Binomial combination

To understand what a binomial combination is, let’s do an example.

Suppose we are expanding the expression (a+b)^2, into a^2 + 2ab + b^2.

THe coefficient of the ab term is 2. The reason for that is because, when we multiply out all the brackets, there are two times where we multiply a and b together:

a from the first bracket and b from the second bracket
b from the first bracket and a from the second bracket

That means that the binomial combinations for the ab term are 2.

We write that as ^2C_1.

That’s what ^nC_r means!

On a calculator

There’s a button on a calculator to calculate ^nC_r. Often it’s above the divide button (press shift), but on some of the new ‘upgraded™®’ casio calculators you have to:

Press the Catalog button
Scroll down and click Probability
Scroll down to Combination C and click it
Then put the n value before it and the r value after it, e.g. 6 C 3 to calculate ^6C_3.

Casio CW calculator rant over.

Finding the coefficient of a term

Let’s say we want to find the coefficient of the a^2b^3 term of the expansion of (a+b)^5. That would be ^5C_2 because the a^2b^3 term is the term where a has power 2.

Terms with coefficients

If we want to find the coefficient of the x^3 term of the expansion of (2+3x)^5:

Find ^5C_3:
- ^5C_3=10
The extra coefficients are:
- (3x)^3 = 27x^3
- 2^2 = 4
The extra coefficients are 4 \times 27x^3 = 108x^3
If we times everything together, the coefficient of the x^3 term is 10 \times 108 = 1080.

Find the coefficient of the x^4 term of the expansion of (1+2x)^6.

^6C_4=15
The extra coefficients are:
- (2x)^4 = 16x^4
- 1^2 = 1
The extra coefficient is 1 \times 16x^4 = 16x^4
If we times everything together, the coefficient of the x^4 term is 15 \times 16 = 240.
Answer: 240.

Find the first 3 terms of the expansion of (3+x)^5.

For the x^0 term:
- ^5C_0=1
- The extra coefficients are:
  - (3)^5 = 243
  - x^0 = 1
- The extra coefficient is 243 \times 1 = 243
- The coefficient of the x^0 term is 1 \times 243 = 243.
For the x^1 term:
- ^5C_1=5
- The extra coefficients are:
  - (3)^4 = 81
  - x^1 = x
- The extra coefficient is 81 \times x = 81x
- The coefficient of the x^1 term is 5 \times 81 = 405.
For the x^2 term:
- ^5C_2=10
- The extra coefficients are:
  - (3)^3 = 27
  - x^2 = x^2
- The extra coefficient is 27 \times x^2 = 27x^2
- The coefficient of the x^2 term is 10 \times 27 = 270.
Answer: 243 + 405x + 270x^2 + ....

flashcards

Question	Answer
What does ^{n}C_{r} represent in a binomial expansion?	The binomial combination for the term where one variable has power r and the other has power n-r, for example in (a+b)^n, the coefficient of a^r b^{n-r} is ^{n}C_{r}.
How do you calculate ^{6}C_{3} on a ‘upgraded’ Casio CW calculator?	Press the Catalog button, scroll down and click Probability, scroll down to Combination C and click it, then put 6 before it and 3 after it, e.g. 6 C 3.
How do you find the coefficient of the a^{2}b^{3} term in the expansion of (a+b)^{5}?	It is ^{5}C_{2} because the term has a to the power of 2.
How do you find the coefficient of the x^{3} term in the expansion of (2+3x)^{5}?	Compute ^{5}C_{3}=10, multiply by the extra coefficients: (3x)^{3}=27x^{3} and 2^{2}=4, so 10\times4\times27=1080.
Find the coefficient of the x^{4} term in the expansion of (1+2x)^{6}.	^{6}C_{4}=15, extra coefficients: (2x)^{4}=16x^{4} and 1^{2}=1, so 15\times16=240.
Find the first 3 terms of the expansion of (3+x)^{5}.	x^{0} term: ^{5}C_{0}=1, 3^{5}=243, x^{0}=1, so 1\times243=243; x^{1} term: ^{5}C_{1}=5, 3^{4}=81, x^{1}=x, so 5\times81=405; x^{2} term: ^{5}C_{2}=10, 3^{3}=27, x^{2}=x^{2}, so 10\times27=270; answer: 243 + 405x + 270x^{2} + ...