Factorials, combinations and binomial coefficients
Введено| English | Значение |
|---|---|
| combination/ˌkɒmbɪˈneɪʃn/ | сочетание |
Why is the x² coefficient not just the middle number in Pascal’s triangle?
- A binomial coefficient counts which factors contribute an x term.
- This lesson studies combination 组合: A selection in which order does not distinguish different choices.
The method
- For a nonnegative integer n, n!=n(n−1)…1 and 0!=1.
- For 0≤r≤n, nCr=n!/[r!(n−r)!] counts unordered selections.
- In (a+bx)^n with positive integer n, the x^r coefficient is nCr × a^(n−r) × b^r.
- The term number is r+1 when the expansion is written in ascending powers of x.
Why is the x² coefficient in (2+3x)⁴ not just 4C2=6?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
The binomial coefficient counts the arrangements; the x² term is 4C2×2²×(3x)²=6×4×9x²=216x². Omitting the a and b powers is the standard error.
A worked case
4!=24 and 4C2=24/(2×2)=6. The coefficients for n=4 are 1,4,6,4,1. In (2+3x)^4, the x² term is 4C2×2²×(3x)²=216x²; the constant is 2⁴=16 and the x term is 4×2³×3x=96x. For four independent trials each with success probability 1/3, exactly two successes have probability 4C2×(1/3)²×(2/3)²=8/27. The coefficient 6 counts the possible success positions; each arrangement has the same probability.
Factorials, combinations and binomial coefficients
For a nonnegative integer n, n!=n(n−1)…1 and 0!=1
Connect a factored expression to its graph or expansion coefficients.
Evaluate 4!.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
4!=4×3×2×1=24.
Where it goes wrong
- Do not omit the powers of a or b when using a combination coefficient.
- The x² term is the third term, not the second.
- The probability expression needs a fixed number of trials, independent binary outcomes and the same success probability; a changing probability invalidates this simple model.
Not true: The x² coefficient of (a+bx)^n is just nC2. The x² term is nC2·a^(n−2)·b²x², so a and b contribute too.
Evaluate 4C2.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
4C2=4!/(2!2!)=24/4=6.
The x² term is nC2·a^(n−2)·b²x², so a and b contribute too.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
True. The x² term is nC2·a^(n−2)·b²x², so a and b contribute too.
Applying it
- Use symmetry nCr=nC(n−r) and cancel factorial factors before calculating.
- To find the x³ coefficient in (2+3x)^4, use 4C3×2×3³=216.
- The finite polynomial formula here is for positive integer n; a rational exponent instead gives the separate general binomial series with its stated validity interval.
Find the coefficient of x² in (2+3x)^4.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
The coefficient is 6×2²×3²=216.
Match each binomial quantity to its value or rule.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
nCr counts unordered selections; in an expansion it multiplies a^(n-r)b^r, and ascending powers start at r=0.
In your course
- AQA A-level Mathematics (7357), specification section D.
- Every paper allows a calculator, but method marks still need the expression or equation written before its value.
- Combination: A selection in which order does not distinguish different choices.
- For a nonnegative integer n, n!=n(n−1)…1 and 0!=1.
- Watch for: do not omit the powers of a or b when using a combination coefficient.