A. Counting and Probability · Permutations and CombinationsBinomial Coefficients and the Binomial Theorem
A2.4About 65 minutes
Learning goals- I can explain why the coefficients in the expansion of (a + b)ⁿ are the combinations C(n, k).
- I can expand binomials using the binomial theorem.
- I can find a particular term or coefficient without expanding everything.
- I can connect binomial coefficients to counting sequences of coin flips.
Words to know
binomial
A polynomial with two terms, such as a + b or 2x − 3.
binomial theorem
(a + b)ⁿ = C(n, 0)aⁿ + C(n, 1)aⁿ⁻¹b + C(n, 2)aⁿ⁻²b² + … + C(n, n)bⁿ.
binomial coefficient
The number C(n, k), the coefficient of aⁿ⁻ᵏbᵏ in (a + b)ⁿ.
general term
The term tₖ₊₁ = C(n, k)aⁿ⁻ᵏbᵏ of the expansion of (a + b)ⁿ. Note that k starts at 0, so the first term has k = 0.
Remember from Grade 11
In Sequences and Series (MCR3U) you expanded powers like (x + 2)⁵ using row n of Pascal's triangle for the coefficients, with the powers of the first term going down and the powers of the second term going up. Now you know that row n of Pascal's triangle is C(n, 0), C(n, 1), …, C(n, n). So the coefficients must be combinations. But why should choosing have anything to do with multiplying brackets?
Expanding is choosing
Write (a + b)⁵ as five brackets multiplied together:
(a + b)(a + b)(a + b)(a + b)(a + b)
When you multiply out, each term of the answer comes from picking one letter from each bracket and multiplying them. If you pick b from exactly 3 brackets and a from the other 2, you get a²b³.
How many ways are there to pick b from exactly 3 of the 5 brackets? That is C(5, 3) = 10. So after collecting like terms, the expansion contains 10a²b³.
The same reasoning works for every term and every n.
The binomial theorem
(a + b)ⁿ = C(n, 0)aⁿ + C(n, 1)aⁿ⁻¹b + C(n, 2)aⁿ⁻²b² + … + C(n, k)aⁿ⁻ᵏbᵏ + … + C(n, n)bⁿ
- There are n + 1 terms.
- In each term the exponents add to n.
- The general term is tₖ₊₁ = C(n, k) aⁿ⁻ᵏ bᵏ, for k = 0, 1, …, n. The term number is one more than k because k starts at 0.
This is why C(n, k) is called a binomial coefficient. In the unit Discrete Probability Distributions you will see these same coefficients in the binomial distribution, which gives the probability of k successes in n trials.
Worked exampleExample 1: Expanding
- Here a = x, b = 2 and n = 4. The coefficients are C(4, 0) to C(4, 4): 1, 4, 6, 4, 1.
- Terms: 1(x⁴)(2⁰) + 4(x³)(2¹) + 6(x²)(2²) + 4(x)(2³) + 1(2⁴).
- Simplify: x⁴ + 8x³ + 24x² + 32x + 16.
- Check with x = 1: (1 + 2)⁴ = 81 and 1 + 8 + 24 + 32 + 16 = 81 ✓.
x⁴ + 8x³ + 24x² + 32x + 16
Worked exampleExample 2: A negative term
- a = 2x and b = −3. Keep the brackets so the signs and powers are right.
- C(3, 0)(2x)³ + C(3, 1)(2x)²(−3) + C(3, 2)(2x)(−3)² + C(3, 3)(−3)³
- = 8x³ + 3(4x²)(−3) + 3(2x)(9) + (−27)
- = 8x³ − 36x² + 54x − 27. The signs alternate because odd powers of −3 are negative.
Worked exampleExample 3: One term only
Find the term containing x³ in the expansion of (x + 2)¹⁰.
- General term: C(10, k) x¹⁰⁻ᵏ 2ᵏ.
- For x³ we need 10 − k = 3, so k = 7.
- C(10, 7) × 2⁷ = 120 × 128 = 15 360.
15 360x³ (it is the 8th term, since k = 7).
Worked exampleExample 4: The constant term
Find the constant term (the term with no x) in (x + 1/x)⁶.
- General term: C(6, k) x⁶⁻ᵏ (1/x)ᵏ = C(6, k) x⁶⁻²ᵏ.
- No x means the exponent is 0: 6 − 2k = 0, so k = 3.
- C(6, 3) = 20.
Binomial coefficients and coin flips
Write the result of one coin flip as H + T. For 4 flips, (H + T)⁴ = H⁴ + 4H³T + 6H²T² + 4HT³ + T⁴. Each term counts the sequences of flips with that many heads: for example 6 sequences have exactly 2 heads and 2 tails (C(4, 2) = 6, as in lesson 6). Adding the coefficients, 1 + 4 + 6 + 4 + 1 = 16 = 2⁴, the total number of sequences.
Just for funIsaac Newton extended the binomial theorem to exponents that are not whole numbers, like (1 + x)^(1/2). The catch is that the expansion then goes on forever. You may be relieved that this course sticks to whole-number exponents.
Hands-on activityCoins and coefficients
You need: one coin (or four coins), paper and a pencil.
