Derivations of formulas and their connections

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

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.

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 \(n-r+1\).

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

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

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

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

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

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

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

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

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.

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.

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

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

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.

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

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.

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

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

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.

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

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

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.

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).