Home / IBDP Maths AHL 1.10 permutations and combinations AA HL Paper 1- Exam Style Questions

IBDP Maths AHL 1.10 permutations and combinations AA HL Paper 1- Exam Style Questions

IBDP Maths AHL 1.10 permutations and combinations AA HL Paper 1- Exam Style Questions- New Syllabus

IBDP Maths AA HL - Paper 1 - All Topics
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) The number of ways to assign the students is given by the binomial coefficient:

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

Explanation:

  • We must choose exactly 3 students out of $n$ for the first group
  • The remaining $n-3$ students form the second group
  • Since $n-3 \geq 3$ (from the problem statement), we have $n \geq 6$

(b) With the restriction that two specific students must be in different groups:

Method 1: Count valid assignments directly

  1. Choose which of the two special students goes in Group 1: $^2C_1$ ways
  2. Choose 2 more students from the remaining $n-2$ for Group 1: $^{n-2}C_2$ ways
  3. Total valid assignments: $^2C_1 \times ^{n-2}C_2$

Since this is half the original number of assignments:

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

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

$n^2 – 13n + 36 = 0$

$(n-9)(n-4) = 0$

Since $n \geq 6$, the solution is $n = 9$.

Method 2: Alternative approach

Alternatively, we could consider cases where the two students are together or apart:

$\frac{1}{2}\binom{n}{3} = \binom{n-2}{3} + \binom{n-2}{1}$

This also leads to the same solution of $n = 9$.

Final Answer: $n = \boxed{9}$

Question:

A farmer has six sheep pens, arranged in a grid with three rows and two columns.
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]
(b) Each pen may only contain one sheep. Amber and Brownie must not be placed in pens which share a boundary. [4]

▶️ Answer/Explanation
Solution

(a) Each sheep chooses one of 6 pens. Total ways without restriction: \( 6^5 \).
Ways Amber (A) and Brownie (B) are together: Both in same pen (6 choices), others in any pen: \( 6 \cdot 6^3 \).
Ways A and B are apart: \( 6^5 – 6 \cdot 6^3 = 6^4 (6 – 1) = 5 \cdot 6^4 = 6480 \).

(b) 5 sheep in 5 of 6 pens (1 pen empty), each with 1 sheep. Total ways: \( 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 = 720 \).
Ways A and B share a boundary (adjacent pens):
– 4 vertical edges (e.g., top row left-right): Choose edge (4), place A/B (2 ways), others in 4 pens (4!): \( 4 \cdot 2 \cdot 4! = 192 \).
– 3 horizontal edges (e.g., left column top-middle): Choose edge (3), place A/B (2), others (4!): \( 3 \cdot 2 \cdot 4! = 144 \).
Total boundary ways: \( 192 + 144 = 336 \).
Ways A and B not adjacent: \( 720 – 336 = 384 \).

Markscheme
(a)
Total ways: \( 6^5 \quad \mathbf{A1} \)
Ways A and B together: \( 6 \cdot 6^3 \quad \mathbf{A1} \)
Subtract: \( 6^5 – 6 \cdot 6^3 \quad \mathbf{M1} \)
Result: \( 5 \cdot 6^4 = 6480 \quad \mathbf{A1} \)
(b)
Total ways: \( 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 = 720 \quad \mathbf{A1} \)
Ways A and B adjacent: \( (4 \cdot 2 + 3 \cdot 2) \cdot 4! = 336 \quad \mathbf{A1} \)
Subtract: \( 720 – 336 \quad \mathbf{M1} \)
Result: \( 384 \quad \mathbf{A1} \)
[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]
(b)
(i) Explain why the total number of possible outcomes for the results of the 6 games is 64. [1]
(ii) By expanding \((1 + x)^6\) and choosing a suitable value for \(x\), prove
\[ 64 = \binom{6}{0} + \binom{6}{1} + \binom{6}{2} + \binom{6}{3} + \binom{6}{4} + \binom{6}{5} + \binom{6}{6} \] [2]
(iii) State the meaning of this equality in the context of the 6 games played. [1]
(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 \(\binom{6}{r}^2 \left( \frac{2}{3} \right)^s \left( \frac{1}{3} \right)^t\), where the values of \(r\), \(s\), and \(t\) are to be found. [3]
(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. [3]
(iii) Hence prove that \(\binom{12}{6} = \binom{6}{0}^2 + \binom{6}{1}^2 + \binom{6}{2}^2 + \binom{6}{3}^2 + \binom{6}{4}^2 + \binom{6}{5}^2 + \binom{6}{6}^2\). [2]
(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 \(\text{E}(A) = \sum_{r=0}^n r \binom{n}{r} \frac{a^r}{b^n}\).
(i) Find the values of \(a\) and \(b\). [2]
(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
Solution

