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3.4.1 Sequences and Series (A-level only)

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Question 38

Using the formula $nCr=n!r!(n−r)!{}^nC_r = \frac{n!}{r!(n-r)!}$ , prove that $nC4=n(n−1)(n−2)(n−3)24{}^nC_4 = \frac{n(n-1)(n-2)(n-3)}{24}$ .

[2]

bi.

A cybersecurity firm is testing $n$ distinct encryption keys. A 'Quad-Lock' configuration is formed by selecting a subset of 4 keys, while a 'Dual-Lock' configuration is formed by selecting a subset of 2 keys.

Given that the number of possible Quad-Lock configurations is exactly 11 times the number of possible Dual-Lock configurations, show that $n^2 - 5n - 126 = 0$ .

[3]

bii.

Hence, determine the number of encryption keys $n$ .

[2]

Markscheme

3.4.1 Sequences and Series (A-level only) Questions

A Level
Maths
3.4.1 Sequences and Series (A-level only)

43 exam-style questions on WJEC A Level Maths 3.4.1 Sequences and Series (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.

3.4.1 Sequences and Series (A-level only)

Question bank

3.4.1 Sequences and Series (A-level only)

Easy Medium

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42

Question 38

Using the formula $nCr=n!r!(n−r)!{}^nC_r = \frac{n!}{r!(n-r)!}$ , prove that $nC4=n(n−1)(n−2)(n−3)24{}^nC_4 = \frac{n(n-1)(n-2)(n-3)}{24}$ .

[2]

bi.

Given that the number of possible Quad-Lock configurations is exactly 11 times the number of possible Dual-Lock configurations, show that $n^2 - 5n - 126 = 0$ .

[3]

bii.

Hence, determine the number of encryption keys $n$ .

[2]

Markscheme

3.4.1 Sequences and Series (A-level only) Questions

A Level
Maths
3.4.1 Sequences and Series (A-level only)

43 exam-style questions on WJEC A Level Maths 3.4.1 Sequences and Series (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.