Derivations of formulas and their connections

Correct answer: B. (^{n}P_r=^{n}C_r\times r!)

Explanation: Each selected group can be arranged in (r!) ways. In exams treat permutation as selection followed by arrangement.

Correct answer: A. Because arrangements of (r) objects are not counted

Explanation: In combinations order is not important so (r!) duplicate arrangements are removed. In exams divide when order is ignored.

Correct answer: B. (^{n}P_r=\frac{n!}{(n-r)!})

Explanation: After (r) decreasing factors the remaining tail is ((n-r)!). In exams put the missing tail in the denominator.

Correct answer: B. Choosing (r) and leaving (n-r) are equivalent decisions

Explanation: Choosing (r) objects is equivalent to not choosing (n-r) objects. In exams use complement selection to identify symmetry.

Correct answer: A. A fixed object is included or not included

Explanation: Make cases on one fixed object: include it or exclude it. In exams derive such identities by inclusion cases.

Correct answer: B. (\frac{^{n}C_{r+1}}{^{n}C_r}=\frac{n-r}{r+1})

Explanation: Writing factorial forms and canceling common terms gives the ratio. In exams ratio method is fast for consecutive combinations.

Correct answer: B. (^{n}C_r=\frac{^{n}P_r}{r!})

Explanation: A permutation includes (r!) orders for each combination. In exams divide ordered count by (r!) to get unordered count.

Correct answer: B. Because (E) appears twice

Explanation: Interchanging identical (E)'s does not give a new arrangement. In exams divide by factorials of repeated identical items.

Correct answer: B. (^{7}P_3)

Explanation: Both selection and order are needed so (^{7}P_3=^{7}C_3\cdot3!). In exams use permutation when order matters.

Correct answer: A. Because the two posts are different

Explanation: Switching captain and vice-captain changes the outcome. In exams treat different roles as order important.

Correct answer: C. Because only a group is needed

Explanation: Only membership is counted in a committee, not order. In exams use combination for a group without posts.

Correct answer: B. Choices are (n), then (n-1), then decreasing

Explanation: After each chosen object available choices decrease. In exams write decreasing products for distinct arrangements.

Correct answer: C. (^{n}C_r=^{n}C_{n-r})

Explanation: This is the symmetry of choosing (r) and leaving (n-r). In exams use this relation when opposite indices sum to (n).

Correct answer: B. \(2!\)

Explanation: The relation between permutations and combinations is \(^{n}P_r=^{n}C_r\cdot r!\). Here \(r=2\), so \(^{n}P_2=^{n}C_2\cdot 2!\), giving \(k=2!\). Note that \(2\) is numerically equal to \(2!\), so listing both as separate options would create two correct answers; always check equivalent expressions in MCQs. Exam tip: order matters in permutations, so each combination is multiplied by \(r!\).

Correct answer: A. There is one way to choose no object

Explanation: The empty selection is also a valid selection. In exams remember both (0!) and empty choice as (1).

Correct answer: A. There is exactly one way to choose all (n) objects

Explanation: When all objects are chosen the selection is fixed. In exams quickly identify extreme cases (^{n}C_0) and (^{n}C_n).

Correct answer: B. Choosing r objects from n distinct objects and arranging them in r distinct positions

Explanation: First choose r objects in \(\binom{n}{r}\) ways, then arrange them in \(r!\) orders. Thus it represents selection followed by ordering; a committee has no order. Exam tip: multiply by \(r!\) whenever positions or order matter.

Correct answer: A. \(4!\)

Explanation: The relation is \(^{n}P_r=^{n}C_r\cdot r!\), because the \(r\) selected objects can be arranged in \(r!\) orders. Here \(r=4\), so \(x=4!\). \(\frac{12!}{8!}\) is \(^{12}P_4\) itself, while \(\frac{12!}{4!8!}\) is \(^{12}C_4\) itself. Exam tip: in the relation between permutations and combinations, the multiplying factor is always \(r!\).

Correct answer: B. (^{6}P_4=^{6}C_4\cdot4!)

Explanation: Changing order changes the code. In exams usually use permutation relation for codes or passwords.

Correct answer: A. There are (n) independent choices for selecting one object

Explanation: In single selection order does not arise and choices are (n). In exams both (^{n}C_1) and (^{n}P_1) equal (n).

Correct answer: A. Because the order of (1) object creates no different result

Explanation: One object can be arranged in (1!) way. In exams permutation and combination give the same value for (r=1).

Correct answer: A. Because ordered pairs are divided by (2!) to become unordered pairs

Explanation: (AB) and (BA) are the same pair so repetition is removed. In exams (\frac{n(n-1)}{2}) is often useful for pairs.

Correct answer: B. Pascal identity

Explanation: This is a shifted form of Pascal identity. In exams add adjacent lower-row combinations to form the upper-row combination.

Correct answer: B. (5)

Explanation: The direct answer is option B: r = 5. The key idea is the symmetry property of combinations: for a fixed n, \\(^{n}C_k=^{n}C_{n-k}\\). Thus, two combination values with the same upper number are equal when their lower numbers are complementary, so r+(r+4)=14. Now solve step by step: r+r+4=14; 2r+4=14; 2r=10; r=5. Therefore, \\(^{14}C_5=^{14}C_9\\), and r+4=9, confirming the result. Option A, r=4, would give lower indices 4 and 8; these are not complementary because 4+8=12, not 14, so the values are not equal. Option B, r=5, gives 5 and 9, whose sum is 14, so it is correct. Option C, r=6, gives 6 and 10; their sum is 16, so it does not satisfy the symmetry condition. Option D, r=7, gives 7 and 11; their sum is 18, so it is also wrong. Remember: when \\(^{n}C_a=^{n}C_b\\) in this standard situation, use a+b=n, then solve carefully.

Correct answer: A. (3!)

Explanation: First choose (3) objects and arrange them in (3!) ways. In exams remember (^{n}P_3=^{n}C_3\cdot3!).