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How many 3-digit numbers can be made with digits 1, 2, and 3?

Number System is a mathematical value used for counting and measuring objects, and for performing arithmetic calculations. It is a system of writing for expressing numbers. It gives a special representation to every number and represents the arithmetic and algebraic form of the number. It allows us to operate arithmetic operations like addition, subtraction, multiplication, and division.

An equation is a statement that connects two algebraic expressions of the same values with the ‘=’ sign. For example: In equation 9x + 4 = 7, 9x + 4 is the left-hand side expression and 7 is the right-hand side expression connected with the ‘=’ sign.



What is a Number?

A word or symbol that indicates a quantity is known as a number. The numbers 2, 4, 6, etc. are even numbers and 1, 3, 5, etc. are odd numbers. A number is a value created by the merger of integers. These numbers are used to represent algebraic quantities. An integer is a sign from a set of 10 characters ranging from 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Any combination of integers represents a number. The size of a Number depends on the count of digits that are used for its formation. For Example: 126, 128, 0.356, -12, 78, 94 etc.



Types of Numbers

Numbers are of various types depending upon the patterns of digits that are used for their creation. Various symbols and rules are also applied to Numbers which classifies them into a variety of different types:

Permutations And Combinations 

Permutation is the different arrangements of a given number of components taken one by one, or some, or all at a time. For example, if we have two components X and Y, then there are two possible arrangements, XY and YX.

The number of permutations when ‘r’ elements are arranged out of a total of ‘n’ elements is 

nPr = n!/(n – r)! 

For example, let n = 4 (A, B, C and D) and r = 2 (All permutations of size 2). The answer is 4!/(4-2)! = 12. So, the twelve permutations are AB, AC, AD, BA, BC, BD, CA, CB, CD, DA, DB, and DC.

Combination is the different selections of a given number of components taken one by one, or some, or all at a time. For example, if we have two components A and B, then there is only one way to select two items, we select both of them.

Number of combinations when ‘r’ elements are selected out of a total of ‘n’ elements is 

nCr = n!/[(r!) × (n – r)!]

For example, let n = 4 (A, B, C and D) and r = 2 (All combinations of size 2). The answer is 4!/((4-2)! × 2!) = 6. So, the six combinations are AB, AC, AD, BC, BD, CD.

nCr = nC(n –r)

Note: In the same example, we have different cases for permutation and combination. If we talk about permutation, AB and BA are two different things but for selection, AB and BA are same.

How many 3-digit numbers can be made with digits 1, 2, and 3?

(i) repetition of the digits is allowed?

Solution:

Answer: 27

Method:

Here, Total number of digits = 3

Let us assume the 3-digit number be ABC.

Now the number of digit available for A=3

As repetition is allowed,

So the number of digits available for B and C will also be 3 (each).

Thus, The total number of 3-digit numbers that can be formed = 3 × 3 × 3 = 27

(ii) repetition of the digits is not allowed?  

Solution:

Answer: 6

Method:

Here, Total number of digits = 3

Let us assume 3-digit number be ABC.

Now the number of digits available for A = 3,

As repetition is not allowed,

So the number of digits available for B = 2 (As one digit has already been chosen at A),

Similarly, the number of digits available for C = 1.

Thus, the total number of 3-digit numbers that can be formed = 3 × 2 × 1 = 6.

Similar Questions

Question 1: How many 3 digit numbers can be made with 4 digits 1, 2, 3, 4?

(i) repetition of the digits is allowed?

Solution:

Answer: 64

Method:

Here, Total number of digits = 4

Let us assume the 3-digit number be ABC.

Now the number of digit available for A=4

As repetition is allowed,

So the number of digits available for B and C will also be 4 (each).

Thus, The total number of 3-digit numbers that can be formed = 4 × 4 × 4 =64

(ii) repetition of the digits is not allowed?  

Solution:

Answer: 24

Method:

Here, Total number of digits = 4

Let us assume 3-digit number be ABC.

Now the number of digits available for A = 4,

As repetition is not allowed,

So the number of digits available for B = 3 (As one digit has already been chosen at A),

Similarly, the number of digits available for C = 2.

Thus, The total number of 3-digit numbers that can be formed = 4 × 3 × 2 = 24

Question 2: How many 3-digit numbers can be formed from the digits 1, 2, 3, 4, and 5 assuming that –  

(i) repetition of the digits is allowed?

Solution:  

Answer: 125

Method:

Here, Total number of digits = 5

Let 3-digit number be ABC.

Now the number of digits available for A = 5,

As repetition is allowed,

So the number of digits available for B and C will also be 5 (each).

Thus, The total number of 3-digit numbers that can be formed = 5 × 5 × 5 = 125.

(ii) repetition of the digits is not allowed?  

Solution:

Answer: 60

Method:

Here, Total number of digits = 5

Let 3-digit number be ABC.

Now the number of digits available for A = 5,

As repetition is not allowed,

So the number of digits available for B = 4 (As one digit has already been chosen at A),

Similarly, the number of digits available for C = 3.

Thus, The total number of 3-digit numbers that can be formed = 5 × 4 × 3 = 60.

Question 3: How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?

Solution:

Answer: 108

Method:

Here, Total number of digits = 6

Let 3-digit number be ABC.

Now, as the number should be even so the digits at unit place must be even, so number 

of digits available for C = 3 (As 2, 4, 6 are even digits here),

As the repetition is allowed,

So the number of digits available for A = 6,

Similarly, the number of digits available for B = 6.

Thus, The total number of 3-digit even numbers that can be formed = 6 × 6 × 3 = 108

Question 4: How many 4-letter code can be formed using the first 10 letters of the English alphabet if no letter can be repeated?  

Solution:

Answer: 5040

Method:

Here, Total number of letters = 10

Let the 4-letter code be 1234.

Now, the number of letters available for 1st place = 10,

As repetition is not allowed,

So the number of letters possible at 2nd place = 9 (As one letter has already been 

chosen at 1st place),

Similarly, the number of letters available for 3rd place = 8,

and the number of letters available for 4th place = 7.

Thus, The total number of 4-letter code that can be formed = 10 × 9 × 8 × 7 = 5040


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