Expands a binomial raised to a whole power, showing that the coefficients are the combination counts of the previous section and explaining why. Covers Pascal's triangle as a generating device and the general term formula for extracting a single term without expanding.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 11 — Sequences, Probability and Counting Theory
§11.6 Binomial Theorem, pp. 1358-1365
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §11.6 Binomial Theorem §11.6, pp. 1358-1365 — the pages these objectives are drawn from
Warm-up
Squaring a binomial gives a familiar middle coefficient.
Discussion prompt
Expanding a binomial cubed gives a middle coefficient of 3. Where does that number come from?
Hint: Think about picking one term from each bracket.
Answer:
Expanding means multiplying three brackets, choosing a or b from each and adding all the products.
To get a squared times b, exactly one bracket must supply the b — and there are three brackets it could be.
So the coefficient counts which bracket supplies the b, which is a combination. The coefficients are counting problems, which is why the previous section came first.
Concept
Expanding a binomial power means choosing one term from each bracket. The coefficient of each product counts how many ways that selection can be made.
\[ (a+b)^n=\sum_{k=0}^{n}C(n,k)a^{n-k}b^k \]
This makes the binomial theorem a consequence of §11.5 rather than a new formula. The combination count appearing in it is the same count, applied to which brackets contribute which term.
Figure (svg): A diagram showing that expanding a binomial power means choosing one term from each factor, so the coefficient counts the ways to make each selection
OpenStax, Precalculus, §11.6 Binomial Theorem §11.6, pp. 1358-1360
Section
Section 1
Concept
Each product in the expansion comes from picking one term from every bracket, and the coefficient counts how many such picks give that product.
The order of the brackets does not matter for the resulting product — picking b from the first and third brackets gives the same product as picking it from the third and first. That is exactly why the count is a combination rather than a permutation.
Figure (svg): A diagram showing that expanding a binomial power means choosing one term from each factor, so the coefficient counts the ways to make each selection
OpenStax, Precalculus, §11.6 Binomial Theorem §11.6, pp. 1358-1361
Picture it
Counting which bracket supplies the b.
Figure (svg): A diagram showing that expanding a binomial power means choosing one term from each factor, so the coefficient counts the ways to make each selection
The coefficient is a selection count, which is why the section follows the counting principles rather than preceding them. Nothing new has to be introduced.
Worked example
Which brackets supply the second term.
\[ \text{In a fifth power, what is the coefficient of the term with } b^2? \]
Note how many brackets there are
Why: The power.
\[ 5 \]
Note how many supply the second term
Why: The exponent on b.
\[ 2 \]
Count the selections
Why: Which two of the five.
\[ C(5, 2) \]
Compute
Why: The combination.
\[ 10 \]
Figure (svg): A diagram showing that expanding a binomial power means choosing one term from each factor, so the coefficient counts the ways to make each selection
\[ C(5,2)=10 \]
Verify: check against Pascal's triangle
Why: The fifth row reads 1, 5, 10, 10, 5, 1, and the third entry — corresponding to two b's — is 10. The triangle and the combination count agree, as they must, since the triangle's entries are those counts.
OpenStax, Precalculus, §11.6 Binomial Theorem §11.6, pp. 1359-1360
Prediction
In a sixth power, the term with three of the second variable.
Predict first
What is its coefficient?
Correct: Twenty.
Why: Three of the six brackets supply the second term, and the number of ways to choose which three is the combination of three from six, which is twenty. The coefficient is that count.
Worked example
The order of the brackets is irrelevant.
\[ \text{Why is the coefficient a combination rather than a permutation?} \]
Pick two brackets
Why: Say the first and third.
Pick them in the other order
Why: Third and first.
Compare the products
Why: Identical.
Conclude
Why: Orderings are pooled.
Figure (svg): The solution to Worked example why not a permutation shown as a ladder of expressions, one row per legal move
\[ \text{a combination} \]
Verify: check the count difference
Why: Using a permutation would give 20 rather than 10 for two from five — double, since each selection has two orderings. Comparing against Pascal's triangle would immediately show the discrepancy.
