Factorial Numbers
Factorial value of a number is the product of a given number and all smaller positive numbers. The factorial of n is written as n! and is read aloud "n factorial".
By definition, 0! = 1.
Factorial formula: n! = n · (n – 1) · (n – 2) · · · 3 · 2 · 1.
For example: 5! = 5 · 4 · 3 · 2 · 1 = 120.
Factorial Numbers - Examples
Example 1: Addition 4! + 5!
Solution:
4! + 5! = (4 ·3 · 2 · 1) + (5 · 4 · 3 · 2 · 1) = 24 + 120 = 144
Therefore 4! + 5! = 144
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Example 2: Subtraction 5! - 4!
Solution:
5! - 4! = (5 · 4 ·3 · 2 · 1) - (4 · 3 · 2 · 1) = 120 - 24 = 96
Therefore 5! + 4! = 96
Example 3: Multiplication 6! · 3!
Solution:
6! · 3! = (6 · 5 · 4 ·3 · 2 · 1) · (3 · 2 · 1) = 720 · 6 = 4320
Therefore 6! · 3! = 4320
Example 4: Division 6! ÷ 3!
Solution:
6! ÷ 3! = (6 · 5 · 4 ·3 · 2 · 1) ÷ (3 · 2 · 1) = 720 ÷ 6 = 120
Therefore 6! ÷ 3! = 120
Example 5: In how many ways can a person, 3 numbers be chosen from among 7 numbers?
Solution:
The problem involves 7 numbers taken 3 at a time.
Using permutations formula
P(7,3) = 7! / (7 - 3)! = 7! / 4! = 7 · 6 · 5 · 4! / 4! = 210
There are 210 possible ways to choose a person, 3 numbers be chosen from among 7 numbers.
Between, if you have problem on these topics Numbers in Ascending Order please browse expert math related websites for more help on prime number 1-100.
Example 6: In how many ways can a person, 5 apples be chosen from among 10 apples?
Solution:
The problem involves 10 apples taken 5 at a time.
Using permutations formula
P(10,5) = 10! / (10 - 5)! = 10! / 5! = 10 · 9 · 8 · 7 · 6 · 5! / 5! = 30240
There are 30240 possible ways to choose a person, 5 apples be chosen from among 10 apples.
Example 7: A word contains 5 letters. How many different words can be made with the 9 letters (A – I) if no letter is used more than once?
Solution:
By the reasoning
For the first position, there are 9 possible choices. Then, there are 8 choices, 7 choices and 6 possible choices.
9 × 8 × 7 × 6 = 3024
By using permutation formula
There are 9 possible choices for the first position. For the next 4 positions, we are selecting from 9 letters.
P(9,5) = 9! / (9 – 5)! = 9! / 4! = 3024
Example 8: A number plate contains 4 digits. How many different digits can be made with the 9 digits (1 – 9) if no digit is used more than once?
Solution:
By the reasoning
For the first position, there are 9 possible choices. Then, there are 8 possible choices, 7 possible choices, 6 possible choices and 5 possible choices.
9 × 8 × 7 × 6 × 5 = 15120
By using permutation formula
There are 9 possible choices for the first position. For the next 5 positions, we are selecting from 9 digits.
P(9,4) = 9! / (9 – 4)! = 9! / 5! = 15120
Factorial Numbers - Practice
Problem 1: In how many ways can 6 balls be chosen from among 12 balls?
Answer: 924
Example 2: In how many ways can a person, 4 marbles be chosen from among 8 marbles?
Answer: 70
By definition, 0! = 1.
Factorial formula: n! = n · (n – 1) · (n – 2) · · · 3 · 2 · 1.
For example: 5! = 5 · 4 · 3 · 2 · 1 = 120.
Factorial Numbers - Examples
Example 1: Addition 4! + 5!
Solution:
4! + 5! = (4 ·3 · 2 · 1) + (5 · 4 · 3 · 2 · 1) = 24 + 120 = 144
Therefore 4! + 5! = 144
I have recently faced lot of problem while learning Properties of Numbers But thank to online resources of math which helped me to learn myself easily on net.
Example 2: Subtraction 5! - 4!
Solution:
5! - 4! = (5 · 4 ·3 · 2 · 1) - (4 · 3 · 2 · 1) = 120 - 24 = 96
Therefore 5! + 4! = 96
Example 3: Multiplication 6! · 3!
Solution:
6! · 3! = (6 · 5 · 4 ·3 · 2 · 1) · (3 · 2 · 1) = 720 · 6 = 4320
Therefore 6! · 3! = 4320
Example 4: Division 6! ÷ 3!
Solution:
6! ÷ 3! = (6 · 5 · 4 ·3 · 2 · 1) ÷ (3 · 2 · 1) = 720 ÷ 6 = 120
Therefore 6! ÷ 3! = 120
Example 5: In how many ways can a person, 3 numbers be chosen from among 7 numbers?
Solution:
The problem involves 7 numbers taken 3 at a time.
Using permutations formula
P(7,3) = 7! / (7 - 3)! = 7! / 4! = 7 · 6 · 5 · 4! / 4! = 210
There are 210 possible ways to choose a person, 3 numbers be chosen from among 7 numbers.
Between, if you have problem on these topics Numbers in Ascending Order please browse expert math related websites for more help on prime number 1-100.
Example 6: In how many ways can a person, 5 apples be chosen from among 10 apples?
Solution:
The problem involves 10 apples taken 5 at a time.
Using permutations formula
P(10,5) = 10! / (10 - 5)! = 10! / 5! = 10 · 9 · 8 · 7 · 6 · 5! / 5! = 30240
There are 30240 possible ways to choose a person, 5 apples be chosen from among 10 apples.
Example 7: A word contains 5 letters. How many different words can be made with the 9 letters (A – I) if no letter is used more than once?
Solution:
By the reasoning
For the first position, there are 9 possible choices. Then, there are 8 choices, 7 choices and 6 possible choices.
9 × 8 × 7 × 6 = 3024
By using permutation formula
There are 9 possible choices for the first position. For the next 4 positions, we are selecting from 9 letters.
P(9,5) = 9! / (9 – 5)! = 9! / 4! = 3024
Example 8: A number plate contains 4 digits. How many different digits can be made with the 9 digits (1 – 9) if no digit is used more than once?
Solution:
By the reasoning
For the first position, there are 9 possible choices. Then, there are 8 possible choices, 7 possible choices, 6 possible choices and 5 possible choices.
9 × 8 × 7 × 6 × 5 = 15120
By using permutation formula
There are 9 possible choices for the first position. For the next 5 positions, we are selecting from 9 digits.
P(9,4) = 9! / (9 – 4)! = 9! / 5! = 15120
Factorial Numbers - Practice
Problem 1: In how many ways can 6 balls be chosen from among 12 balls?
Answer: 924
Example 2: In how many ways can a person, 4 marbles be chosen from among 8 marbles?
Answer: 70