- List all 16 possible sequences of 4 coin flips (HHHH, HHHT, …, TTTT). Work systematically so you miss none.
- Sort the sequences by the number of heads: 4, 3, 2, 1 or 0. Count each group.
- Compare your counts with the coefficients in (H + T)⁴ = H⁴ + 4H³T + 6H²T² + 4HT³ + T⁴.
- Now flip 4 coins (or one coin 4 times) 32 times and tally the number of heads each time. Which number of heads happens most often? Why?
What to look for: group sizes 1, 4, 6, 4, 1. In 32 trials, 2 heads should come up most often (expected about 32 × 6/16 = 12 times), because 6 of the 16 equally likely sequences have exactly 2 heads. Your tallies will vary from this.
Your turn
1Expand (a + b)⁴ using combinations.
C(4, 0)a⁴ + C(4, 1)a³b + C(4, 2)a²b² + C(4, 3)ab³ + C(4, 4)b⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴.
x³ + 3(x²)(3) + 3(x)(9) + 27 = x³ + 9x² + 27x + 27.
16x⁴ + 4(8x³)(−1) + 6(4x²)(1) + 4(2x)(−1) + 1 = 16x⁴ − 32x³ + 24x² − 8x + 1.
4Find the coefficient of a³b⁵ in (a + b)⁸.
5Find the 4th term of (x + 2)⁷.
4th term: k = 3. C(7, 3)x⁴(2³) = 35 × 8x⁴ = 280x⁴.
6How many terms are in the expansion of (x + y)¹²?
7Find the coefficient of x² in (1 + x)¹⁰.
More practice: warm-up
1What are the coefficients of (a + b)³?
3What is the coefficient of a²b² in (a + b)⁴?
4How many terms are in the expansion of (a + b)⁷?
5What is the sum of the coefficients of (a + b)⁶?
More practice: on your own
Coefficients C(5, k)(−2)ᵏ: 1, −10, 40, −80, 80, −32. x⁵ − 10x⁴ + 40x³ − 80x² + 80x − 32.
2Find the coefficient of x³ in (2x + 1)⁶.
Term C(6, k)(2x)⁶⁻ᵏ(1)ᵏ; need 6 − k = 3, so k = 3: C(6, 3) × 2³ = 20 × 8 = 160.
3Find the constant term in (x + 2/x)⁶.
C(6, k)x⁶⁻ᵏ(2/x)ᵏ = C(6, k)2ᵏx⁶⁻²ᵏ. k = 3: 20 × 8 = 160.
4Find the middle term of (a − b)⁸.
9 terms, so the middle is the 5th (k = 4): C(8, 4)a⁴(−b)⁴ = 70a⁴b⁴.
5Use the first three terms of the binomial theorem to estimate 1.01⁵ to four decimal places.
(1 + 0.01)⁵ ≈ 1 + 5(0.01) + 10(0.0001) = 1 + 0.05 + 0.001 = 1.0510. (The exact value is 1.0510100501.)
Challenge
1Find the coefficient of x⁴ in (1 + x)³(1 + x)⁵ in two ways, and explain what identity this shows.
Way 1: (1 + x)⁸, so C(8, 4) = 70. Way 2: multiply the expansions; x⁴ comes from xᵏ in the first and x⁴⁻ᵏ in the second: C(3, 0)C(5, 4) + C(3, 1)C(5, 3) + C(3, 2)C(5, 2) + C(3, 3)C(5, 1) = 5 + 30 + 30 + 5 = 70 ✓. This shows that choosing 4 from a group of 3 plus a group of 5 can be counted by cases on how many come from the first group.
2Use the binomial theorem to show that C(n, 0) − C(n, 1) + C(n, 2) − … + (−1)ⁿC(n, n) = 0 for n ≥ 1.
Put a = 1 and b = −1 in (a + b)ⁿ: the expansion gives exactly the alternating sum, and (1 − 1)ⁿ = 0ⁿ = 0. ✓
3In the expansion of (1 + x)ⁿ, the coefficient of x² is 36. Find n.
C(n, 2) = 36, so n(n − 1) = 72 = 9 × 8. n = 9.
Quiz
1. What is the coefficient of a²b³ in the expansion of (a + b)⁵?
2. Which is the general term of (a + b)ⁿ?
3. How many terms are in the expansion of (x + y)⁹?
4. What is the constant term in (x + 1/x)⁶?
5. Which is the expansion of (x − 1)³?
6. What is the sum of the coefficients in the expansion of (a + b)⁷?
Big idea (A2.4): multiplying out (a + b)ⁿ means choosing a or b from each bracket, so the coefficient of aⁿ⁻ᵏbᵏ is the number of ways to choose the k brackets that give b: C(n, k). This explains why Pascal's triangle gives binomial coefficients, a fact your teen used without proof in Grade 11.
How to help: Ask your teen to explain where the 10 in 10a²b³ comes from. For any expansion, have them check the answer by substituting x = 1 into both sides.
Common difficulties:
- Forgetting brackets: (2x)³ = 8x³, not 2x³, and (−3)² = 9.
- Off-by-one errors with the general term: the 4th term has k = 3.
- Losing negative signs; with a negative second term the signs alternate.
- Writing P(n, k) instead of C(n, k) (the order in which brackets are chosen does not matter).