OpenStax, Precalculus, §11.6 Binomial Theorem §11.6, pp. 1360-1361
Trap
\[ \text{memorise the coefficients for each power} \]
Learn the numbers as a table
Why: The counting explanation is not made.
Powers beyond the memorised range become inaccessible.
Each coefficient is a combination count, computable for any power at all.
The count answers a concrete question: how many of the brackets supply the second term.
That makes any power reachable, where a memorised table stops at whatever was learned.
Faded example
Two second terms from five brackets.
Fill in the blanks
C(5,2)=\frac10___=___
Why: The count is which two of the five brackets supply the second term. Order is irrelevant since the product is the same either way, so it is a combination.
Sorting
It answers a concrete question.
Sort into buckets
Sort each description.
Explain it to yourself
They look like arbitrary numbers.
Discussion prompt
Explain what they are counting.
Hint: What does expanding actually do?
Answer:
Expanding multiplies n brackets, and each product in the result picks one term from each bracket.
Products with the same exponents are then collected, and the coefficient records how many products were collected together.
So it counts which brackets supplied the second term — a selection, since the order of the brackets does not change the product. A good explanation notes that this makes the binomial theorem a consequence of §11.5 rather than an independent formula.
Section
Section 2
Concept
Arranging the combination counts in rows produces a triangle where every entry is the sum of the two directly above it, which makes generating the next row trivial.
The addition rule has a counting explanation: selecting k brackets from n either includes the last bracket or does not, and those two cases are the two entries above. So the triangle's rule is a fact about combinations rather than a numerical coincidence.
Figure (svg): Pascal's triangle to six rows, with each entry the sum of the two above it
OpenStax, Precalculus, §11.6 Binomial Theorem §11.6, pp. 1361-1363
Picture it
Six rows, each generated from the one above.
Figure (svg): Pascal's triangle to six rows, with each entry the sum of the two above it
Reaching row twenty this way would mean writing twenty rows, which is why the formula is preferred for large powers. For small ones the triangle is quicker than any computation.
Worked example
Add adjacent pairs.
\[ \text{Generate row } 6 \text{ from row } 5. \]
Write row five
Why: From the triangle.
\[ 1, 5, 10, 10, 5, 1 \]
Start with one
Why: Every row does.
\[ 1 \]
Add adjacent pairs
Why: Six, fifteen, twenty, and so on.
\[ 6, 15, 20, 15, 6 \]
End with one
Why: Every row does.
\[ 1 \]
Figure (svg): Pascal's triangle to six rows, with each entry the sum of the two above it
\[ 1,\;6,\;15,\;20,\;15,\;6,\;1 \]
Verify: check the entry count
Why: Row six should have seven entries, and it does. Checking the count catches a dropped or duplicated addition, which is the usual slip when generating rows by hand.
OpenStax, Precalculus, §11.6 Binomial Theorem §11.6, pp. 1362-1363
Prediction
Row seven of Pascal's triangle.
Predict first
How many entries does it have?
Correct: Eight.
Why: Row n has n plus one entries, matching the number of terms in the expansion of the nth power. Counting them is the quickest check that the right row is in use.
Worked example
A counting argument.
\[ \text{Why is each entry the sum of the two above it?} \]
Consider selecting k from n
Why: The entry itself.
\[ C(n, k) \]
Split by the last item
Why: Either included or not.
Count the including case
Why: Choose k-1 from the rest.
\[ C(n - 1, k - 1) \]
Count the excluding case
Why: Choose k from the rest.
\[ C(n - 1, k) \]
Figure (svg): The solution to Worked example why the addition rule holds shown as a ladder of expressions, one row per legal move
\[ C(n,k)=C(n-1,k-1)+C(n-1,k) \]
Verify: check the two cases are exhaustive and separate
Why: Every selection either contains the last item or does not, and none does both — so adding the two counts gives the total exactly once. That is why the rule is an equality rather than an approximation.
OpenStax, Precalculus, §11.6 Binomial Theorem §11.6, pp. 1363-1363
Error analysis
A student expands a fourth power.
Annotate
On: \( \text{coefficients } 1,\;5,\;10,\;10,\;5,\;1 \)
The rows are numbered from zero, so the top row is row zero. Counting the entries — which must be one more than the power — is the quickest check that the right row was used.
