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×nC2308 exam-style questions on OCR A Level Maths 1.4 Sequences and Series, covering 1.4.1 Binomial expansion for positive integer n, 1.4.2 Link to binomial probabilities, 1.4.3 Binomial expansion for rational n (A-level only), 1.4.4 Validity of the expansion (A-level only), 1.4.5 Sequences (A-level only), 1.4.6 Increasing, decreasing and periodic sequences (A-level only), 1.4.7 Sigma notation (A-level only), 1.4.8 Arithmetic sequences and series (A-level only), 1.4.9 Geometric sequences and series (A-level only), 1.4.10 Sum to infinity of a geometric series (A-level only), and 1.4.11 Modelling with sequences and series (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.