Permutations, Combinations, and the Binomial Theorem
Permutations, Combinations, and the Binomial Theorem
Counting problems hinge on one decision: does order matter? If yes, use permutations; if no, use combinations.
Concepts. The fundamental counting principle multiplies choices across independent stages. Permutations of $n$ objects taken $r$ at a time are ${}_nP_r=\dfrac{n!}{(n-r)!}$; combinations are ${}nC_r=\dfrac{n!}{r!(n-r)!}$. Permutations with identical items divide by the factorial of each repetition count, as in the number of distinct arrangements of the letters of a word. Circular arrangements of $n$ distinct objects number $(n-1)!$. The binomial theorem expands $(a+b)^n=\sum{r=0}^{n}{}_nC_r,a^{n-r}b^r$, and the general term ${}_nC_r a^{n-r}b^r$ locates any single term without expanding everything.
Traps. The most frequent error is choosing a permutation when the selection is unordered, or vice versa. "At least one" problems are usually easiest by complementary counting: total minus none. Students forget that ${}_nC_r={}nC{n-r}$, which can simplify arithmetic. In binomial expansions, the exponents of $a$ and $b$ must always sum to $n$, and the term containing a specific power is found by solving for $r$, not by counting terms from the front. Remember the first term uses $r=0$.
Rule of thumb. Ask "would swapping two chosen items create a different outcome?" If yes, it is a permutation.
Worked example. From 8 students, choose a committee of 3: ${}_8C_3=\dfrac{8!}{3!5!}=56$. If instead the 3 are president, secretary, and treasurer, use ${}_8P_3=8\cdot7\cdot6=336$. For the $x^4$ term of $(2x+3)^6$, solve $6-r=4$, so $r=2$: ${}_6C_2(2x)^4(3)^2=15\cdot16x^4\cdot9=2160x^4$.