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How many 4 digit numbers can be formed using the numbers 1, 2, 3, 4, 5 with digits repeated?

  • Last Updated : 18 Jan, 2022

In mathematics, permutation relates to the function of ordering all the members of a group into some series or arrangement. In other words, if the group is already directed, then the redirecting of its components is called the process of permuting. Permutations take place, in more or less important ways, in almost every district of mathematics. They frequently appear when different commands on certain limited places are observed.

Permutation

A permutation is known as the process of organizing the group, body, or numbers in order, selecting the or numbers from the set, is known as combinations in such a way that the sequence of the integer does not bother.

Permutation Formula

In permutation, r items are collected from a set of n items without any replacement. In this sequence of collecting matter.

nPr = (n!)/(n – r)!

Here,

n = set dimensions, the total number of object in the set

r = subset dimensions, the number of objects to be choose from the set

Combination

The combination is a way of choosing objects from a group, such that (unlike permutations) the sequence of choosing does not matter. In smaller cases, it is imaginable, to sum up, the number of combinations. Combination refers to the combination of n objects taken k at a time without repetition. To mention combinations in which repetition is allowed, the expressions k-selection or k-combination with repetition are frequently used.

Combination Formula

In combination, r objects are selected from a group of n objects and where the sequence of selecting does not matter.

nCr = n!⁄((n – r)! r!)

Here,

n = Number of objects in group

r = Number of objects selected from the group

How many 4 digit numbers can be formed using the numbers 1, 2, 3, 4, 5 with digits repeated?

Solution:

Repetition of digit is allowed. So, for the ones place we have 5 option i.e., 1,2,3,4,5 similarly for tens place we have again 5 option i.e., 1,2,3,4,5 for the hundredth place we have 5 option i.e., 1,2,3,4,5 similarly, for the thousandth place we have 5 option i.e., 1,2,3,4,5.

Total no. of four digit number = 5 × 5 × 5 ×5 

                                               = 625

Similar Questions

Question 1: How many 6 digit numbers can be formed by using the digit 0,1,2,3,4,5. If repetition of digits is allowed?

Answer:

Repetition of digit is allowed. So, for the first place we have 6 option i.e., 0,1,2,3,4,5 similarly for second place we have again 6 option i.e., 0,1,2,3,4,5  for the third place we have 6 option i.e., 0,1,2,3,4,5 for the fourth place we have 6 option i.e., 0,1,2,3,4,5 and for the fifth thousandth place we have 6 option i.e., 0,1,2,3,4,5 and for the sixth place we have 5 option i.e., 1,2,3,4,5 we can’t take 0 at last  place because if 0 will be filled at last place it will not become 6 digit number it will be taken as 5 digit number.

Total no. of six digit number = 4 × 5 × 5 × 5 × 5 × 5

                                             = 12500

Question 2: How many 4-digit even numbers can be set up using the integer (3,5,7,9,1,0) if repetition of digits is not allowed?

Answer:

For even number unit integer must be 0, Now the endure integers are 5 i.e., 3,5,7,9,1 now for the thousand place we have 5 choices for the hundredth place we have 4 choices and for the tens place we have 3 choices

Total no. of 4 digits even number can be found = 5 × 4 × 3

                                                                          = 60 

Question 3: How many 8 digit numbers can be found using the digits 1, 2, 3,4,5,6 and 7 (repetitions allowed) such that the number reads the same from left to right or from right to left?

Solution:

A eight digit number which reads the same left to right and right to left, means the last four digits are same as the first four digits but in the contrasting direction. So it is a four digit number. 

Repetition of digit is allowed. So, for the first number we have 7 option similarly for second number we have again 7 option for the third number we have 7 option and for the fourth number we have 7 option.

 So the possible numbers = 7 × 7 × 7 × 7

                                        = 2,401

Question 4: The number of six-digit numbers that can be established from the digits 1,2,3,4,5,6 and 7 so that digits do not repeat and the last digits are even is

Solution:

Since, last digits are even.  

Therefore, For 1st place can be permeate in 3 ways and last place can be permeate in 2 ways and remaining places can be permeate in  

5P4 ​= 120 ways  

Hence, the number of six digit number, so that the last digits are even, is 3 × 120 × 2 = 720.  

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