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

The study of permutations and combinations involves counting principles to determine the number of ways objects can be arranged or selected. This section explores the fundamentals, advanced concepts, and extensions to the binomial theorem.

Counting Principles

Counting principles are the foundation for permutations and combinations:

Permutations: The arrangement of \( n \) objects in a specific order.
- Total permutations of \( n \) objects: \( n! \).
- Permutations of \( n \) objects taken \( r \) at a time: \( P(n, r) = \frac{n!}{(n-r)!} \).
Combinations: The selection of \( r \) objects from \( n \) without regard to order.
- Total combinations: \( C(n, r) = \frac{n!}{r!(n-r)!} \).

Binomial Theorem Extensions

The binomial theorem provides a method to expand \( (a+b)^n \) for any rational \( n \in \mathbb{Q} \):

For positive integers \( n \), the expansion is:
\( (a + b)^n = \sum_{k=0}^{n} C(n, k) a^{n-k} b^k \).
For fractional or negative \( n \), the expansion is extended using power series:
\( (a + b)^n = a^n \left(1 + \frac{b}{a}\right)^n \), where:
- \( \left(1 + x\right)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots \).
- The series converges for \( |x| < 1 \).

Note: The proof of the binomial theorem is not required for this syllabus.

Applications and Connections

The binomial theorem links directly to power series expansions and advanced calculus topics (e.g., AHL 5.19).

Examples

Example 1: Permutations

Find the number of ways to arrange 4 books on a shelf:

Solution: Total arrangements = \( 4! = 24 \).

Example 2: Combinations

Find the number of ways to select 3 students from a group of 5:

Solution: \( C(5, 3) = \frac{5!}{3!(5-3)!} = 10 \).

Example 3: Binomial Expansion

Expand \( (2+x)^3 \):

Solution:
\( (2+x)^3 = C(3, 0)(2)^3(x)^0 + C(3, 1)(2)^2(x)^1 + C(3, 2)(2)^1(x)^2 + C(3, 3)(2)^0(x)^3 \).
\( = 8 + 12x + 6x^2 + x^3 \).

Example 4: Fractional Binomial Expansion

Expand \( (1+x)^{1/2} \) up to the \( x^3 \) term:

Solution:
\( (1+x)^{1/2} = 1 + \frac{1}{2}x – \frac{1}{8}x^2 + \frac{1}{16}x^3 + \dots \).

Question

A farmer has six sheep pens, arranged in a grid with three rows and two columns as shown in the following diagram.

Five sheep called Amber, Brownie, Curly, Daisy and Eden are to be placed in the pens. Each pen is large enough to hold all of the sheep. Amber and Brownie are known to fight. Find the number of ways of placing the sheep in the pens in each of the following cases:

(a) Each pen is large enough to contain five sheep. Amber and Brownie must not be placed in the same pen. [4]

▶️Answer/Explanation

Ans: METHOD 1
B has one less pen to select (M1)
EITHER
A and B can be placed in 6×5 ways (A1)
C, D, E have 6 choices each (A1)
OR
A (or B), C, D, E have 6 choices each (A1)
B (or A) has only 5 choices (A1)
THEN
5×6⁴ =6480
METHOD 2
total number of ways = 6⁵ (A1)
number of ways with Amber and Brownie together 46 = (A1)
attempt to subtract (may be seen in words) (M1)
6⁵-6⁴ = 5×6⁴ =6480 [4 marks]

(b) Each pen may only contain one sheep. Amber and Brownie must not be placed in pens which share a boundary. [4]

▶️Answer/Explanation

Ans METHOD 1
total number of ways = 6!(= 720) (A1)
number of ways with Amber and Brownie sharing a boundary
= 2x7x4!(= 336) (A1)
attempt to subtract (may be seen in words) (M1)
720 – 336 =384 A1
METHOD 2
case 1: number of ways of placing A in corner pen
3x4x3x2x1
Four corners total no of ways is 4x(3x4x3x2x1) =12×4!(= 288) (A1)
case 2: number of ways of placing A in the middle pen
2x4x3x2x1
two middle pens so 2x(2x4x3x2x1) = 4×4!(= 96) (A1)
attempt to add (may be seen in words) (M1)
total no of ways = 288+96
=16×4!(= 384) A1
[4 marks] Total [8 marks]

Question

On Saturday, Alfred and Beatrice play 6 different games against each other. In each game, one of the two wins. The probability that Alfred wins any one of these games is \(\frac{2}{3}\).

