Derivations of formulas and their connections

If the special object is chosen the remaining (r-1) objects must be chosen from (n-1). In exams write the included case separately.

Put (n=6) and (r=2) in (^{n}C_r=^{n-1}C_r+^{n-1}C_{r-1}). In exams (n) decreases by (1) in both terms.

Each object has two choices so total subsets are (2^n). In exams think of subsets using include or exclude.

All selections include the empty selection too. In exams subtract the empty case when at least one appears.

The direct answer is option C, \(^{n}C_r\). A subset is a group selected from a larger set, and its order does not matter: choosing A, B is the same subset as choosing B, A. The notation \(^{n}C_r=\frac{n!}{r!(n-r)!}\) counts the ways to select r distinct objects from n objects without arranging them. Option A, \(^{n}P_r\), counts ordered arrangements, so it counts the same group repeatedly in different orders. Option B, \(n!\), arranges all n objects and is not restricted to selecting r. Option D, \(r!\), only counts arrangements of the selected objects and does not choose them from n. Thus C is the required subset formula. Exam cue: subset means unordered selection, so think combination.

Two independent selections are needed together so the multiplication principle applies. In exams the word and often means multiply.

First choose (3) students and then arrange them in (3!) ways. In exams multiply when arrangement follows selection.

Rotations are considered same in a circle so fixing one object gives ((n-1)!) arrangements. In exams fix (1) object in circular arrangement.

Rotating a circle does not create a new arrangement so (n!) is divided by (n). In exams treat rotation as duplicate counting.

Interchanging identical letters does not create a new word. In exams divide by the factorial of identical letters.

Interchanges inside two identical groups are not counted separately so we divide by (p!q!). In exams place the factorial of each repeated group in the denominator.

Each position independently has (n) choices so the product is (n\times n\times\cdots\times n=n^r). In exams choices do not decrease when repetition is allowed.

Without repetition one object cannot be taken again. In exams pay attention to allowed and not allowed words.

Combination is obtained by removing (r!) internal orders from permutation. In exams make the ordered count first and then divide.

The direct answer is option A, \(\frac{n!}{(n-r+1)!}\). Begin with the given rule \(^{n}P_k=\frac{n!}{(n-k)!}\). To find \(^{n}P_{r-1}\), replace every occurrence of \(k") with \(r-1\): \(^{n}P_{r-1}=\frac{n!}{(n-(r-1))!}\). Simplify the denominator: \(n-(r-1)=n-r+1\). Therefore \(^{n}P_{r-1}=\frac{n!}{(n-r+1)!}\). Option A has exactly this denominator and is correct. Option B, \(\frac{n!}{(n-r-1)!}\), comes from changing the sign incorrectly. Option C reverses numerator and denominator and is not the permutation formula. Option D uses \(r!\), which is unrelated to the required substitution. The safe exam method is to put parentheses around \(r-1\) before subtracting: \(n-(r-1)\), then simplify carefully.

Simplifying the factorial formula gives the ratio (\frac{n-r+1}{r}). In exams ratio method is useful for adjacent combinations.

For the (r)th position (n-r+1) choices are added as a factor. In exams identify the factor of the next position.

In each (r)-selection one of the (r) chosen objects can be marked. In exams understand this relation through marked object counting.

It is obtained by simplifying the ratio of (^{n}C_r) and (^{n}C_{r-1}). In exams use ratio for consecutive (C) terms.

There are (^{n}C_r) ways to choose (r) objects and then (r!) arrangements. In exams write ordered selection as choose and arrange.

Choosing (4) means leaving (3) so (^{7}C_4=^{7}C_3). In exams look at both chosen and not chosen.

The direct answer is option C, 10. For a permutation, the first position has 12 choices, the second has 11 choices because one object has already been used, and the third has 10 choices because two objects have been used. Thus, \(^{12}P_3=12\times11\times10\), so the last factor is 10. Option A, 12, is the first factor, not the last. Option B, 11, is the second factor. Option C, 10, is correct because it equals \(12-3+1\). Option D, 9, would be one factor too far; only three factors are required. The useful general pattern is \(^{n}P_r=n(n-1)(n-2)\cdots(n-r+1)\). Memory cue: for three selections, count down from \(n\) three times; the last number is \(n-2\).

The governing identity is the symmetry property of combinations: nCr = nC(n-r). Selecting r objects from n is equivalent to deciding which n-r objects are left out, so both selections have the same count. Applying this to 11C8 gives 11C8 = 11C(11-8) = 11C3. The smaller lower index makes calculation easier, and 11C3 can be evaluated as 11 × 10 × 9 / (3 × 2 × 1) = 165. Thus option A is correct. Option B changes the total number from 11 to 8, so it is not equivalent. Option C is a permutation and counts ordered arrangements, while option D asks for 11 selections from only 3 objects and is not valid in the usual combinatorial setting.

The row arrangement of all (6) distinct books is (^{6}P_6=6!). In exams connect all distinct arrangement with factorial.

Order is important when placing from top to bottom so the chosen (3) colours are arranged in (3!) ways. In exams add arrangement when positions appear.