Binomial Coefficients & Pascal's Triangle (DP IB Analysis & Approaches (AA)): Revision Note

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The Binomial Coefficient nCr

What is the formula for scriptbase bold C subscript r end scriptbase presubscript blank presuperscript bold italic n?

  • The formula for is scriptbase straight C subscript r space end scriptbase presubscript blank presuperscript n is

scriptbase straight C subscript r end scriptbase presubscript blank presuperscript n equals blank fraction numerator n factorial over denominator r factorial open parentheses n minus r close parentheses factorial end fraction

  • where the factorial symbol factorial means, for example:

    • 4 factorial equals 4 cross times 3 cross times 2 cross times 1 equals 24

  • e.g. C presubscript blank presuperscript 5 subscript 2 equals fraction numerator 5 factorial over denominator 2 factorial open parentheses 5 minus 2 close parentheses factorial end fraction equals fraction numerator 5 factorial over denominator 2 factorial 3 factorial end fraction equals fraction numerator 5 cross times 4 cross times up diagonal strike 3 cross times 2 cross times 1 end strike over denominator 2 cross times 1 cross times up diagonal strike 3 cross times 2 cross times 1 end strike end fraction equals 20 over 2 equals 10

Examiner Tips and Tricks

You can find the values of C presubscript blank presuperscript n subscript r on your GDC.

What properties of scriptbase bold C subscript r end scriptbase presubscript blank presuperscript bold italic n do I need to know?

  • Useful properties of scriptbase straight C subscript r space end scriptbase presubscript blank presuperscript n to know are

    • 0 less or equal than r less or equal than n

    • scriptbase straight C subscript 0 space end scriptbase presubscript blank presuperscript n equals 1

      • note that 0 factorial equals 1

    • scriptbase straight C subscript n space end scriptbase presubscript blank presuperscript n equals 1

    • The numbers scriptbase straight C subscript 0 space end scriptbase presubscript blank presuperscript n comma space scriptbase straight C subscript 1 space end scriptbase presubscript blank presuperscript n comma space scriptbase straight C subscript 2 space end scriptbase presubscript blank presuperscript n comma space... comma space scriptbase straight C subscript n minus 2 end subscript space end scriptbase presubscript blank presuperscript n comma space scriptbase straight C subscript n minus 1 end subscript space end scriptbase presubscript blank presuperscript n comma space scriptbase straight C subscript n space end scriptbase presubscript blank presuperscript n are symmetric

      • i.e. scriptbase straight C subscript r end scriptbase presubscript blank presuperscript n space equals space scriptbase straight C subscript n space minus space r end subscript end scriptbase presubscript blank presuperscript n

How does scriptbase bold C subscript r end scriptbase presubscript blank presuperscript bold italic n relate to the binomial theorem?

  • The binomial theorem gives you the expansion of open parentheses a plus b close parentheses to the power of n for different positive integer powers of n:

table row cell open parentheses a plus b close parentheses to the power of n end cell equals cell a to the power of n plus scriptbase straight C subscript 1 end scriptbase presubscript blank presuperscript n a to the power of n minus 1 end exponent b plus... plus scriptbase straight C subscript r end scriptbase presubscript blank presuperscript n a to the power of n minus r end exponent b to the power of r plus... plus b to the power of n end cell row cell scriptbase straight C subscript r end scriptbase presubscript blank presuperscript n end cell equals cell fraction numerator n factorial over denominator r factorial open parentheses n minus r close parentheses factorial end fraction end cell end table

  • where C presubscript blank presuperscript n subscript r is the binomial coefficient

Examiner Tips and Tricks

The binomial theorem and binomial coefficient formula are given in the formula booklet.

How does scriptbase bold C subscript r end scriptbase presubscript blank presuperscript bold italic n relate to counting principles?

  • The value of scriptbase straight C subscript r space end scriptbase presubscript blank presuperscript n represents the number of ways to choose r objects out of n different objects

    • This is an example of a counting principle

    • e.g. how many ways can you choose 2 people out of 5 people?

      • There are C presubscript blank presuperscript 5 subscript 2 equals fraction numerator 5 factorial over denominator 2 factorial open parentheses 5 minus 2 close parentheses factorial end fraction equals 10 ways to do this

  • You can relate scriptbase straight C subscript r space end scriptbase presubscript blank presuperscript n in the binomial theorem to the number of ways to choose r objects out of n different objects

  • open parentheses a plus b close parentheses to the power of n means multiply open parentheses a plus b close parentheses by itself n times

    • i.e. open parentheses a plus b close parentheses to the power of n equals open parentheses a plus b close parentheses open parentheses a plus b close parentheses... open parentheses a plus b close parentheses

    • Without using the binomial theorem, expand the right-hand side

      • i.e. multiplying all combinations of taking a letter from each bracket

  • e.g. one possibility is a cross times a cross times a... cross times a cross times b cross times b where the last 2 are b's

    • The first open parentheses n minus 2 close parentheses are a's

    • This simplifies to a to the power of n minus 2 end exponent b squared

    • But you can also choose 2 b's out of another 2 brackets (not just the last 2)

      • How many ways can you chose 2 b's out of the n brackets?

    • This is number of ways to choose r objects out of n different objects

      • i.e. there are C presubscript blank presuperscript n subscript 2 ways to do this

    • so the full term is C presubscript blank presuperscript n subscript 2 a to the power of n minus 2 end exponent b squared

Worked Example

Without using a calculator, find the coefficient of the term in x cubed in the expansion of left parenthesis 1 plus x right parenthesis to the power of 9.