Faded example
From two entries in the row above.
Fill in the blanks
5+10=15, \text___
Why: Each entry is the sum of the two directly above it, which follows from splitting a selection by whether it includes the last item. That is why the rule is exact rather than a pattern.
Sorting
The rows are numbered from zero.
Sort into buckets
Sort each pairing.
Explain it
Each entry is the sum of the two above.
Discussion prompt
Explain why to a classmate.
Hint: Split the selections into two cases.
Answer:
Every selection of k items from n either includes the last item or does not. Those two cases cover everything and overlap nowhere.
If it includes the last, the remaining k minus one come from the other n minus one — one of the entries above. If it excludes it, all k come from those n minus one — the other entry above.
So the total is the sum of the two. A good explanation notes that this makes the rule a fact about combinations rather than a numerical pattern that happens to hold.
Section
Section 3
Concept
The expansion has one more term than the power, with the first variable's exponent descending from the power to zero while the second ascends.
The exponent sum is the fastest check on an expansion. Every term picks one factor from each of n brackets, so the two exponents must total n — and a term violating that has been written wrongly.
Figure (svg): A contrast between correct and incorrect binomial terms, distinguished by whether the exponents sum to the power
OpenStax, Precalculus, §11.6 Binomial Theorem §11.6, pp. 1363-1364
Picture it
The left column is what a correct expansion looks like.
Figure (svg): A contrast between correct and incorrect binomial terms, distinguished by whether the exponents sum to the power
The caption gives the reason. Each term takes one factor from each bracket, so the exponents cannot sum to anything but the power.
Worked example
Coefficients and exponents together.
\[ \text{Expand } (a+b)^4. \]
Get the coefficients
Why: Row four.
\[ 1, 4, 6, 4, 1 \]
Descend the first exponent
Why: From four to zero.
\[ 4, 3, 2, 1, 0 \]
Ascend the second
Why: From zero to four.
\[ 0, 1, 2, 3, 4 \]
Assemble
Why: Five terms.
Figure (svg): A contrast between correct and incorrect binomial terms, distinguished by whether the exponents sum to the power
\[ a^4+4a^3b+6a^2b^2+4ab^3+b^4 \]
Verify: check every exponent sum
Why: Each term's two exponents add to four: four plus zero, three plus one, two plus two, and so on. Any term failing that check has been written wrongly, and the check takes one glance per term.
OpenStax, Precalculus, §11.6 Binomial Theorem §11.6, pp. 1363-1364
Prediction
Expanding a seventh power.
Predict first
How many terms are there?
Correct: Eight.
Why: The exponent on the first variable runs from seven down to zero inclusive, which is eight values. That matches row seven of Pascal's triangle having eight entries.
Worked example
The variables carry their own factors.
\[ \text{Expand } (2x+3)^3. \]
Get the coefficients
Why: Row three.
\[ 1, 3, 3, 1 \]
Raise the first part
Why: Descending powers of 2x.
\[ 8 x ^{3}, 4 x ^{2}, 2 x, 1 \]
Raise the second
Why: Ascending powers of 3.
\[ 1, 3, 9, 27 \]
Multiply all three together
Why: Term by term.
Figure (svg): The solution to Worked example a binomial with coefficients shown as a ladder of expressions, one row per legal move
\[ 8x^3+36x^2+54x+27 \]
Verify: check at a convenient value
Why: At x equal to one the original is five cubed, which is 125, and the expansion gives 8 plus 36 plus 54 plus 27, also 125. Substituting one value confirms the whole expansion in a line.
OpenStax, Precalculus, §11.6 Binomial Theorem §11.6, pp. 1364-1364
Trap
\[ (2x+3)^3: \text{ first term }2x^3 \]
Raise only the variable
Why: The numerical coefficient is left unraised.
The first term should be eight x cubed, since the whole of 2x is cubed.
The exponent applies to the entire term, coefficient included. Cubing 2x gives 8 times x cubed.
The same applies to the second part: the ascending powers are of 3, giving 1, 3, 9 and 27.
Substituting a convenient value checks the whole expansion, and it catches this error immediately.