(a) Binomial distribution: \( A \sim B\left(6, \frac{2}{3}\right) \).
Probability Alfred wins 4 games: \( P(A=4) = \binom{6}{4} \left( \frac{2}{3} \right)^4 \left( \frac{1}{3} \right)^2 \).
Calculate: \( \binom{6}{4} = 15 \), \( \left( \frac{2}{3} \right)^4 = \frac{16}{81} \), \( \left( \frac{1}{3} \right)^2 = \frac{1}{9} \).
\( P(A=4) = 15 \cdot \frac{16}{81} \cdot \frac{1}{9} = 15 \cdot \frac{16}{729} = \frac{240}{729} = \frac{80}{243} \).

(b)
(i) Each game has 2 outcomes (Alfred wins or loses). For 6 games: \( 2^6 = 64 \) outcomes.
(ii) Expand: \( (1 + x)^6 = \sum_{r=0}^6 \binom{6}{r} x^r \).
Set \( x = 1 \): \( (1 + 1)^6 = 2^6 = 64 = \binom{6}{0} + \binom{6}{1} + \binom{6}{2} + \binom{6}{3} + \binom{6}{4} + \binom{6}{5} + \binom{6}{6} \).
(iii) The equality represents the total number of possible outcomes (64), as the sum of ways Alfred wins \( r \) games for \( r = 0 \) to 6.

(c)
(i) Probability Alfred wins 4 games on day 1: \( \binom{6}{4} \left( \frac{2}{3} \right)^4 \left( \frac{1}{3} \right)^2 \).
Probability 2 games on day 2: \( \binom{6}{2} \left( \frac{2}{3} \right)^2 \left( \frac{1}{3} \right)^4 \).
Combined: \( \binom{6}{4} \binom{6}{2} \left( \frac{2}{3} \right)^{4+2} \left( \frac{1}{3} \right)^{2+4} = \binom{6}{2}^2 \left( \frac{2}{3} \right)^6 \left( \frac{1}{3} \right)^6 \).
Parameters: \( r = 2 \), \( s = 6 \), \( t = 6 \).
(ii) Probability of 6 total wins: Sum over pairs \( (k, 6-k) \):
\( P(0,6) + P(1,5) + P(2,4) + P(3,3) + P(4,2) + P(5,1) + P(6,0) = \sum_{k=0}^6 \binom{6}{k}^2 \left( \frac{2}{3} \right)^6 \left( \frac{1}{3} \right)^6 = \frac{2^6}{3^{12}} \sum_{k=0}^6 \binom{6}{k}^2 \).
(iii) Probability of 6 wins in 12 games: \( \binom{12}{6} \left( \frac{2}{3} \right)^6 \left( \frac{1}{3} \right)^6 = \frac{2^6}{3^{12}} \binom{12}{6} \).
From (c)(ii): \( \frac{2^6}{3^{12}} \sum_{k=0}^6 \binom{6}{k}^2 = \frac{2^6}{3^{12}} \binom{12}{6} \implies \binom{12}{6} = \sum_{k=0}^6 \binom{6}{k}^2 \).

(d)
(i) \( \text{E}(A) = \sum_{r=0}^n r \binom{n}{r} \left( \frac{2}{3} \right)^r \left( \frac{1}{3} \right)^{n-r} = \sum_{r=0}^n r \binom{n}{r} \frac{2^r}{3^n} \).
Thus, \( a = 2 \), \( b = 3 \).
(ii) Differentiate: \( (1 + x)^n = \sum_{r=0}^n \binom{n}{r} x^r \implies n (1 + x)^{n-1} = \sum_{r=1}^n r \binom{n}{r} x^{r-1} \).
Set \( x = 2 \): \( n (1 + 2)^{n-1} = n \cdot 3^{n-1} = \sum_{r=1}^n r \binom{n}{r} 2^{r-1} \).
Multiply by 2, divide by \( 3^n \): \( \frac{2 \cdot n \cdot 3^{n-1}}{3^n} = \frac{2n}{3} = \sum_{r=1}^n r \binom{n}{r} \frac{2^r}{3^n} = \text{E}(A) \).

