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3.3 Sequences and series (A-level only)

3.3 Sequences and series (A-level only)

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Question 98
a.

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

[2]
bi.

A cybersecurity firm is testing nnn 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 n2−5n−126=0n^2 - 5n - 126 = 0n2−5n−126=0.

[3]
bii.

Hence, determine the number of encryption keys nnn.

[2]
Markscheme

3.3 Sequences and series (A-level only) Questions

  1. A Level
  2. /Maths
  3. /3.3 Sequences and series (A-level only)

200 exam-style questions on CCEA A Level Maths 3.3 Sequences and series (A-level only), covering 3.3.1 Sequences and series (A-level only), 3.3.2 Sequences and series (A-level only), 3.3.3 Sequences and series (A-level only), 3.3.4 Sequences and series (A-level only), 3.3.5 Sequences and series (A-level only), 3.3.6 Sequences and series (A-level only), 3.3.7 Sequences and series (A-level only), and 3.3.8 Sequences and series (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.

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