IBDP Maths analysis and approaches Topic: AHL 1.10 permutations and combinations HL Paper 1

Question

A teacher takes $n$ students on a field trip. The students are assigned randomly into two groups.

For safety reasons there must be exactly three students in the first group and at least three students in the second group.

The teacher will randomly assign three students to the first group and the other students to the second group.

(a) Write down an expression for the number of ways that the students could be assigned.

Two of the students ask the teacher not to work in the same group.

The teacher agrees and now finds that the number of ways to assign the students is halved.

(b) Determine the value of $n$.

▶️Answer/Explanation

Solution:-

(a) \binom{n}{3}

(b) EITHER

finding the number of ways to assign the students with the two students apart

valid attempt to eliminate all factorials from their equation

$\frac{n(n-1)(n-2)}{3 \times 2} = 2 \times \frac{(n-2)(n-3)}{2}$ or equivalent with no factorials

$n(n-1) = 12(n-3)$

finding the number of ways to assign the students with the two students together

number of ways to assign two students and one other to the first group: $^{n}C_{3}$
(seen anywhere)

number of ways to assign three other students the first group: $^{n-2}C_{3}$ (seen anywhere)

number of ways = $^{n}C_{3} + ^{n-2}C_{3}$

attempt to set up an equation involving either half of their answer to part (a) and their number of ways or their answer to part (a) is twice their number of ways
$\frac{1}{2}^{n}C_{3} = ^{n-2}C_{3} + ^{n-2}C_{3}$ OR $^{n}C_{3} = 2(^{n-2}C_{1} + ^{n-2}C_{3})$

valid attempt to eliminate all factorials from their equation
$\frac{n(n-1)(n-2)}{3 \times 2} = 2(n-2) + 2\frac{(n-2)(n-3)(n-4)}{3 \times 2}$
$n(n-1) = 12 + 2(n-3)(n-4)$
$n^{2} – 13n + 36 = 0$
$(n-9)(n-4) = 0$

THEN

$n = 9$

Solution

(a) Given data

- Total students: n

- First group: Exactly 3 students

- Second group: Remaining students (), with at least 3 students

- Assignment: Randomly choose 3 students for the first group; the rest go to the second group

For the second group to have at least 3 students:

n – 3 \geq 3

n \geq 6

So, n must be at least 6, which makes sense for a field trip with safety constraints Since the groups are distinct (first group vs. second group), we’re choosing 3 students out of n for the first group, and the remaining n − 3 automatically form the second group. The number of ways to choose 3 students from n is given by the binomial coefficient:

(\binom{n}{3}) = \frac{n!}{3! (n - 3)!} = \frac{n (n - 1) (n - 2)}{6}

- Order within each group doesn’t matter (it’s a group assignment, not a seating arrangement).

The condition
$n - 3 \geq 3ensures the second group’s size is valid.$

(b) First we find the number of ways to assign the students with the two students apart

Solving the equation we get

$\frac{n(n-1)(n-2)}{3 \times 2} = 2 \times\frac{(n-2)(n-3)}{2}$

$n(n-1) = 12(n-3)$

This equation is true for n = 4 and n=9, but from case (a) n≥6, therefore n=9 is the answer.

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]

Question

A set of positive integers {$1,2,3,4,5,6,7,8,9$} is used to form a pack of nine cards.

Each card displays one positive integer without repetition from this set. Grace wishes to select four cards at random from this pack of nine cards.

a.Find the number of selections Grace could make if the largest integer drawn among the four cards is either a 5, a 6 or a 7.[3]

▶️Answer/Explanation

Ans: use of the addition principle with 3 terms (M1)

to obtain $^4{C_3}{ + ^5}{C_3}{ + ^6}{C_3}{\text{ }}( = 4 + 10 + 20)$ A1

number of possible selections is $34$ A1

[3 marks]

b. Find the number of selections Grace could make if at least two of the four integers drawn are even.[4]

▶️Answer/Explanation

EITHER

recognition of three cases: ($2$ odd and $2$ even or $1$ odd and $3$ even or $0$ odd and $4$ even) (M1)

$\left( {^5{C_2}{ \times ^4}{C_2}} \right) + \left( {^5{C_1}{ \times ^4}{C_3}} \right) + \left( {^5{C_0}{ \times ^4}{C_4}} \right)\;\;\;( = 60 + 20 + 1)$ (M1)A1

recognition to subtract the sum of $4$ odd and $3$ odd and $1$ even from the total (M1)

$^9{C_4}{ – ^5}{C_4} – \left( {^5{C_3}{ \times ^4}{C_1}} \right)\;\;\;( = 126 – 5 – 40)$ (M1)A1

THEN

number of possible selections is $81$ A1

[4 marks]

Total [7 marks]

IBDP Maths analysis and approaches Topic: AHL 1.10 permutations and combinations HL Paper 1

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