Derivations of formulas and their connections

Correct answer: C. (n=2r-3)

Explanation: Different lower indices are complementary, so (r+(r-3)=n). In exams solve equal combinations using the complement rule.

Correct answer: B. (n-r=3(r+1))

Explanation: The ratio is (\frac{n-r}{r+1}), and it is set equal to (3). In exams form the equation directly from the ratio formula.

Correct answer: A. \(n-r+1=4r\)

Explanation: For adjacent combinations, \(\frac{^{n}C_r}{^{n}C_{r-1}}=\frac{n-r+1}{r}\). Hence, \(\frac{n-r+1}{r}=4\) gives \(n-r+1=4r\). Therefore, option A is correct. Option D incorrectly omits the \(+1\), which arises from the \(^{n}C_{r-1}\) term. Exam tip: in ratios of adjacent combinations, carefully check that the numerator is \(n-r+1\).

Correct answer: C. (^{12}C_5)

Explanation: A pentagon needs an unordered selection of (5) points. In exams do not count the order of points when forming a shape.

Correct answer: D. (^{12}P_2)

Explanation: In a directed segment, changing start and end changes the object. In exams use permutation when direction exists.

Correct answer: C. (^{12}C_2)

Explanation: In an ordinary segment, the order of endpoints does not matter. In exams use (^{n}C_2) for unordered pairs.

Correct answer: A. Circular count (=\frac{9!}{9})

Explanation: Each circular arrangement is counted (9) times in the linear count due to rotations. In exams treat rotations as extra count in a circle.

Correct answer: B. (\frac{7!}{2})

Explanation: First removing rotations gives (7!), then mirror images being the same makes us divide by (2). In exams always check reflection in necklace problems.

Correct answer: A. (4!2!)

Explanation: Internal interchanges of repeated letters do not create new arrangements. In exams divide by the factorial of each repeated group.

Correct answer: B. (\frac{10!}{4!3!})

Explanation: Internal orders of two identical groups are not different, so divide by (4!3!). In exams multiply the factorials in the denominator.

Correct answer: B. This is impossible because (r!>1)

Explanation: (^{n}P_r=^{n}C_r r!), and for (r>1), (r!>1). In exams verify equality of (P) and (C) using factorials.

Correct answer: B. 7

Explanation: The relation between permutations and combinations is \(^{n}P_r=\,^{n}C_r\times r!\). Hence, \(^{n}C_r=\frac{^{n}P_r}{r!}\). Here the denominator is 5040, so \(r!=5040\). Since \(7!=5040\), \(r=7\). For instance, \(6!=720\), so 6 is not correct. Exam tip: in such questions, equate the given denominator directly to \(r!\) and identify the factorial.

Correct answer: B. (5)

Explanation: The direct answer is option B, \(r=5\). Use the identity \(^{n}P_r=^{n}C_r\,r!\). The question gives \(^{n}P_r=120\times{}^{n}C_r\). After comparing the common factor \(^{n}C_r\), we get \(r!=120\). Since \(5!=5\cdot4\cdot3\cdot2\cdot1=120\), the value is \(r=5\). Option A, 4, would give \(4!=24\), not 120. Option B, 5, gives exactly 120 and is correct. Option C, 6, gives \(6!=720\). Option D, 7, gives \(7!=5040\). The identity comes from first choosing r objects, counted by the combination, and then arranging those chosen objects in \(r!\) orders. Exam cue: the numerical multiplier between \(^{n}P_r\) and \(^{n}C_r\) is always \(r!\), provided the common combination factor is nonzero.

Correct answer: B. (7)

Explanation: Complementary indices have sum (20), so (a+a+6=20) and (a=7). In exams form a linear equation from equal combinations.

Correct answer: C. (7)

Explanation: Different lower indices are complementary, so (x+2x=21) and (x=7). In exams solve equal (C) terms using the complement rule.

Correct answer: B. (7)

Explanation: This is the sum of non-empty selections (2^n-1), and (2^7-1=127). In exams subtract (1) when the empty selection is removed.

Correct answer: C. (9)

Explanation: The total sum is (2^n), and (2^9=512). In exams connect the sum of all combinations with a power of (2).

Correct answer: B. (9)

Explanation: The even indexed sum is (2^{n-1}), and (2^8=256) gives (n=9). In exams remember that even and odd sums are equal.

Correct answer: A. To remove internal orders of chosen and unchosen groups

Explanation: In the (n!) count, the orders inside both groups are counted extra. In exams understand both corrections (r!) and ((n-r)!).

Correct answer: A. (\frac{n!}{a!b!c!})

Explanation: Order inside each labelled group is irrelevant, so divide by (a!b!c!). In exams do not divide extra among labelled groups themselves.

Correct answer: A. (\frac{^{10}C_5}{2!})

Explanation: After choosing the first (5)-group the second is fixed, and interchanging the two groups creates duplicate count. In exams divide by (2!) for equal groups.

Correct answer: A. False, because the correct multiplier is (4!=24)

Explanation: (^{n}P_r=^{n}C_r r!), so for (r=4) the multiplier is (24). In exams verify the multiplier using factorials.

Correct answer: A. (^{n}P_r) counts selection along with arrangement of selected objects

Explanation: Permutation is (r!) times combination because the selected objects must be ordered. In exams remember (P=C\times r!) conceptually.

Correct answer: B. (n-3=2\times4)

Explanation: The ratio is (\frac{^{n}C_r}{^{n}C_{r-1}}=\frac{n-r+1}{r}), so for (r=4), (\frac{n-3}{4}=2). In exams use the ratio formula for adjacent combinations.

Correct answer: C. (4!)

Explanation: The relation is (^{n}P_r=^{n}C_r\times r!), so here the factor is (4!). In exams add the arrangements of selected objects when going from (C) to (P).