Permutations, Combinations and Binomial Theorem
The exam checks counting with arrangement and selection formulae and the general term of a binomial expansion, especially finding a specified coefficient.
Counting rules
- nPr = n!/(n−r)!, nCr = n!/(r!(n−r)!), with nCr = nC(n−r) and nCr + nC(r−1) = (n+1)Cr.
- Arrangements of n objects with repeats p, q: n!/(p!q!). Circular arrangements of n distinct objects: (n−1)!.
- "Arrangement/order matters" → permutation; "selection/committee" → combination.
Binomial theorem
(a+b)ⁿ = Σ nCr a^(n−r) b^r, with general term T(r+1) = nCr a^(n−r) b^r. Sum of binomial coefficients = 2ⁿ. For the middle term: one middle term when n is even, at r = n/2; two when n is odd.
Worked example
Find the term independent of x in (2x² + 1/x)⁹. General term = 9Cr (2x²)^(9−r) (1/x)^r = 9Cr 2^(9−r) x^(18−3r). For the term to be free of x, 18 − 3r = 0, so r = 6. Coefficient = 9C6 · 2³ = 84 × 8 = 672.
Traps that cost marks
- Omitting the constant factor: forgetting 2^(9−r) or the coefficient of x inside the bracket.
- Solving the exponent equation for n instead of r.
- Using nCr where order matters (seating in a row is nPr, a handshake count is nCr).
- "At least one" questions done case-by-case instead of by complement (total − none).
- nCr is defined only for 0 ≤ r ≤ n.