Faded example
In a fourth-power expansion.
Fill in the blanks
a^3b^1}: \; 3+1=4\;\checkmark
Why: Every term takes one factor from each of four brackets, so its two exponents must sum to four. A term failing that check cannot arise from the expansion.
Sorting
In the expansion of a fifth power.
Sort into buckets
Sort each term.
Step zero
You are asked to expand a binomial power.
Discussion prompt
What do you write down first?
Hint: Two lists.
Answer:
The coefficients from the right row, and the exponent pattern — descending for one variable and ascending for the other.
Writing both before assembling any terms means the structure is fixed and only the arithmetic remains.
Then check every exponent sum equals the power. That single check catches a mis-assembled term, and it takes one glance per term.
Section
Section 4
Concept
Any one term of an expansion can be written directly, using an index that counts how many second variables it contains — and that index starts at zero.
The index starting at zero is this section's recurring difficulty, and it is the same off-by-one that appeared in §11.2 and §11.3 in a different guise. Writing the index explicitly before substituting is the defence.
Figure (svg): A card giving the general term of a binomial expansion, with each part labelled
OpenStax, Precalculus, §11.6 Binomial Theorem §11.6, pp. 1364-1365
Picture it
Each part labelled, with the zero-indexing flagged.
Figure (svg): A card giving the general term of a binomial expansion, with each part labelled
The red line at the bottom is the recurring error. Asking for the fifth term and substituting five gives the sixth, which is a plausible-looking wrong answer.
Worked example
Convert the position to an index.
\[ \text{Find the fourth term of } (x+2)^6. \]
Convert position to index
Why: One less.
\[ k = 3 \]
Find the coefficient
Why: The combination.
\[ C(6, 3) = 20 \]
Find the exponents
Why: Six minus three, and three.
\[ x ^{3}\text{ and } 2 ^{3} \]
Assemble
Why: Multiply.
\[ 20 \times 8 \times x\text{ cubed} \]
Figure (svg): A card giving the general term of a binomial expansion, with each part labelled
\[ 160x^3 \]
Verify: check the exponent sum
Why: The exponents are three and three, summing to six — the power, as required. And the position-to-index conversion was the one place an error could enter, which the sum check would not catch, so it is worth writing explicitly.
OpenStax, Precalculus, §11.6 Binomial Theorem §11.6, pp. 1364-1365
Prediction
Finding the seventh term.
Predict first
What index do you use?
Correct: Six.
Why: The index counts the second variable's exponent and starts at zero, so the first term has index zero and the seventh has index six. Using seven would give the eighth term.
Worked example
The exponent gives the index directly.
\[ \text{Find the term containing } x^2 \text{ in } (x+3)^5. \]
Read the first exponent
Why: Two.
\[ n - k = 2 \]
Solve for the index
Why: Five minus two.
\[ k = 3 \]
Find the coefficient
Why: The combination.
\[ C(5, 3) = 10 \]
Assemble
Why: With three factors of three.
\[ 10 \times 27 \times x\text{ squared} \]
Figure (svg): The solution to Worked example find a term by its variable power shown as a ladder of expressions, one row per legal move
\[ 270x^2 \]
Verify: check the exponent sum
Why: Two and three sum to five, the power. Working from a stated variable power avoids the position-to-index conversion entirely, which is why this phrasing of the question is less error-prone than asking for a numbered term.
OpenStax, Precalculus, §11.6 Binomial Theorem §11.6, pp. 1365-1365
Error analysis
A student finds the fourth term.
Annotate
On: \( k=4: \; C(6,4)x^2 2^4 \)
The answer produced is a legitimate term of the expansion, just the wrong one — so nothing about it looks incorrect. Writing the index conversion as its own step is the only reliable defence.
Faded example
For the fourth term.
Fill in the blanks
k=4-1=3
Why: The index runs from zero, so it is one less than the term's position. Writing this conversion as its own step is what prevents using the position directly.
Sorting
The index counts the second variable's exponent.
Sort into buckets
Sort each term of a fifth power.
Explain it
The index starts at zero rather than one.
Discussion prompt
Explain to a classmate why, and what to do about it.
