Derivations of formulas and their connections

First place the (m) special members in the second part and give free choices to the remaining (n-m) members. In exams choose the fixed marked set first in such sums.

Choose the (3) marked members first and let each of the remaining (n-3) members enter the subset or not. In exams count the inner selection first.

Differentiate ((1+x)^n) and multiply by (x) to obtain this sum. In exams think of the derivative method when a factor (r) appears.

Two differentiations produce the factor (r(r-1)), and (x^2) restores the power. In exams use the second derivative for (r(r-1)).

Breaking powers into falling factorials allows standard binomial sums. In exams write (r^3) in falling form instead of expanding directly.

Both sides count the same selection with (s) special members inside an (r)-group. In exams change the order of nested selection.

In sequential selection the internal order of each selected block is removed. In exams form a multinomial denominator when many labelled groups appear.

Interchanging the two size (2) groups and the two size (4) groups gives duplicates. In exams divide extra by factorials of equal-size unlabelled groups.

All three groups have the same size and are unlabelled, so group order (3!) is also removed. In exams remember the extra (3!) for equal unlabelled groups.

First choose the empty boxes, then distribute onto the remaining boxes. In exams use selection plus inclusion-exclusion for exactly empty boxes.

Empty boxes are subtracted and added by inclusion-exclusion. In exams interpret onto as every labelled box being non-empty.

First count (3) labelled non-empty groups, then remove the (3!) label permutations. In exams divide the labelled count for unlabelled groups.

After removing the minimum sum (14), (16) remains, so ({}^{16+4-1}C_{3}) is obtained. In exams subtract unequal lower bounds first.

Since the total is (24) and the maximum of each of the three variables is (8), only ((8,8,8)) is possible. In exams check extreme feasibility before inclusion-exclusion.

A violation starts at (x_i\geq6), so inclusion-exclusion applies. In exams use shift (6) for upper bound (5).

Choose the (3) positive variables first, then split (12) into (3) positive parts. In exams use choose variables plus positive stars and bars for exactly positive variables.

First choose the empty boxes, then distribute positively into the remaining (3) boxes. In exams split exactly empty into box selection and positive distribution.

Comparing the alternating factorial expression for derangements gives this recurrence. In exams remember the inclusion-exclusion form for derangement recurrence.

First choose the (2) objects that stay fixed and derange the remaining (5). In exams use choose fixed plus derange rest for exactly fixed points.

This is a derangement of (6) objects and (D_6=265). In exams treat letter-envelope mismatch as derangement.

The positions of (A,B) are at distance (k+1), and there are (2) choices for order. In exams count positions first in fixed-gap problems.

There are (10-3-1=6) position choices and (2) choices for the order of (A,B). In exams count position pairs for between conditions.

Only (1) of the (3!) relative orders of those (3) books is allowed. In exams divide total by (k!) for fixed relative order.

Three independent before-after restrictions divide the count by (2^3). In exams divide by a power of (2) for independent pairs.

After removing circular rotation, positions with fixed distance are counted. In exams fix one object first for circular distance.