Derivations of formulas and their connections

This is the hockey-stick identity and the largest selected element creates cases. In exams use the last-element method for such staircase sums.

The upper and lower indices increase together so it is a diagonal sum. In exams think of hockey-stick when a diagonal combination sum appears.

Choose the marked pair first and choose the remaining elements freely. In exams think of pair marking when ({}^{r}C_2) appears.

Choose the (3) marked members first and freely choose the remaining (n-3) members. In exams count the inner combination first.

If a pair is marked and then one member is marked, cases occur inside or outside the pair. In exams split products of marks into cases.

Choose the innermost (c) elements first and then add layers. In exams count nested choices from inside to outside.

The smaller selected set must lie inside the larger selected set. In exams first check valid index order in nested combinations.

Differentiate ((1+x)^n) and put (x=-1) to get zero. In exams use derivative plus (x=-1) for alternating weighted sums.

After the second derivative and putting (x=-1), ((1-1)^{n-2}) appears. In exams check zero when the falling factor is below the degree.

This split connects first and second weighted sums. In exams convert powers into falling factorials.

Using (r^2=r(r-1)+r) adds two standard sums. In exams simplify the final form to (n(n+1)2^{n-2}).

Using (r^3=r(r-1)(r-2)+3r(r-1)+r) gives this form. In exams use falling factorial decomposition for cubic sums.

Two groups have the same size (a), so interchanging them gives duplicates. In exams divide extra by factorial for equal-size unlabelled groups.

Both internal orders and equal-size group swaps are removed. In exams write both group sizes and equal-group factorials in the denominator.

Choose the empty box first and then distribute onto the remaining (r-1) boxes. In exams use choose empty plus onto count for exactly non-empty boxes.

Choose the empty boxes first and then the remaining two boxes must both be non-empty. In exams apply onto to the remaining boxes in exactly-empty cases.

Empty-box cases are removed by inclusion-exclusion. In exams count non-empty labelled boxes like onto functions.

After removing the minimum sum (10), (12) remains and is distributed among (4) variables. In exams subtract lower bounds and use stars and bars.

The upper violation is (x_i\geq8) and inclusion-exclusion is applied. In exams use shift (8) for upper bound (7).

Choose the zero variable and split positive sum (16) among the remaining (3) variables. In exams convert exactly-zero cases into positive distribution.

First choose the (3) non-empty boxes and then split (18) balls into (3) positive parts. In exams use choose boxes plus positive stars and bars for exactly non-empty.

If two objects swap with each other, the remaining (n-2) objects are deranged. In exams separate swap and non-swap cases in derangement recurrence.

Using the derangement recurrence, (D_7=6(D_6+D_5)=1854). In exams remember small (D_n) values through recurrence.

First choose the (3) fixed objects and derange the remaining (5) objects. In exams use choose fixed plus derange rest for exactly fixed points.

Choose the correct letters and derange the remaining letters into wrong envelopes. In exams solve exactly correct letters using the fixed-point formula.