Factorial Formula

The factorial is one of the most fundamental mathematical operations in combinatorics, algebra, and number theory. Represented by an exclamation mark (!), the factorial of a non-negative integer n, denoted as n!, is the product of all positive integers less than or equal to n. It plays a crucial role in permutations, combinations, binomial expansions, and more

The formula for the factorial is:

Factorial of n = n! = n × (n - 1) × (n - 2) × ... × 1

Examples:

• 0! = 1
• 1! = 1
• 3! = 3 x 2 x 1 = 6
• 4! = 4 x 3 x 2 x 1 = 24

Factorial Zero (0!)

The value of Zero factorial is equal to 1. It its represented as 0! = 1, to read more about why zero factorial is zero check this article. [Read Here!]

Formula for Counting Trailing 0s in Factorial

Trailing 0s in n! = Count of 5s in prime factors of n! = floor⌊n/5⌋ + floor⌊n/25⌋ + floor⌊n/125⌋ + …

How does this work? A 0 is formed by the multiplication of 5 and 2 (10). So if we consider all prime factors of all numbers from 1 to n, there would be more 2s than 5s. So the number of 0s is limited by a number of 5s. If we count a number of 5s in prime factors, we get the result. Consider the following examples:

Count the number of training zeros in 100!

1. ⌊100/5⌋ = 20 (There are 20 multiples of 5 between 1 and 100).
2. ⌊100/25⌋ = 4 (There are 4 multiples of 25 between 1 and 100).
3. ⌊100/125⌋ = 0 (There are no multiples of 125 between 1 and 100).

So, the total number of trailing zeros in 100! is: 20 + 4 = 24 trailing zeros.

To read more about floor function check - Floor Function ⌊ ⌋

Factorials of Numbers 1 to 10 Table

Here's a clear and concise Factorial Table for Numbers 1 to 10, including both the expansion and the computed value:

Applications of Factorial

The factorial formula is used in many areas, specifically in permutations and combinations of mathematics. For example,

Permutations:

A permutation is an arrangement of objects in a definite order. The members or elements of sets are arranged here in a sequence or linear order

• The number of ways n distinct objects can be arranged in a row is equal to n!
• Permutation gives the number of ways to select r elements from n elements when order matters. It is given using the formula ⁿP_r.

ⁿP_r = n! / (n - r)!

Combinations

The combination is defined as “An arrangement of objects where the order in which the objects are selected does not matter.” The combination means “Selection of things”

• Combination gives the number of ways to select r elements from n elements where order does not matter. It is given as ⁿC_r.

ⁿC_r = n! / r! (n - r)!

Binomial Theorem:

The Binomial Theorem provides a systematic method for expanding expressions raised to a positive integer power. It is a powerful tool in algebra with wide-ranging applications in areas such as probability, combinatorics, and calculus.

(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k

Properties of Factorial

• A factorial of any number is a whole number.
• A factorial can also be represented as a recursive function.

n! = n × (n - 1) × (n - 2) × ... × 1 = n × (n - 1)!

• Factorial of zero is 1, that is 0! = 1.
• Factorial of negative numbers is not defined.

Solved question on Factorial

Question 1: Find the value of factorial of 5.
Solution:

To find the factorial of 5, we need to multiply all the whole numbers smaller than or equal to 5.

5! = 5 × 4 × 3 × 2 × 1 = 120
Hence, 5! = 120

Question 2: Find the value of a number x, given factorial of x is 720.
Solution:

Apply the recursive property of factorial to find x. Until and unless we get 720 as our result, we will proceed recursively.

1! = 1
2! = 2 × 1! = 2
3! = 3 × 2! =6
4! = 4 × 3! = 4 × 6 = 24
5! = 5 × 4! = 5 × 24 = 120
6! = 6 × 5! = 6 × 120 = 720

Since 720 is obtained as the factorial of 6, one can compare the value of x with 6.
Thus, the value of x = 6

Question 3: Find the number of ways 5 distinct objects can be arranged in a row.
Solution:

Use the property that the number of ways n distinct objects can be arranged in a row is equal to n!

Thus, 5 distinct objects can be arranged in 5! = 5 × 4 × 3 × 2 × 1 = 120.
So, the number of ways is equal to 120.

Question 4: Find the number of ways 3 students can be selected from a class of 50 students.
Solution:

To find the number of ways 3 students can be selected from a class of 50 students, we can use the formula for Combination, since the order of the selected three students does not matter here.

Thus, the total number of ways = ⁵⁰C₃
So, this can be simplified as ⁵⁰C₃ = 50! / (3! × 47!) = (50 × 49 × 48 × 47!) / (3! × 47!) = 50 ×49 × 48 / 6 = 19,600

So, there are a total of 19,600 ways.

Question 5: Three different fruits are to be distributed among a group of 10 people. Find the total number of ways this can be possible.
Solution:

Since, in this case, the order of how the fruits are distributed matters, we need to implementPermutation.
So, the total number of ways is given as ¹⁰P₃.

Simplifying, this can be written as,
¹⁰P₃ = 10! / (10 - 3) ! = 10! / 7! = 10 × 9 × 8 × 7! / 7! = 10 × 9 × 8 = 720

Thus, there are a total of 720 ways possible.

Practice Problems on Factorial Formula

Question 1: Find the value of 7!.

Question 2: Determine the value of x if the factorial of x is 5040.

Question 3: How many ways can 6 distinct objects be arranged in a row?

Question 4: Calculate the number of ways 4 students can be selected from a class of 30 students.

Question 5: Find the number of permutations of the word "MATH".

Question 6: How many different ways can 3 identical red balls and 2 identical blue balls be arranged in a row?

Question 7: In how many ways can a committee of 4 members be chosen from a group of 12 people?

Question 8: A password consists of 3 distinct letters chosen from the alphabet. How many possible passwords can be formed?

Question 9: Calculate the number of ways to arrange the letters in the word "COMPUTER" such that the vowels always come together.

Question 10: How many ways can 5 people be seated in a row if two specific people must sit next to each other?