Using nCr=n!r!(n−r)!{}^nC_r = \frac{n!}{r!(n-r)!}nCr=r!(n−r)!n!, show 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 researcher is selecting distinct plant species from a population of nnn available species for a DNA sequencing study. Show that the equation
2×nC4=15×nC2 2 \times {}^nC_4 = 15 \times {}^nC_2 2×nC4=15×nC2simplifies to
n2−5n−84=0 n^2 - 5n - 84 = 0 n2−5n−84=0Hence, solve the equation
2×nC4=15×nC2 2 \times {}^nC_4 = 15 \times {}^nC_2 2×nC4=15×nC2Practise AQA A Level Maths 1.7 D: Sequences and series with exam-style questions for A Level Maths. 267 questions covering 1.7.1 Binomial expansion, 1.7.2 Types of sequence (A-level only), 1.7.3 Sigma notation (A-level only), 1.7.4 Arithmetic sequences and series (A-level only), 1.7.5 Geometric sequences and series (A-level only), and 1.7.6 Sequences and series in modelling (A-level only), matched to the AQA A Level Maths (7357) specification and written in Paper 1, Paper 2 and Paper 3 style. Every question includes a full worked solution and mark scheme, so you can see where marks are awarded rather than just whether you got the answer right.