1-5-1-binomial-coefficient-we-solution-2

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Pascal's Triangle

What is Pascal’s triangle?

  • Pascal’s triangle is formed as follows:

    • The first row and second row form a triangle of 1s

    • Every row below starts and ends with a 1

    • The middle terms are found by

      • adding the two terms above it

4.1.1-Binomial-Expansion-Notes-Diagram-3-1024x868

How does Pascal’s triangle relate to binomial coefficients scriptbase bold C subscript r end scriptbase presubscript blank presuperscript bold italic n?

  • The binomial coefficients scriptbase straight C subscript 0 space end scriptbase presubscript blank presuperscript n comma space scriptbase straight C subscript 1 space end scriptbase presubscript blank presuperscript n comma space scriptbase straight C subscript 2 space end scriptbase presubscript blank presuperscript n comma space... comma space scriptbase straight C subscript n minus 2 end subscript space end scriptbase presubscript blank presuperscript n comma space scriptbase straight C subscript n minus 1 end subscript space end scriptbase presubscript blank presuperscript n comma space scriptbase straight C subscript n space end scriptbase presubscript blank presuperscript n are rows in Pascal’s triangle

  • e.g. calculate C presubscript blank presuperscript 5 subscript 1 comma space C presubscript blank presuperscript 5 subscript 2 comma space C presubscript blank presuperscript 5 subscript 3 comma space C presubscript blank presuperscript 5 subscript 4 comma space C presubscript blank presuperscript 5 subscript 5

    • If you use the formula table row cell scriptbase straight C subscript r end scriptbase presubscript blank presuperscript n end cell equals cell fraction numerator n factorial over denominator r factorial open parentheses n minus r close parentheses factorial end fraction end cell end table

      • e.g. C presubscript blank presuperscript 5 subscript 0 equals fraction numerator 5 factorial over denominator 0 factorial open parentheses 5 minus 0 close parentheses factorial end fraction equals fraction numerator 5 factorial over denominator 5 factorial end fraction equals 1 (as 0 factorial equals 1)

      • e.g. C presubscript blank presuperscript 5 subscript 1 equals fraction numerator 5 factorial over denominator 1 factorial open parentheses 5 minus 1 close parentheses factorial end fraction equals fraction numerator 5 factorial over denominator 4 factorial end fraction equals fraction numerator 5 cross times up diagonal strike 4 cross times up diagonal strike 3 cross times up diagonal strike 2 cross times up diagonal strike 1 over denominator up diagonal strike 4 cross times up diagonal strike 3 cross times up diagonal strike 2 cross times up diagonal strike 1 end fraction equals 5

      • etc

    • you get 1 comma space 5 comma space 10 comma space 10 comma space 5 comma space 1

    • This is the same as the 1 comma space 5 comma space 10 comma space 10 comma space 5 comma space 1 row in Pascal's triangle!

How does Pascal’s triangle relate to the binomial theorem?

  • The binomial theorem gives you the expansion of open parentheses a plus b close parentheses to the power of n for different positive integer powers of n:

table row cell open parentheses a plus b close parentheses to the power of n end cell equals cell a to the power of n plus scriptbase straight C subscript 1 end scriptbase presubscript blank presuperscript n a to the power of n minus 1 end exponent b plus... plus scriptbase straight C subscript r end scriptbase presubscript blank presuperscript n a to the power of n minus r end exponent b to the power of r plus... plus b to the power of n end cell row cell scriptbase straight C subscript r end scriptbase presubscript blank presuperscript n end cell equals cell fraction numerator n factorial over denominator r factorial open parentheses n minus r close parentheses factorial end fraction end cell end table

  • Instead of using table row cell scriptbase straight C subscript r end scriptbase presubscript blank presuperscript n end cell equals cell fraction numerator n factorial over denominator r factorial open parentheses n minus r close parentheses factorial end fraction end cell end table to calculate

    • scriptbase straight C subscript 0 space end scriptbase presubscript blank presuperscript n comma space scriptbase straight C subscript 1 space end scriptbase presubscript blank presuperscript n comma space scriptbase straight C subscript 2 space end scriptbase presubscript blank presuperscript n comma space... comma space scriptbase straight C subscript n minus 2 end subscript space end scriptbase presubscript blank presuperscript n comma space scriptbase straight C subscript n minus 1 end subscript space end scriptbase presubscript blank presuperscript n comma space scriptbase straight C subscript n space end scriptbase presubscript blank presuperscript n

    • just read the values off the relevant row in Pascal's triangle!

Examiner Tips and Tricks

A lot of students finding sketching Pascal's triangle a quicker method to find coefficients than the C presubscript blank presuperscript n subscript r formula, especially for powers of n that are not too big.

Worked Example

Find the row beginning 1, 6, ... in Pascal’s triangle and use it to find the value of  scriptbase straight C subscript 4 end scriptbase presubscript blank presuperscript 6 space end presuperscript.

1-5-1-pascals-triangle-we-solution-3

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Amber

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Amber gained a first class degree in Mathematics & Meteorology from the University of Reading before training to become a teacher. She is passionate about teaching, having spent 8 years teaching GCSE and A Level Mathematics both in the UK and internationally. Amber loves creating bright and informative resources to help students reach their potential.

Mark Curtis

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Expertise: Maths Content Creator

Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.