What is a Factorial Notation?

Sometimes, to find order, arrangements, or combinations of objects are required. Combinatorics is that branch of mathematics that focuses on the study of counting. So the fundamental counting principle was introduced which states that if one event has m possible outcomes and the second event has n possible outcomes, then there are m × n outcomes for two events together. Here, the order is not important. So to focus on the order or arrangement of objects permutations were introduced and for the selection of objects, combinations were introduced. 

Permutation and Combination

A permutation is the arrangement of objects in a specific order. Here, order plays a very vital role. Permutations are very helpful as they help to place the items in sequence. For example, there are 3 letters X, Y, Z. If permutations of 3 letters are required, they are XYZ, XZY, YXZ, YZX, ZXY, ZYX. 

Suppose there are n objects and in order to arrange k objects at a time, the formula for permutation:

P(n, k) = n! /(n – k)! 

Combinations, on the other hand, means the selection of objects. Here the order does not matter. For example, there are three Girls G₁, G₂, G_3, a team of two girls is required. It can be G₁-G₂, G₃-G₁, G₂-G₃. Here G₂-G₃ and G₃-G₂ are the same. 

Suppose there are n objects and in order to find combination for k objects. The formula for combination is:

C(n, k) = n! / (k! x (n – k)!) 

What is a Factorial Notation?

Answer:

Let n be a positive integer. The factorial of n is denoted by n!. It is the product of all positive integers less than Or equal to one. In the factorial, decrease the number by 1, 2, 3, and so on and keep on multiplying till the value becomes 1. The formula is:

n! = n(n – 1)(n – 2)(n – 3)(n – 4)… 

For example:

Let’s calculate the value of 6! 

6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

Let’s calculate the value of 9! 

9! = 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 362880

Note: The value of 0! is 1

Similar Problems

Question 1: Evaluate the value of 8! /(3! × 2!) 

Solution: 

The value of 8! /(3! × 2!) = (8 × 7 × 6 × 5 × 4 × 3!) /(3! × 2!) 

= 8 × 7 × 6 × 5 × 4/(3 × 2 × 1) 

= 1120

Question 2: Evaluate, n! /(n + 3)! 

Solution: 

The value is n! /[(n + 3) × (n + 2) × (n + 1) × (n!)]

Reducing the expression,

1/(n + 3) × (n + 2) × (n + 1) 

Question 3: Find the number of words with or without meaning that can be formed from the word “WORST” (no repetition of letters). 

Solution: 

Since meaning is not important and letters do not repeat in a word we can arrange the letters in any way. 

Number of letters = 5

Therefore the number of words that can be formed is 5! = 5 × 4 × 3 × 2 × 1 = 120

Question 4: Evaluate the value of 3! + 5! 

Solution: 

Let’s find the value of 3! and 5! separately

3! = 3 × 2 × 1 = 6

5! = 5 × 4 × 3 × 2 × 1= 120

Adding the values, answer will be 126

Question 5: Find the value of x, 1/4! + 1/3! = x/5! 

Solution:

Take 3! common from the denominator and reduce the 5! 

1/3! (1 + 1/4) = x/5! 

5/4 = x/20

x = 25