Derivations of formulas and their connections

There are (n-k-1) position pairs for (A,B) and (2) choices for their order. In exams count positions first in fixed-gap problems.

There are (11-4-1=6) position pairs and (A,B) can be ordered in (2) ways. In exams there is no need to separately choose the people between them.

After fixing (A), the position of (B) is fixed and the remaining (8) people are arranged. In exams fix one person in one-direction circular gap problems.

Total circular arrangements are (8!) and the adjacent block gives (2\cdot7!) ways. In exams handle circular not-adjacent by complement.

Seat the men circularly in (7!) ways and place the women in the gaps in (8!) ways. In exams do not add an extra factor (2) in circular alternation.

Only rotations are duplicates, so the circular count is ((10-1)!). In exams divide by (2) only after reading the reflection condition.

In a bracelet, both rotations and reflections are considered the same. In exams use (\frac{(n-1)!}{2}) for bracelets.

When the last digit is fixed as (0), there is no leading-zero issue and (5) places are filled from (9) non-zero digits. In exams separate the zero-last case.

There are (4) non-zero even choices for the last digit and (8) remaining non-zero choices for the first digit. In exams handle first and last restrictions together.

Choose the even positions and then multiply even and odd choices independently. In exams choose positions first in exactly digit-type problems.

Every symbol appearing is an onto condition and missing symbols are removed. In exams solve at least once by inclusion-exclusion.

First choose (4) symbols and then form onto strings on them. In exams use choose set plus onto count for exactly distinct symbols.

The Stirling part partitions positions into non-empty groups and (s!) assigns groups to symbols. In exams treat exactly used symbols as onto mapping.

The exponents sum to (10) and the coefficient comes from the multinomial form. In exams treat powers as group sizes.

The cases are: choose three (x)'s or choose one (x^2) and one (x). In exams add all cases that form the same power.

All disjoint choices that form exponent (4) must be added. In exams make exponent partitions for polynomial coefficients.

The cube roots of unity filter is useful for separating modulo (3) classes. In exams identify three-step coefficient sums with an advanced filter.

The ratio (\frac{n-r}{r+1}) must be greater than (1). In exams identify the increasing region by ratio.

When the consecutive ratio is less than (1), the coefficients start decreasing. In exams the inequality reverses after the peak.

Unequal equal-combination indices are complementary. In exams set the sum of lower indices equal to the upper index.

The complementary condition gives (3r-2+r+8=30), so (r=6). In exams solve same-index and complement cases separately.

({}^{n}P_5=(n-4){}^{n}P_4), so (n-4=15). In exams apply the consecutive permutation relation directly.

The ratio (\frac{n-r}{r+1}=\frac{4}{5}) gives (5n-5r=4r+4). In exams cross-multiply combination ratios.

(\frac{{}^{n}C_r}{{}^{n}C_{r-1}}=\frac{n-r+1}{r}), and cross multiplication gives the relation. In exams keep the ratio direction correct.

The cases for at least (2) special are exactly (2) and exactly (3). In exams direct cases are clear for a small special group.