1.7.1 Binomial expansion

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Question 77

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]

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]

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

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

[2]

1.7.1 Binomial expansion

Question 77

Markscheme

1.7.1 Binomial expansion

Question 77

Markscheme

1.7.1 Binomial expansion

Question 77

1.7.1 Binomial expansion Questions

1.7.1 Binomial expansion

Question 77

1.7.1 Binomial expansion Questions