Everything from the 2 lessons of this topic on a few pages: the key idea of each lesson, the diagram to remember, the points to learn and the traps examiners set. Read it, cover it, test yourself.
\(n\) different objects can be arranged in a line in \(n!\) ways. Divide by \(p!\) for each group of \(p\) identical objects. Glue objects that must be together into one unit; for objects that must be apart, subtract the 'together' count from the total or slot them into gaps.
Dividing by \(3! + 2!\) instead of \(3! \times 2!\). Each repeated group contributes its own factorial to the product in the denominator.
Forgetting the internal arrangements of the block. With different objects inside, multiply by their factorial; with identical objects inside (EEE), there is nothing to multiply.
Using 'total minus together' for three or more people who must all be apart. Subtracting 'all three together' leaves arrangements where two of them are still next to each other. Use the gaps method instead.
When two restrictions overlap (first digit and last digit both even), split into cases. A single product \(3 \times 3 \times 4 \times 3\) double-uses digits like 4 and 6.
The number of ways of choosing \(r\) objects from \(n\) different objects, when order does not matter, is \({}^{n}C_{r} = \dbinom{n}{r} = \dfrac{n!}{r!\,(n - r)!}\). Since each selection can be ordered in \(r!\) ways, \({}^{n}P_{r} = {}^{n}C_{r} \times r!\).
Using \({}^{n}P_{r}\) for a committee. Ordered counting counts each committee \(r!\) times over: \({}^{10}P_{4} = 5040\) is 24 times too big.
Counting 'at least one woman' as \({}^{4}C_{1} \times {}^{8}C_{3}\): choose one woman, then any 3 of the rest. A group with two women is then counted twice, once for each woman picked 'first'.
Subtracting only the committees with neither Khan (462). The forbidden case is BOTH Khans, not neither.
Fixing C in the first position (\({}^{7}P_{3} = 210\)). 'Includes C' means C may be in any of the four places.