Markscheme
(a)
Binomial model: \( B\left(6, \frac{2}{3}\right) \quad \mathbf{M1} \)
Probability: \( \binom{6}{4} \left( \frac{2}{3} \right)^4 \left( \frac{1}{3} \right)^2 \quad \mathbf{A1} \)
Result: \( \frac{80}{243} \quad \mathbf{A1} \)
(b)
(i) \( 2^6 = 64 \quad \mathbf{R1} \)
(ii) Expansion: \( (1 + x)^6 = \sum_{r=0}^6 \binom{6}{r} x^r \), set \( x = 1 \quad \mathbf{A1} \)
Result: \( 64 = \sum_{r=0}^6 \binom{6}{r} \quad \mathbf{R1} \)
(iii) Sum of outcomes for 0 to 6 wins \quad \mathbf{R1} \)
(c)
(i) Day 1 and day 2 probabilities: \( \binom{6}{4} \left( \frac{2}{3} \right)^4 \left( \frac{1}{3} \right)^2 \cdot \binom{6}{2} \left( \frac{2}{3} \right)^2 \left( \frac{1}{3} \right)^4 \quad \mathbf{M1A1} \)
Form: \( \binom{6}{2}^2 \left( \frac{2}{3} \right)^6 \left( \frac{1}{3} \right)^6 \), \( r=2 \), \( s=t=6 \quad \mathbf{A1} \)
(ii) Sum: \( P(0,6) + \dots + P(6,0) = \frac{2^6}{3^{12}} \sum_{k=0}^6 \binom{6}{k}^2 \quad \mathbf{M1A2} \)
(iii) Equate: \( \frac{2^6}{3^{12}} \binom{12}{6} = \frac{2^6}{3^{12}} \sum_{k=0}^6 \binom{6}{k}^2 \quad \mathbf{A1} \)
Result: \( \binom{12}{6} = \sum_{k=0}^6 \binom{6}{k}^2 \quad \mathbf{A1} \)
(d)
(i) \( \text{E}(A) = \sum_{r=0}^n r \binom{n}{r} \frac{2^r}{3^n} \), \( a=2 \), \( b=3 \quad \mathbf{M1A1} \)
(ii) Differentiate: \( n (1 + x)^{n-1} = \sum_{r=1}^n r \binom{n}{r} x^{r-1} \quad \mathbf{A1A1} \)
Set \( x = 2 \): \( n \cdot 3^{n-1} = \sum_{r=1}^n r \binom{n}{r} 2^{r-1} \quad \mathbf{M1} \)
Adjust: \( \frac{2 \cdot n \cdot 3^{n-1}}{3^n} = \frac{2n}{3} \quad \mathbf{M1} \)
Result: \( \text{E}(A) = \frac{2n}{3} \quad \mathbf{A2} \)
[20 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]

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

▶️ Answer/Explanation
Solution

(a) [3 marks]

To find the number of selections where the largest number is 5, 6, or 7:

  1. Largest is 5: We must choose 3 additional numbers from {1,2,3,4} ⇒ \(^4C_3 = 4\) ways
  2. Largest is 6: Choose 3 from {1,2,3,4,5} ⇒ \(^5C_3 = 10\) ways
  3. Largest is 7: Choose 3 from {1,2,3,4,5,6} ⇒ \(^6C_3 = 20\) ways

Total selections = \(4 + 10 + 20 = \boxed{34}\)

(b) [4 marks]

To find selections with at least two even numbers (even numbers in the set: {2,4,6,8}):

Method 1: Direct Counting

Count all valid cases:

  • 2 even + 2 odd: \(^4C_2 \times ^5C_2 = 6 \times 10 = 60\)
  • 3 even + 1 odd: \(^4C_3 \times ^5C_1 = 4 \times 5 = 20\)
  • 4 even + 0 odd: \(^4C_4 = 1\)

Total = \(60 + 20 + 1 = \boxed{81}\)

Method 2: Complementary Counting

Subtract invalid cases from total:

  • Total selections: \(^9C_4 = 126\)
  • 0 even + 4 odd: \(^5C_4 = 5\)
  • 1 even + 3 odd: \(^4C_1 \times ^5C_3 = 4 \times 10 = 40\)

Valid selections = \(126 – 5 – 40 = \boxed{81}\)

Total marks: [7]

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