Derivations of formulas and their connections

Correct answer: A. Fixing the largest chosen element

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

Correct answer: B. Treat it as a shifted form of the hockey-stick identity

Explanation: 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.

Correct answer: B. Choosing a subset and selecting an unordered marked pair inside it

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

Correct answer: C. (^{n}C_3 2^{n-3})

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

Correct answer: D. (2{}^{r}C_2+3{}^{r}C_3)

Explanation: 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.

Correct answer: A. (^{n}C_c{}^{n-c}C_{b-c}{}^{n-b}C_{a-b})

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

Correct answer: A. (s\leq r\leq n)

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

Correct answer: B. (0)

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

Correct answer: C. (0)

Explanation: 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.

Correct answer: A. (r^2=r(r-1)+r)

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

Correct answer: C. (n(n+1)2^{n-2})

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

Correct answer: A. (n^2(n+3)2^{n-3})

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

Correct answer: C. (2!)

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

Correct answer: A. (3!3!4!4!2!2!)

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

Correct answer: A. (^{r}C_1\left((r-1)^n-{}^{r-1}C_1(r-2)^n+\cdots\right)) / (^{r}C_1\left((r-1)^n-{}^{r-1}C_1(r-2)^n+\cdots\right)

Explanation: 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.

Correct answer: A. (^{4}C_2(2^n-2))

Explanation: 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.

Correct answer: A. (3^n-3\cdot2^n+3)

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

Correct answer: B. (^{15}C_3)

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

Correct answer: A. (^{21}C_2-3{}^{13}C_2+3{}^{5}C_2)

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

Correct answer: A. (^{4}C_1{}^{15}C_2)

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

Correct answer: A. (^{6}C_3{}^{17}C_2)

Explanation: 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.

Correct answer: B. When the first object and its target object swap places

Explanation: 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.

Correct answer: A. (1854)

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

Correct answer: A. (^{8}C_3D_5)

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

Correct answer: A. (^{7}C_2D_5)

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