Hint: What is the index counting?
Answer:
The index counts the second variable's exponent, and the first term has none of that variable — so its exponent, and therefore its index, is zero.
That means the index is always one less than the position. Asking for the fifth term means using an index of four.
The error is dangerous because the wrong index still gives a legitimate term, just the wrong one — nothing about the answer looks incorrect. A good explanation stresses writing the conversion as its own step rather than doing it mentally.
Section
Section 5
Concept
An expansion can be verified without redoing it: count the terms, check every exponent sum, and substitute a convenient value.
The substitution check is the most complete, since it tests the coefficients and the exponents together. But it gives no indication of where an error is, so the structural checks are worth running first.
Figure (svg): A contrast between correct and incorrect binomial terms, distinguished by whether the exponents sum to the power
OpenStax, Precalculus, §11.6 Binomial Theorem §11.6, pp. 1360-1365
Picture it
The left column is the structure to check against.
Figure (svg): A contrast between correct and incorrect binomial terms, distinguished by whether the exponents sum to the power
The first three rows are the three structural checks: the term count, the exponent sums and the coefficients coming from the right row.
Worked example
One value tests everything.
\[ \text{Check } (a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4 \text{ by substituting ones.} \]
Substitute into the original
Why: Two to the fourth.
\[ 16 \]
Substitute into the expansion
Why: All the variables become one.
\[ 1 + 4 + 6 + 4 + 1 \]
Add
Why: The coefficient total.
\[ 16 \]
Compare
Why: They match.
Figure (svg): A contrast between correct and incorrect binomial terms, distinguished by whether the exponents sum to the power
\[ 16=16 \]
Verify: note what this check covers
Why: Substituting ones makes every term equal its coefficient, so this checks the coefficients sum correctly — which they must, since the coefficients of the nth power always total two to the n. It does not check the exponents, which is why the structural checks matter too.
OpenStax, Precalculus, §11.6 Binomial Theorem §11.6, pp. 1362-1364
Prediction
Substituting ones into a fifth-power expansion.
Predict first
What do the coefficients sum to?
Correct: Thirty-two.
Why: Substituting ones makes the original two to the fifth, which is 32, and every term equal its coefficient. So the coefficients of the nth power always total two to the n.
Worked example
The coefficients read the same both ways.
\[ \text{Why are the coefficients symmetric?} \]
Consider choosing k brackets
Why: For the second variable.
\[ C(n, k) \]
Consider choosing the rest
Why: For the first variable.
\[ C(n, n - k) \]
Compare
Why: Choosing which to include or exclude.
Conclude
Why: The row is symmetric.
Figure (svg): The solution to Worked example check the symmetry shown as a ladder of expressions, one row per legal move
\[ C(n,k)=C(n,n-k) \]
Verify: check on a row
Why: Row five reads 1, 5, 10, 10, 5, 1 — the same forwards and backwards. A row that is not symmetric has been generated wrongly, which makes this a fast structural check on the coefficients.
OpenStax, Precalculus, §11.6 Binomial Theorem §11.6, pp. 1364-1365
Trap
\[ \text{the substitution works, so the expansion is right} \]
Run a single check and stop
Why: Structural errors are assumed to be caught by it.
Two errors can cancel in a single substitution, leaving both undetected.
Run the structural checks too: the term count, the exponent sums and the coefficient symmetry.
Each catches a different kind of error, and a substitution alone cannot say where a problem is.
All three together take about ten seconds and between them cover the ways an expansion can go wrong.
Sorting
Each covers a different failure.
Sort into buckets
Sort each check.
Faded example
Ones into a fourth-power expansion.
Fill in the blanks
1+4+6+4+1=16=2^4
Why: Substituting ones makes every term equal its coefficient, so the sum should equal two raised to the power. That is a fast check on the whole set of coefficients.
Explain it to yourself
An expansion can be verified without redoing it.
Discussion prompt
Explain what each check covers.
Hint: Structure and value.
Answer:
Counting the terms catches a missing or extra term, and it must equal the power plus one.
Checking the exponent sums catches a mis-assembled term, since every one must total the power. Checking the coefficients' symmetry catches a badly generated row.