a. Show that the probability that Alfred wins exactly 4 of the games is \(\frac{{80}}{{243}}\).[3]

▶️Answer/Explanation

Ans:

\(B\left( {6,\frac{2}{3}} \right)\) (M1)

\(p(4) = \left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right){\left( {\frac{2}{3}} \right)^4}{\left( {\frac{1}{3}} \right)^2}\) A1

\(\left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right) = 15\) A1

\( = 15 \times \frac{{{2^4}}}{{{3^6}}} = \frac{{80}}{{243}}\) AG

[3 marks]

b. (i) Explain why the total number of possible outcomes for the results of the 6 games is 64.

▶️Answer/Explanation

Ans: 2 outcomes for each of the 6 games or \({2^6} = 64\) R1

(ii) By expanding \({(1 + x)^6}\) and choosing a suitable value for x, prove

\[64 = \left( {\begin{array}{*{20}{c}}
6 \\
0
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
1
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
2
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
3
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
5
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
6
\end{array}} \right)\]

▶️Answer/Explanation

Ans:

\({(1 + x)^6} = \left( {\begin{array}{*{20}{c}}
6 \\
0
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
1
\end{array}} \right)x + \left( {\begin{array}{*{20}{c}}
6 \\
2
\end{array}} \right){x^2} + \left( {\begin{array}{*{20}{c}}
6 \\
3
\end{array}} \right){x^3} + \left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right){x^4} + \left( {\begin{array}{*{20}{c}}
6 \\
5
\end{array}} \right){x^5} + \left( {\begin{array}{*{20}{c}}
6 \\
6
\end{array}} \right){x^6}\) A1

Note: Accept \(^n{C_r}\) notation or \(1 + 6x + 15{x^2} + 20{x^3} + 15{x^4} + 6{x^5} + {x^6}\)

setting x = 1 in both sides of the expression R1

Note: Do not award R1 if the right hand side is not in the correct form.

\(64 = \left( {\begin{array}{*{20}{c}}
6 \\
0
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
1
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
2
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
3
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
5
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
6
\end{array}} \right)\) AG

(iii) State the meaning of this equality in the context of the 6 games played.[4]

▶️Answer/Explanation

Ans: the total number of outcomes = number of ways Alfred can win no games, plus the number of ways he can win one game etc. R1

[4 marks]

c. The following day Alfred and Beatrice play the 6 games again. Assume that the probability that Alfred wins any one of these games is still \(\frac{2}{3}\).

(i) Find an expression for the probability Alfred wins 4 games on the first day and 2 on the second day. Give your answer in the form \({\left( {\begin{array}{*{20}{c}}
6 \\
r
\end{array}} \right)^2}{\left( {\frac{2}{3}} \right)^s}{\left( {\frac{1}{3}} \right)^t}\) where the values of r, s and t are to be found.

▶️Answer/Explanation

Ans:

Let \({\text{P}}(x,{\text{ }}y)\) be the probability that Alfred wins x games on the first day and y on the second.

\({\text{P(4, 2)}} = \left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right) \times {\left( {\frac{2}{3}} \right)^4} \times {\left( {\frac{1}{3}} \right)^2} \times \left( {\begin{array}{*{20}{c}}
6 \\
2
\end{array}} \right) \times {\left( {\frac{2}{3}} \right)^2} \times {\left( {\frac{1}{3}} \right)^4}\) M1A1

\({\left( {\begin{array}{*{20}{c}}
6 \\
2
\end{array}} \right)^2}{\left( {\frac{2}{3}} \right)^6}{\left( {\frac{1}{3}} \right)^6}\) or \({\left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right)^2}{\left( {\frac{2}{3}} \right)^6}{\left( {\frac{1}{3}} \right)^6}\) A1

r = 2 or 4, s = t = 6

(ii) Using your answer to (c) (i) and 6 similar expressions write down the probability that Alfred wins a total of 6 games over the two days as the sum of 7 probabilities.

