Combinatorics: Permutations, Combinations and the Binomial Theorem
Core concepts
Know the multiplication and addition principles, factorials, $^nP_r=\frac{n!}{(n-r)!}$ for ordered arrangements and $\binom{n}{r}=\frac{n!}{r!(n-r)!}$ for unordered selections. Understand complementary counting, arrangements with identical items ($\frac{n!}{a!b!\cdots}$), circular arrangements ($(n-1)!$), and restricted arrangements handled by cases or by the gap method. The binomial theorem gives $(a+b)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r$, with general term $\binom{n}{r}a^{n-r}b^r$.
Traps
- Using permutations when order does not matter (or vice versa).
- Forgetting to divide by factorials for repeated letters in a word.
- "At least one" problems are usually easiest by complement: total minus none.
- When two people must not sit together, count total minus arrangements with them together.
- In the general term, the exponent of $a$ is $n-r$, not $r$ - check which term is requested.
Rule of thumb
Ask two questions: does order matter, and can items repeat? That decides between $^nP_r$, $\binom{n}{r}$ and powers. For "at least one", compute $1-P(\text{none})$.
Worked example
From 8 students choose a committee of 3 with at least one prefect, given 2 prefects. Total committees $\binom{8}{3}=56$. Committees with no prefect: $\binom{6}{3}=20$. So the answer is $56-20=36$. Alternatively sum cases: $\binom{2}{1}\binom{6}{2}+\binom{2}{2}\binom{6}{1}=30+6=36$, confirming the complement method.