Derivations of formulas and their connections

Choose (k) from the first group and (r-k) from the second, then add all cases. In exams identify Vandermonde in two-group selection.

Choosing (n-r) from the second group equals ({}^{n}C_{n-r}={}^{n}C_r). In exams connect square sums with two equal groups.

Choose (r) objects first, then arrange both chosen and unchosen lists. In exams factorial cancellation also gives this quickly.

The left side chooses a group and marks one member, while the right side chooses the marked member first. In exams treat the factor (r) as a marked choice.

In an (r)-group, the first and second marked members are chosen with order. In exams think of ordered marking when you see (r(r-1)).

There are (n) choices for the marked member and two choices for each remaining member. In exams connect the extra (r) in binomial sums with marking.

There are (n(n-1)) ways to choose two ordered marked members and the remaining (n-2) are chosen freely. In exams two marks lead to (2^{n-2}).

Write (r^2=r(r-1)+r) and add two standard sums. In exams splitting (r^2) is the fastest method.

Choose the (k) marked members first, then each of the remaining (n-k) members may or may not enter the subset. In exams choose the marked set first in nested selection.

First choose the innermost (t) members, then add (s-t), then (r-s) members. In exams count nested choices backward by layers.

Labelled groups are fixed and order inside each group is not counted. In exams put group-size factorials in the multinomial denominator.

All three groups have equal size and are unlabelled, so the (3!) interchange of groups is also removed. In exams divide extra by factorial for equal unlabelled groups.

Choose (a) first, then (b) from the remaining, and (c) is automatically fixed. In exams match sequential selection with factorial form.

Cases with empty boxes are removed from total functions by inclusion-exclusion. In exams read onto as every box being non-empty.

Three empty-box choices are subtracted, then over-subtraction of two empty boxes is added. In exams apply inclusion-exclusion carefully for onto distribution.

From total (2^n) assignments, the two all-in-one empty-box cases are invalid. In exams subtract empty cases separately for labelled boxes.

In the labelled count, interchanging the two groups counts each division twice. In exams divide by (2) for two unlabelled groups.

This is multiset selection and the answer is ({}^{n+r-1}C_r). In exams handle repetition with no order by the bars method.

Give (2) first to the four variables, leaving (17). In exams subtract the lower bound and apply non-negative stars and bars.

The governing concept is the stars-and-bars method with lower bounds. First remove the compulsory minimum values by setting x₁=3+y₁, x₂=4+y₂, and x₃=5+y₃, where y₁,y₂,y₃ are non-negative integers. Substitution gives y₁+y₂+y₃=20−3−4−5=8. The number of non-negative integer solutions of y₁+y₂+y₃=8 is C(8+3−1,3−1)=C(10,2). Therefore option A is correct. Option B incorrectly uses 8 as the top number without adding the two separators. Options C and D incorrectly add the lower bounds instead of subtracting the compulsory amounts. The count assumes integer solutions, as stated.

Choose the zero variables first, then the remaining two variables form a positive sum of (18). In exams use positive distribution for exactly-zero cases.

Cases with (x_i\geq7) are subtracted and double violations are added. In exams shift by (7) for bounded solutions.

A violation begins with (x_i\geq5), so inclusion-exclusion applies. In exams use a subtract shift of (5) for upper limit (4).

The first letter can go to (n-1) positions other than its original position. In exams watch the first object's wrong choice in derangement recurrence.

Permutations with fixed points are subtracted and added alternately. In exams think of inclusion-exclusion when no object is in its correct place.