Concept	Formula	Key Note
Factorial Notation	$n! = n(n – 1)(n – 2) … 3 \cdot 2 \cdot 1$	$0! = 1$
Permutations ($n$ things, $r$ at a time)	$^nP_r = \frac{n!}{(n – r)!}$	Order matters (e.g., $ab \neq ba$)
Permutations (all $n$ things)	$n!$	Taken all at a time
Permutations (with identical items)	$\frac{n!}{(p_1!)(p_2!)…(p_r!)}$	Where $p_1, p_2…$ are alike items
Combinations ($n$ things, $r$ at a time)	$^nC_r = \frac{n!}{(r!)(n – r)!}$	Order does not matter (e.g., $ab = ba$)

Property	Formula / Result
Symmetry Property	$^nC_r = ^nC_{(n – r)}$
Full Selection	$^nC_n = 1$
Zero Selection	$^nC_0 = 1$

Type	Problem	Calculation	Result
Factorial	$4!$	$4 \times 3 \times 2 \times 1$	24
Permutation	$^6P_2$	$6 \times 5$	30
Permutation	$^7P_3$	$7 \times 6 \times 5$	210
Combination	$^{11}C_4$	$\frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1}$	330
Combination	$^{16}C_{13}$	same as $^{16}C_3 = \frac{16 \times 15 \times 14}{3 \times 2 \times 1}$	560

1. Factorial Notation

Let n be a positive integer. Then, factorial n, denoted n! is defined as:

$n! = n(n – 1)(n – 2) … 3 \cdot 2 \cdot 1$

• 󰪚 We define $0! = 1$.
• 󰪚 $4! = (4 \times 3 \times 2 \times 1) = 24$.
• 󰪚 $5! = (5 \times 4 \times 3 \times 2 \times 1) = 120$.

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2. Permutations

The different arrangements of a given number of things by taking some or all at a time, are called permutations.

• 󰪚 All permutations made with the letters a, b, c by taking two at a time are (ab, ba, ac, ca, bc, cb).
• 󰪚 All permutations made with the letters a, b, c taking all at a time are: (abc, acb, bac, bca, cab, cba).

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3. Number of Permutations

Number of all permutations of n things, taken r at a time, is given by:

$^nP_r = n(n – 1)(n – 2) … (n – r + 1) = \frac{n!}{(n – r)!}$

• 󰪚 $^6P_2 = (6 \times 5) = 30$.
• 󰪚 $^7P_3 = (7 \times 6 \times 5) = 210$.
• 󰪚 Cor. number of all permutations of n things, taken all at a time = $n!$.

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4. An Important Result

If there are n subjects of which $p_1$ are alike of one kind; $p_2$ are alike of another kind; $p_3$ are alike of third kind and so on, the number of permutations is:

$= \frac{n!}{(p_1!)(p_2!)…..(p_r!)}$

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5. Combinations

Each of the different groups or selections which can be formed by taking some or all of a number of objects is called a combination.

• 󰪚 Selection of two out of three boys A, B, C: AB, BC, and CA. (Note: AB and BA are the same selection).
• 󰪚 All combinations formed by a, b, c taking two at a time: ab, bc, ca.
• 󰪚 The only combination of a, b, c taken all at a time: abc.

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6. Number of Combinations

The number of all combinations of n things, taken r at a time is:

$^nC_r = \frac{n!}{(r!)(n – r)!} = \frac{n(n – 1)(n – 2) … \text{to } r \text{ factors}}{r!}$

• 󰪚 Note 1: $^nC_n = 1$ and $^nC_0 = 1$.
• 󰪚 Note 2: $^nC_r = ^nC_{(n – r)}$.
• 󰪚 $^{11}C_4 = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330$.
• 󰪚 $^{16}C_{13} = ^{16}C_3 = \frac{16 \times 15 \times 14}{3 \times 2 \times 1} = 560$.