▶️Answer/Explanation

Ans:

P(Total = 6) =

P(0, 6) + P(1, 5) + P(2, 4) + P(3, 3) + P(4, 2) + P(5, 1) + P(6, 0) (M1)

\( = {\left( {\begin{array}{*{20}{c}}
6 \\
0
\end{array}} \right)^2}{\left( {\frac{2}{3}} \right)^6}{\left( {\frac{1}{3}} \right)^6} + {\left( {\begin{array}{*{20}{c}}
6 \\
1
\end{array}} \right)^2}{\left( {\frac{2}{3}} \right)^6}{\left( {\frac{1}{3}} \right)^6} + … + {\left( {\begin{array}{*{20}{c}}
6 \\
6
\end{array}} \right)^2}{\left( {\frac{2}{3}} \right)^6}{\left( {\frac{1}{3}} \right)^6}\) A2

\( = \frac{{{2^6}}}{{{3^{12}}}}\left( {{{\left( {\begin{array}{*{20}{c}}
6 \\
0
\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}
6 \\
1
\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}
6 \\
2
\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}
6 \\
3
\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}
6 \\
5
\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}
6 \\
6
\end{array}} \right)}^2}} \right)\)

Note: Accept any valid sum of 7 probabilities.

(iii) Hence prove that \(\left( {\begin{array}{*{20}{c}}
{12} \\
6
\end{array}} \right) = {\left( {\begin{array}{*{20}{c}}
6 \\
0
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
1
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
2
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
3
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
5
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
6
\end{array}} \right)^2}\).[9]

▶️Answer/Explanation

Ans:

use of \(\left( {\begin{array}{*{20}{c}}
6 \\
i
\end{array}} \right) = \left( {\begin{array}{*{20}{l}}
6 \\
{6 – i}
\end{array}} \right)\) (M1)

(can be used either here or in (c)(ii))

P(wins 6 out of 12) \( = \left( {\begin{array}{*{20}{c}}
{12} \\
6
\end{array}} \right) \times {\left( {\frac{2}{3}} \right)^6} \times {\left( {\frac{1}{3}} \right)^6} = \frac{{{2^6}}}{{{3^{12}}}}\left( {\begin{array}{*{20}{c}}
{12} \\
6
\end{array}} \right)\) A1

therefore \({\left( {\begin{array}{*{20}{c}}
6 \\
0
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
1
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
2
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
3
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
5
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
6
\end{array}} \right)^2} = \left( {\begin{array}{*{20}{c}}
{12} \\
6
\end{array}} \right)\) AG

[9 marks]

d. Alfred and Beatrice play n games. Let A denote the number of games Alfred wins. The expected value of A can be written as \({\text{E}}(A) = \sum\limits_{r = 0}^n {r\left( {\begin{array}{*{20}{c}}

n \\
r
\end{array}} \right)} \frac{{{a^r}}}{{{b^n}}}\).

(i) Find the values of a and b.

▶️Answer/Explanation

Ans:

\({\text{E}}(A) = \sum\limits_{r = 0}^n {r\left( {\begin{array}{*{20}{c}}
n \\
r
\end{array}} \right)} {\left( {\frac{2}{3}} \right)^r}{\left( {\frac{1}{3}} \right)^{n – r}} = \sum\limits_{r = 0}^n {r\left( {\begin{array}{*{20}{c}}
n \\
r
\end{array}} \right)} \frac{{{2^r}}}{{{3^n}}}\)

(a = 2, b = 3) M1A1

Note: M0A0 for a = 2, b = 3 without any method.

(ii) By differentiating the expansion of \({(1 + x)^n}\), prove that the expected number of games Alfred wins is \(\frac{{2n}}{3}\).[6]

▶️Answer/Explanation

Ans:

\(n{(1 + x)^{n – 1}} = \sum\limits_{r = 1}^n {\left( {\begin{array}{*{20}{c}}
n \\
r
\end{array}} \right)} r{x^{r – 1}}\) A1A1

(sigma notation not necessary)

(if sigma notation used also allow lower limit to be r = 0)

let x = 2 M1

\(n{3^{n – 1}} = \sum\limits_{r = 1}^n {\left( {\begin{array}{*{20}{c}}
n \\
r
\end{array}} \right)} r{2^{r – 1}}\)

multiply by 2 and divide by \({3^n}\) (M1)

\(\frac{{2n}}{3} = \sum\limits_{r = 1}^n {\left( {\begin{array}{*{20}{c}}
n \\
r
\end{array}} \right)} r\frac{{{2^r}}}{{{3^n}}}\left( { = \sum\limits_{r = 0}^n {\left( {\begin{array}{*{20}{c}}
n \\
r
\end{array}} \right)} \frac{{{2^r}}}{{{3^n}}}} \right)\) AG

[6 marks]

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