Derivations of formulas and their connections

In (n!), internal orders of chosen and unchosen groups are counted separately. In exams remove internal order in group division.

Once one group is chosen, the other group is fixed, so (^{n}C_r) is enough. In exams count only one selection when the complement is fixed.

First there are (^{8}C_3) ways to form the team and (3) choices for captain. In exams multiply when an extra role follows selection.

One of the three is distinguished as captain, so removing the order of the remaining (2) members gives (^{8}P_3\div2!). In exams reduce overcounting by checking the roles.

Internal orders of each repeated group are not different, so divide by (2!2!3!). In exams treat repeated letters as separate groups.

Interchanging identical letters does not make a new arrangement. In exams write frequencies of repeated letters and divide by factorials.

The circular count is (7!) and the linear count is (8!), so the ratio is (\frac{7!}{8!}=\frac{1}{8}). In exams rotations are removed in circular counting.

Removing rotation first gives (5!), then reflection being same makes us divide by (2). In exams check both rotation and reflection in necklaces.

The selections of boys and girls are independent and order does not matter in a committee. In exams multiply combinations for category-wise selection.

The number of girls can be (2), (3), or (4), and the cases are added. In exams write valid cases separately in at least questions.

Even and odd indexed combination sums are equal and the total sum is (2^n). In exams remember ((1-1)^n) for the even-sum identity.

The sum of odd indexed combinations equals the even indexed sum and is (2^{n-1}). In exams prove it using the alternating identity.

Different lower indices must be complementary, so (2+4=n). In exams quickly check complementary indices in equal (C) terms.

Two objects have (2!) orders and (2!=2). In exams remember the relation between ordered and unordered counts in pairs.

First choose (4) letters and then arrange them in (4!) orders. In exams order is important in words.

The first place cannot have (0), so it has (4) choices and the remaining places have (5) choices. In exams check the leading zero condition first.

There are (5) non-zero choices for the first digit and then the remaining (3) places are filled from (5) remaining digits. In exams treat the first place as a separate case.

Different lower indices are complementary, so (r+(r-1)=n). In exams this relation is useful for adjacent equal combinations.

The ratio is (\frac{n-r+1}{r}), and setting it equal to (1) gives (n-r+1=r). In exams form an equation using the ratio identity.

In the recurrence, the extra factor is the choices for the (r)th place, which is (n-r+1). In exams treat the multiplier as next-place choices.

(^{n}C_{n-r}=\frac{n!}{(n-r)!r!}), which gives the same denominator. In exams verify symmetry using factorials.

In (\frac{n!}{r!(n-r)!}), changing the order of (r!) and ((n-r)!) does not change the value. In exams order does not matter in a product denominator.

Once (4) selected objects are chosen, the (7) rejected objects are fixed. In exams connect selected-rejected split with combination.

In (n!), internal orders of the unchosen ((n-r)) objects are extra, so they are removed. In exams identify the remaining part in the permutation formula.

One selected object has only (1) order. In exams permutation and combination both equal (n) when (r=1).