Derivations of formulas and their connections

The three red balls are identical so their (3!) internal orders are not different. In exams divide by the factorial of the repeated group only.

To remove internal orders of two repeated groups divide by (2!3!). In exams put the factorial of each repeated letter in the denominator.

A mere rotation in a circle does not make a new seating. In exams use the one fixed method in circular arrangement.

First removing rotations gives ((n-1)!) and then same reflection makes us divide by (2). In exams check both rotation and reflection in necklaces.

Because (r!) is (1) only for (r=0) and (r=1). In exams check (^{n}P_r=^{n}C_r r!) for equality.

Permutation counts many orders of the same group. In exams keep the difference between (P) and (C) clear when (r>1).

When making combination from permutation the same selection is counted (r!) times. In exams use division principle when overcounting appears.

Each shirt can pair with (4) pants so there are (3\times4) choices. In exams multiply when independent choices are needed together.

This is an or situation so there are (5+6) choices. In exams look for the addition principle when the word or appears.

Changing rank changes the result so order is important. In exams use permutation when rank or post appears.

After choosing (r) objects (n-r) objects remain. In exams remember this meaning for complement counting.

In permutations, \(^{n}P_r=n(n-1)(n-2)\cdots(n-r+1)\), so the last factor is \(n-r+1\). Here, \(10-4+1=7\), so the correct answer is 7. The value 6 would result from incorrectly omitting \(+1\). Exam tip: write \(n-r+1\) first and then substitute the values.

President and secretary are distinct posts, so interchanging the same two students gives a different outcome; hence order matters and permutations apply. In a committee, order does not matter. Exam tip: look for roles or positions to identify permutations.

Choosing no object gives one empty selection. In exams treat the empty case as valid count (1).

When (r=n) all objects are being arranged so the count is (n!). In exams treat full arrangement as factorial.

Each selection has a unique rejected group. In exams understand the identity by forming a complementary pair.

(n!) creates a larger ordered count which is corrected by (r!(n-r)!). In exams understand the overcounting correction.

A line segment is an unordered pair of two points. In exams do not count the order of endpoints in a segment.

In a directed arrow changing start and end changes the arrow. In exams use ordered pair or permutation when direction exists.

The governing idea is complementary selection, represented by the combination identity ⁿCᵣ = ⁿCₙ₋ᵣ. From a collection of 16 objects, every selection of 7 objects determines exactly one complementary group of 9 objects that was not selected. This one-to-one correspondence proves that the number of 7-element selections equals the number of 9-element selections. Algebraically, 16−7=9, or equivalently 7+9=16, so ¹⁶C₇=¹⁶C₉. Hence option A is correct. Option B is irrelevant because multiplication is not the condition in the symmetry identity. Option C is false because the lower indices are different, and option D confuses combinations, where order is ignored, with permutations, where order matters.