Derivations of formulas and their connections

The first digit cannot be (0), so there are (6) choices and (7) choices for each remaining place. In exams treat leading zero separately.

There are (6) non-zero choices for the first place and then (3) places are filled from the remaining (6) digits. In exams handle the first place separately.

Different lower indices are complementary, so (r+(r-2)=n). In exams solve equal combinations using the complement rule.

The ratio is (\frac{n-r}{r+1}) and it is set equal to (2). In exams form the equation directly from the ratio formula.

For consecutive combinations, \(\frac{^{n}C_r}{^{n}C_{r-1}}=\frac{n-r+1}{r}\). Hence, putting \(\frac{n-r+1}{r}=3\) gives \(n-r+1=3r\). Option D misses the \(+1\), so it is not correct. Exam tip: in ratios involving \(^{n}C_{r-1}\), the numerator is \(n-r+1\).

A quadrilateral needs an unordered selection of (4) 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 line segment the order of endpoints does not matter. In exams use (^{n}C_2) for unordered pairs.

Each circular arrangement is counted (8) times in the linear count due to rotations. In exams treat rotations as extra count in a circle.

First removing rotations gives (6!), 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 (3!2!). In exams multiply the factorials in the denominator.

Because (^{n}P_r=^{n}C_r r!), and (r!=1) only for (r=0) or (r=1). In exams verify equality using factorials.

The relation between permutations and combinations is \(^{n}P_r=^{n}C_r\times r!\). Hence, \(^{n}C_r=\frac{^{n}P_r}{r!}\). Comparing the given denominator \(120\) with \(r!\), we get \(r!=120=5!\); therefore, \(r=5\). Since \(4!=24\) and \(6!=720\), those options are incorrect. Exam tip: in such questions, equate the denominator directly to \(r!\).

The direct answer is option B, \(r=6\). The standard relationship between permutations and combinations is \(^{n}P_r=^{n}C_r\,r!\). The question says \(^{n}P_r=720\times{}^{n}C_r\). Comparing the two expressions, the multiplier must be \(r!=720\). Now calculate factorials: \(5!=120\), \(6!=6\cdot5\cdot4\cdot3\cdot2\cdot1=720\), while \(7!=5040\). Hence \(r=6\). Option A, 5, would give multiplier 120, not 720. Option B, 6, gives exactly 720 and is correct. Option C, 7, gives 5040. Option D, 8, gives \(40320\), also incorrect. This works because a selection of r objects can be arranged in \(r!\) orders. Memory cue: whenever \(^{n}P_r\) is compared with \(^{n}C_r\), their ratio is \(r!\).

Complementary indices have sum (16), so (a+a+4=16) and (a=6). In exams form a linear equation from equal combinations.

Different lower indices are complementary, so (x+2x=18) and (x=6). In exams solve equal (C) terms using the complement rule.

This is the sum of non-empty selections (2^n-1), and (2^6-1=63). In exams subtract (1) when the empty selection is removed.

The total sum is (2^n), and (2^8=256). In exams connect the sum of all combinations with a power of (2).

The even indexed sum is (2^{n-1}), and (2^7=128) gives (n=8). In exams remember that even and odd sums are equal.

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

Permutation needs the order of only the selected (r) objects and not the order of remaining objects. In exams remove the unwanted tail from (n!).

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

The two equal groups are unlabelled, so interchanging the groups counts twice. In exams apply the extra (2!) correction for equal unlabelled groups.

After choosing the first (4)-group the second is fixed, and interchanging the two groups creates duplicate count. In exams divide by (2!) for equal groups.