, or 30 divided by 2.
Check
At least. Subtract the short run.
Check your understanding
In how many ways can you attend at least 3 of 12 games?
Answer: A
Why: Subtract the 0-, 1- and 2-game cases from the 2^12 total.
Check
Binomial theorem. Raise the whole term.
Check your understanding
What is the coefficient of x^4 in the expansion of (3x + 2)^10?
Answer: A
Why: It is 10C6 times 3^4 times 2^6, or 210 times 81 times 64.
Real world
A lottery asks players to choose 6 numbers from 49, with order irrelevant. A ticket costs one unit and there is a single jackpot.
Discussion prompt
Count the possible tickets, and say what the count implies about buying more of them.
Hint: Order does not matter, so this is a combination.
Answer:
\[ \;_{49}C_6 = \frac{49!}{43!\,6!} = 13{,}983{,}816 \]
There are about 14 million possible tickets, so a single ticket wins the jackpot roughly once in fourteen million draws.
Buying a hundred tickets raises the chance to about one in 140,000 — better, but still far smaller than most everyday risks. The arithmetic also explains why syndicates exist and why, very occasionally, a group has bought every combination: at fourteen million units, that is worth doing only when the jackpot has rolled over past that figure, and only if the prize is not shared. Notice how much order mattering would change things: 49P6 is over ten billion, more than seven hundred times larger, which is why the rules say the order of the balls is irrelevant.
Commit first
Answer, then rate your confidence honestly.
Predict first
Is the number of 5-card hands from a 52-card deck equal to 52P5?
Correct: No — a hand is unordered, so it is 52C5, smaller by a factor of 120.
\[ \frac{\;_{52}P_5}{5!} = \frac{311{,}875{,}200}{120} = 2{,}598{,}960 \]
Why: The order in which cards arrive does not change the hand you hold, so the same five cards dealt in any of 5 factorial orders count once. That factor of 120 is large: 52P5 is 311,875,200 while 52C5 is 2,598,960. The question to ask is always whether rearranging the chosen objects gives a different outcome — and for a hand of cards, a committee or a set of pizza toppings, it does not.
Explain it
They have just learned permutations and use them for everything.
Discussion prompt
In four sentences or fewer, explain when to use a combination instead.
Hint: Ask what happens if you shuffle the chosen objects.
Answer:
Ask yourself whether shuffling the things you picked gives you something different. If you are handing out gold, silver and bronze, swapping two winners changes the outcome, so use a permutation.
If you are picking a team or a hand of cards, swapping two of them changes nothing, so use a combination. The combination is just the permutation divided by the number of ways to shuffle what you picked.
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 the first, ask whether shuffling the chosen objects changes anything. For the second, look for AND against OR. For at-least, write the list of included cases before computing anything. For coefficients, set the exponent equal to its target and solve for r before touching a calculator.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a combinations page. Top left: write both formulas side by side, show that the combination is the permutation divided by r factorial, and give one situation of each kind. Top right: work Example 1 in full, marking where the counting principle is used on top of the combinations. Middle: write out Pascal's triangle to row 7, labelling row 6's entries with their combination names, and check that each row is symmetric and sums to a power of 2. Bottom left: work the at-least game problem both ways, adding ten terms and subtracting three, and confirm they agree. Bottom right: expand a binomial to the fourth power using the triangle, then find one coefficient of a tenth power by solving for r, showing that you never expanded the rest.
If your Pascal's triangle rows do not each sum to a power of 2, recheck the additions: every entry is the sum of exactly the two nearest above it.
Recap
Five things, and order has been removed from the picture.
| If you see | Then |
|---|---|
| Shuffling changes the outcome | A permutation |
| Shuffling changes nothing | A combination |
| Two conditions on one selection | Multiply |
| Alternative cases | Add |
| At least, with many cases | Subtract the few unwanted from the total |
| A binomial power | Row n of Pascal's triangle gives the coefficients |
Lesson 10.3 turns counting into probability, dividing the count of favourable outcomes by the count of all outcomes.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.2 Use Combinations and the Binomial Theorem §10.2, pp. 690-695 — everything on these slides traces back here
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