Factorial Notation

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Question 1 of 4

1. Question
On the following events, write $\mathsf{P}$ if it is a Permutation or $\mathsf{C}$ if it is a Combination.
- $(i)$ Forming a doubles team for tennis: (C, c)
  
  $(ii)$ Creating a password for a website: (P, p)
  
  $(iii)$ Selecting the $1$ st, $2$ nd and $3$ rd rankings in a car race: (P, p)
  
  $(iv)$ Being a member of a jury: (C, c)
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Permutation

An arrangement where order is important.

Combination

A grouping where order is not important.

In each item, check whether the order is important or not

$(i)$ Forming a doubles team for tennis:

The order of tennis players in forming a doubles team for tennis is not important.

Hence, this is a Combination

$(ii)$ Creating a password for a website:

The order of characters in creating a password for a website is important.

Hence, this is a Permutation

$(iii)$ Selecting the $1$ st, $2$ nd and $3$ rd rankings in a car race:

The order or rankings in a car race is important.

Hence, this is a Permutation

$(iv)$ Being a member of a jury:

The order of being selected as a member in a jury is not important.

Hence, this is a Combination

$(i)$ $\mathsf{C}$

$(ii)$ $\mathsf{P}$

$(iii)$ $\mathsf{P}$

$(iv)$ $\mathsf{C}$
Question 2 of 4

2. Question
Evaluate

$5!$
- (120)
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Factorial Formula

$n! = n*(n-1)*(n-2)*…*3*2*1$

Substitute the given $n=5$ to the formula

$n!$ $=$ $n*(n-1)*(n-2)*…*3*2*1$ Factorial Formula

$5!$ $=$ $5*4*3*2*1$ Substitute known values

$=$ $120$

$120$

Question 3 of 4

Evaluate

$(12!)/(10!)$

(132)

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Factorial Formula

$n! = n*(n-1)*(n-2)*…*3*2*1$

Apply the formula to the fraction

$n_1$	$=$	$12$
$n_2$	$=$	$10$

$(n_1!)/(n_2!)$	$=$	$(n_1(n_1-1)(n_1-2)…321)/(n_2(n_2-1)(n_2-2)…321)$	Factorial Formula

$(12!)/(10!)$	$=$	$\frac{12\cdot11\cdot\color{#CC0000}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}}{\color{#CC0000}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}}$	Substitute known values

	$=$	$12*11$	Cancel like terms
	$=$	$132$

$132$

Question 4 of 4

Evaluate

$(9!)/(0!)$

(362880)

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Factorial Formula

$n! = n*(n-1)*(n-2)*…*3*2*1$

Zero Factorial

$0! = 1$

Apply the formula to the fraction

$n_1$	$=$	$9$
$n_2$	$=$	$0$

$(n_1!)/(n_2!)$	$=$	$(n_1(n_1-1)(n_1-2)…321)/(n_2(n_2-1)(n_2-2)…321)$	Factorial Formula

$(9!)/(0!)$	$=$	$\frac{9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{1}$	$0! =1$

	$=$	$362 880$

$362 880$

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Quiz summary

Information

1. Question

Keep Going!

Permutation

Combination

2. Question

Hint

Well Done!

Factorial Formula

3. Question

Hint

Nice Job!

Factorial Formula

4. Question

Hint

Excellent!

Factorial Formula

Zero Factorial

Quizzes

$n!$	$=$	$n(n-1)(n-2)…321$	Factorial Formula
$5!$	$=$	$54321$	Substitute known values
	$=$	$120$