IB Mathematics AA HL Flashcards- Permutations and Combinations

PERMUTATIONS// This refers, in simple terms, to the number of ways to arrange a certain number of objects (n), into a certain number of positions (r).

NOTATION// \( nPr \) (n objects into r positions)

L\(_{n}\) or \( nPr \) or \( nPr \) or \( P(n, r) \)

[q]

CALCULATION// Using GDC (see 1.9 for buttons, but press \( nPr \))

L\(_{n}\) or → Formula:

\[
nPr = \frac{n!}{(n-r)!} \quad [n \geq r] \rightarrow \text{in F.B.}
\]

[a]

E.G.1 // Arrange 5 books into 5 spaces. How many ways to do this?

\[
5P5 = \frac{5!}{(5-5)!} = \frac{5!}{1} = 5! = 120
\]

[q]

E.G.2 // 15 cars are in a race. How many ways can top 6 positions be filled?

\[
15P6 = \frac{15!}{(15-6)!} = 3,603,600
\]

[a]

COMBINATIONS// This has come up in *binomial expansion* already, but here we will see questions that involve calculating combinations by themselves → choosing r objects out of a total of n objects, with the order not being an issue.

NOTATION// \( nCr \) or \( \binom{n}{r} \)

[q]

CALCULATION// GDC (see 1.9) or

\[
nCr = \frac{n!}{r!(n-r)!} \quad \text{in F.B.}
\]

[a]

E.G.3 // A lottery involves choosing 6 numbers from 1 to 45. How many possibilities are there to choose from?

\[
45C6 = \binom{45}{6} = 8,145,060
\]

[q]

E.G.4 // Compute the probability of randomly drawing five cards from a deck and getting exactly one king:

Note: we need to know some concepts from topic 4 for this.

[a]

So, the main concept is that we need to do

\[
\frac{\text{no. of ways of getting this}}{\text{total no. of 5-card decks}}
\]

[q]

We need to choose 1 out of the 4 kings, then choose any group of 4 cards out of the remaining 48 cards:

\[
\therefore P(1 \text{ king}) = \frac{(4C1)(48C4)}{52C5} = \frac{778320}{2598960} \approx 0.299
\]

[a]

1// We have seen how to expand \((a+b)^n\), but only with \(n \in \mathbb{Z}^+\), so when \(n\) is a positive integer. In A&A HL, this will be extended to when \(n\) is a fraction or a negative.

[q]

NEGATIVE EXPONENTS// If we consider the expansion of \((1 + x)^{-4}\), you may know that this equals

\[
\frac{1}{(1 + x)^4} = \frac{1}{1 + 4x + 6x^2 + 4x^3 + x^4}
\]

but this is not technically a polynomial, but you can find something slightly closer to one. This requires some complex calculus, which you could look into in 5.19.

[a]

We can also just consider the original way of expanding \((1 + x)^n\), which is:

\[
1^n + \binom{n}{1}x^1 + \binom{2}{1}x^2 + \ldots = 1 + nx + \frac{n(n-1)}{2!} x^2 + \frac{n(n-1)(n-2)}{3!} x^3 + \ldots
\]

This actually still applies for negatives, although this will just be an infinite series.

[q]

E.G.1 // Find the expansion of \(3(1-4x)^{-1}\) up to the \(x^3\) term:

Use the formula above:

\[
3 \left(1 + (-1)(-4x) + \frac{(-1)(-2)}{2!}(-4x)^2 + \frac{(-1)(-2)(-3)}{3!}(-4x)^3 + \ldots\right)
\]

\[
= 3 \left(1 + 4x + 16x^2 + 64x^3 + \ldots\right) = 3 + 12x + 48x^2 + 192x^3 + \ldots
\]

[a]

FRACTIONAL EXPONENTS// We can still use the formula above. Also, you can only have \(|x|<1\), because if \(|x|\geq1\), squaring/cubic accelerates it, so it diverges.

[q]

E.G.2 // Expand \((1-x)^{\frac{3}{2}}\) to \(x^3\) term, stating which values of \(x\) for which it is valid:

Use formula:

\[
(1-x)^{\frac{3}{2}} = 1 + \left(\frac{3}{2}\right)(-x) + \frac{\left(\frac{3}{2}\right)\left(\frac{3}{2}-1\right)}{2!}(-x)^2 + \frac{\left(\frac{3}{2}\right)\left(\frac{3}{2}-1\right)\left(\frac{3}{2}-2\right)}{3!}(-x)^3 + \ldots
\]

\[
= 1 – \frac{3}{2}x – \frac{3}{8}x^2 – \frac{5}{48}x^3 + \ldots
\]

[a]

We need \(1-x < 1\), which just gives us \(|x| < 1\).

NOTE// You need 1 as the constant in the bracket, and if you are given a different constant, you should factor it out to leave you with a 1:

[q]

E.G.3 // Expand \((4 + x)^{\frac{1}{2}}\), stating the valid values for \(x\):

\[
(4 + x)^{\frac{1}{2}} = \left[4\left(1 + \frac{x}{4}\right)\right]^{\frac{1}{2}} = 2\left(1 + \frac{x}{4}\right)^{\frac{1}{2}} = 2\left(1 + \left(\frac{1}{2}\right)\left(\frac{x}{4}\right) – \frac{\left(\frac{1}{2}\right)\left(\frac{-1}{2}\right)}{2!}\left(\frac{x}{4}\right)^2 + \ldots\right)
\]

\[
= 2 \left(1 + \frac{x}{8} – \frac{x^2}{128} + \ldots \right) = 2 + \frac{x}{4} – \frac{x^2}{64} + \ldots
\]

\[
\left|\frac{x}{4}\right| < 1, \quad |x| < 4
\]

[a]

E.G.4 // Expand \((2+3x)^{-2}\):

\[
(2+3x)^{-2} = \left[2\left(1 + \frac{3x}{2}\right)\right]^{-2} = \frac{1}{4}\left(1 + \frac{3x}{2}\right)^{-2} = \frac{1}{4}\left(1 + \left(-2\right)\left(\frac{3x}{2}\right) + \frac{\left(-2\right)\left(-3\right)}{2!}\left(\frac{3x}{2}\right)^2 + \ldots\right)
\]

\[
= \frac{1}{4}\left(1 – 3x + \frac{9x^2}{2} – \frac{27x^3}{4} + \ldots\right) = \frac{1}{4} – \frac{3x}{4} + \frac{9x^2}{8} – \frac{27x^3}{16} + \ldots
\]

\[
\left|\frac{3x}{2}\right| < 1, \quad |x| < \frac{2}{3}
\]

[x] Exit text