And substituting a value tests everything together but says nothing about where a problem is. A good explanation notes that the structural checks localise errors where the substitution only detects them.
Comparison
Fill the blanks from memory. Each suits a different situation.
Comparison matrix
| Pascal's triangle | the combination formula | |
|---|---|---|
| how it works | each entry sums the two above | a factorial quotient |
| best for | small powers | large powers, or one term |
| reaching row twenty | twenty rows to write | one computation |
| finding a single term | still needs the whole row | direct |
The last row is the practical division. The triangle generates whole rows and the formula extracts single entries, which is why both are worth having.
Pattern
Five steps, and the last three are checks.
Step 3's phrase 'each whole part' matters when the binomial has numerical coefficients — the exponent applies to the coefficient as well as the variable.
OpenStax Algebra and Trigonometry 2e, §13.6 Binomial Theorem §13.6
Check
Where the coefficients come from.
Check your understanding
What does a binomial coefficient count?
Answer: A
Why: Expanding means picking one term from each bracket, and products with the same exponents are collected. The coefficient records how many such picks give that product, which is a combination.
Check
The general term.
Check your understanding
To find the fifth term of an expansion, what index do you use?
Answer: A
Why: The index counts the second variable's exponent and starts at zero, so the first term has index zero and the fifth has index four. Using five gives the sixth term, which looks perfectly legitimate.
Check
Checking an expansion.
Check your understanding
In a sixth-power expansion, what must each term's exponents sum to?
Answer: A
Why: Each term takes exactly one factor from each of six brackets, so its two exponents must total six. A term violating that cannot arise from the expansion, which makes this a fast check.
Real world
The coefficients describe the likely outcomes of repeated trials.
Discussion prompt
Ten coins are tossed. Why does the number of ways to get exactly six heads involve a binomial coefficient?
Hint: Which coins show heads?
Answer:
Getting six heads means six of the ten coins show heads, and the count is which six — a combination, and therefore a binomial coefficient.
The whole row gives the counts for every possible number of heads, and their total is two to the tenth — every possible outcome of ten tosses.
So the row's shape is the shape of the outcome distribution: large in the middle, small at the ends. That is why extreme results are rare, and it is the counting fact underlying the bell-shaped curve that appears throughout statistics.
Commit first
State your confidence along with your answer.
Predict first
Why are binomial coefficients combination counts?
Correct: Each term picks one factor per bracket, and the coefficient counts which brackets gave the second.
Why: Expanding multiplies n brackets and collects products with matching exponents. How many products get collected is exactly how many ways the second variable's brackets can be chosen, which is a combination.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
A classmate is memorising Pascal's triangle row by row. Suggest something better.
Hint: What are the entries?
Answer:
The entries are combination counts, computable directly for any row without generating the ones above.
That matters because the triangle becomes impractical for large powers — reaching row twenty means writing twenty rows, where the formula reaches any entry in one computation.
The triangle is still quicker for small powers, so both are worth having. A good explanation notes that knowing what the entries count is what makes the choice between them possible rather than being stuck with one.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The fourth is where the recurring error lives, and it is dangerous because the wrong index still produces a legitimate-looking term. The first is what makes the whole section follow from §11.5.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Write out why the coefficients count which brackets supply the second variable, using a cubed binomial as the example. Beside it, draw six rows of Pascal's triangle and mark the addition rule. Underneath, expand one binomial in full with the three checks marked, and extract one term using the general formula with the index conversion shown explicitly.
If your index conversion appears as its own line and your expansion carries all three checks, the section's recurring error and its verification habits are both on the page.
Recap
Five things, and the first makes the rest follow from §11.5.
| if you remember one thing | it should be this |
|---|---|
| about the coefficients | they count which brackets supply the second variable |
| about the triangle | each entry sums the two above, for a counting reason |
| about expanding | the exponents always sum to the power |
| about single terms | the index starts at zero, so it is one less than the position |
Section 11.7 closes the chapter with probability, where the counting principles of §11.5 supply the numerators and denominators of every ratio.
OpenStax, Precalculus, §11.6 Binomial Theorem §11.6, pp. 1358-1365 — everything on these slides traces back here
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