IB Mathematics AA HL Permutations and Combinations Study Notes | New Syllabus

IB Mathematics AA AHL Permutations and Combinations Study Notes

IB Mathematics AA HL Permutations and Combinations Study Notes

IB Mathematics AA HL Permutations and Combinations Study Notes Offer a clear explanation of Permutations and Combinations , including various formula, rules, exam style questions as example to explain the topics. Worked Out  examples and common problem types provided here will be sufficient to cover for topic Permutations and Combinations.

Permutations and Combinations

Counting Principles: Permutations and Combinations

Counting principles help us determine the number of possible ways to arrange or select items. The two main methods are permutations and combinations.

1. Fundamental Principle of Counting

If one event can happen in \( m \) ways and another in \( n \) ways, then both events together can happen in \( m \times n \) ways.

2. Permutations

Permutations are used when the order matters.

Formula (without repetition):

\( P(n, r) = \frac{n!}{(n – r)!} \)

  • \( n \): Total number of items
  • \( r \): Number of items to arrange
  • \( n! \): “n factorial”, the product of all positive integers up to \( n \)

3. Combinations

Combinations are used when the order does not matter.

Formula:

\( C(n, r) = \binom{n}{r} = \frac{n!}{r!(n – r)!} \)

Use of Technology

Most scientific calculators and GDCs can compute both permutations and combinations using the \( nPr \) and \( nCr \) functions.

Examples

  1. How many ways can 3 students be seated in a row of 5 chairs?
  2. From a group of 10 people, how many ways can you select a committee of 4?
  3. How many 3-digit numbers can be formed using the digits 1 to 5 with no repetition?
▶️ Answer/Explanation

1.Permutations (order matters):

\( P(5, 3) = \frac{5!}{(5 – 3)!} = \frac{120}{2} = 60 \)

Answer: 60 ways

2. Combinations (order doesn’t matter):

\( C(10, 4) = \binom{10}{4} = \frac{10!}{4!6!} = 210 \)

Answer: 210 ways

3. Permutations without repetition:

Choose and arrange 3 out of 5 digits:

\( P(5, 3) = \frac{5!}{2!} = 60 \)

Answer: 60 three-digit numbers

Extension of the Binomial Theorem

Extension of the Binomial Theorem to Fractional and Negative Indices

The binomial theorem can be extended beyond positive integers to include fractional and negative powers, i.e., \( n \in \mathbb{Q} \) (rational numbers). In this case, the expansion becomes an infinite series called a binomial series or power series expansion.

Generalized Binomial Series:

$ (a + b)^n = a^n \left(1 + \frac{b}{a}\right)^n = a^n \sum_{k=0}^{\infty} \binom{n}{k} \left(\frac{b}{a}\right)^k $

where the binomial coefficients for any rational \( n \) are defined as:

$ \binom{n}{k} = \frac{n (n-1) (n-2) \cdots (n – k + 1)}{k!} $

This expansion converges when \( \left| \frac{b}{a} \right| < 1 \).

Key Points:

  • For integer \( n \geq 0 \), the series terminates (finite expansion).
  • For fractional or negative \( n \), the expansion is infinite.
  • This expansion is fundamental in finding approximations using power series.

Power Series Expansions

A power series is an infinite sum of terms involving powers of a variable, typically written as:

$ f(x) = \sum_{n=0}^\infty a_n (x – c)^n = a_0 + a_1 (x – c) + a_2 (x – c)^2 + a_3 (x – c)^3 + \cdots $

  • \( a_n \): coefficients (constants)
  • \( c \): center of the series (often \( c=0 \))
  • \( x \): variable

Examples

Use the binomial series to approximate \( \sqrt{2} \) by expanding \( (1 + x)^{1/2} \) where \( x = 1 \).

  1. Write \( \sqrt{2} = (1 + 1)^{1/2} \).
  2. Use the binomial series to expand \( (1 + x)^{1/2} \) up to 4 terms.
  3. Calculate the approximate value.
▶️ View Solution

1.The binomial series for \( (1 + x)^{1/2} \) is: $ (1 + x)^{1/2} = 1 + \frac{1}{2}x – \frac{1}{8}x^2 + \frac{1}{16}x^3 – \cdots $ where $ \binom{1/2}{0} = 1, \quad \binom{1/2}{1} = \frac{1}{2}, \quad \binom{1/2}{2} = \frac{1/2 \times (-1/2)}{2} = -\frac{1}{8}, \quad \binom{1/2}{3} = \frac{1/2 \times (-1/2) \times (-3/2)}{6} = \frac{1}{16} $

2. Substitute \( x = 1 \): $ \sqrt{2} \approx 1 + \frac{1}{2}(1) – \frac{1}{8}(1)^2 + \frac{1}{16}(1)^3 = 1 + 0.5 – 0.125 + 0.0625 = 1.4375 $

3. The actual value of \( \sqrt{2} \) is approximately 1.4142, so this approximation is close and improves with more terms.

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