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).
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.
Hence, determine the number of encryption keys nnn.
231 exam-style questions on WJEC A Level Maths 3.4 Sequences and Series (A-level only), covering 3.4.1 Sequences and Series (A-level only), 3.4.2 Sequences and Series (A-level only), 3.4.3 Sequences and Series (A-level only), 3.4.4 Sequences and Series (A-level only), 3.4.5 Sequences and Series (A-level only), 3.4.6 Sequences and Series (A-level only), and 3.4.7 Sequences and Series (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.