Derivations of formulas and their connections

To get combinations (3!) arrangements are removed. In exams divide ordered triples by (3!) to get unordered triples.

Rotations in a circle are considered the same so one position is fixed. In exams divide linear (n!) by (n) for circular permutation.

When reflection is also same each circular arrangement is counted twice. In exams remember (\frac{(n-1)!}{2}) for necklace type.

Each of the (r) positions has (n) independent choices. In exams do not write a decreasing product when repetition is allowed.

Without repetition the available count decreases after one choice is used. In exams connect no repetition with a falling product.

In the formula (n!) loses both unwanted tail ((n-r)!) and order (r!). In exams understand both parts of the combination denominator.

Near the maximum (\frac{^{n}C_{r+1}}{^{n}C_r}) is around (1). In exams use the ratio to identify transition from increasing to decreasing.

In the falling product (9\cdot8\cdot7\cdot6), the remaining tail (5!) is removed. In exams write the denominator tail after the last chosen factor.

(^{10}P_4) is an ordered count and (^{10}C_4) is unordered. In exams divide by (r!) to remove order.

To get unordered pairs from ordered pairs (2!) orders are removed. This connection is very useful in pair selection exams.

After choosing (p), the remaining (q) are fixed, so (\frac{n!}{p!q!}). In exams use factorial division for fixed group sizes.

Interchanging identical letters does not create a new arrangement. In exams divide by factorials of repeated counts.

It is obtained by taking the ratio of (^{n}C_r) and (^{n}C_{r-1}). In exams simplify adjacent combinations using factorial ratios.

The denominator in the pair formula is (2!), and (2!=2). In exams simplify pair formulas mentally.

Write (20!) as (20\cdot19\cdot18\cdot17!) and cancel (17!). In exams do not expand large factorials fully.

Complement symmetry is a property of combinations, not permutations. In exams keep identities of (^{n}C_r) and (^{n}P_r) separate.

Combination counts only selection so complement works. In exams do not use complement symmetry when ordered slots appear.

The complement of at least (1) girl is no girl. In exams total minus unwanted is fast for at least conditions.

With exactly (2) women the remaining (2) must be men. In exams use cases or direct product of combinations for exactly conditions.

Treating two together people as one block gives (7) objects to arrange and (2!) ways inside the block. In exams use block method for together conditions.

A triangle is determined by the set of vertices, not their order. In exams use combinations for geometric selection.

Each object has two independent choices, so total subsets are (2^n), and for at least (1) the empty set is removed. In exams use total minus unwanted for at least conditions.

To form (x^r), choose (r) brackets from (n) brackets. In exams connect coefficients with selection.

The left side adds all selection sizes and the right side gives two choices for each object. In exams remember this identity through subset counting.

In total (r) selections, the count from the first group varies as (k). In exams connect two-source selection with Vandermonde identity.