How many numbers greater than 2000 can be formed with the digits 1, 2, 3, 4,5

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How many numbers less than 2000 that can be formed using the digits 1, 2, 3, 4 if repetition is allowed.
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How many numbers less than 2000 that can be formed using the digits 1, 2, 3, 4 if repetition is allowed.

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Hint: First check all possible cases one by one which are less than 2000. Add all cases possible to set the total number of numbers. Check all numbers of digits 1, 2, 3, 4. This total number is the required result in the given question.Complete step-by-step answer:
Given digits which can be used to form the numbers are 1, 2, 3, 4.
The condition in the question, on formed numbers, is given as: The number formed must be less than the number 2000.
Given the condition on repetition of numbers is given in question. Repetition is allowed.
The number of digits possible for the required number is given as 4 digit number, 3 digit number, 2 digit number, 1 digit number.
Case (1): The number of 4 digit numbers possible, are given by: As the number must be less than 2000, the possibilities on the first digit is written as only 1 which is the digit 1. For remaining all the numbers are greater than 2000.
The possibilities for the second digit of the number are all numbers, S0, 4. As repetition is allowed, the possibilities for third digit are also 4. As repetition is allowed, the possibilities for fourth digit are also 4.
Rule of Sum: In combinatorics, the rule of sum or addition principle is basic counting principle. It is simply defined as, if there are A ways of doing P work and B ways of doing Q work. Given P, Q words cannot be done together. Total number of ways to do both P, Q are given by (A+B) ways.
Rule of Product: In combinatorics, the rule of product or multiplication principle is basic counting principle. It is simply defined as, if there are A ways of doing P work and B way of doing Q work given P, Q works can be done at a time. Total number of ways to do both P, Q works are given by (A.B) ways.
As they all are done together we can use product rule:
Total possibilities= $ 1\times 4\times 4\times 4 $
By simplifying the above equation, we can write its value as:
Total possibilities in case (1)=64
Case (3): The number of 3 digit numbers possible are given by:
Any 3 digit number is less than 2000, because of the number of digits.
As repetition allowed, the number of possibilities for the first digit is 4.
The number of possibilities for the second digit are 4.
The number of possibilities for the third digit are 4.
As they all can be done together, we will use product rule:
Total possibilities $ =4\times 4\times 4 $
By simplifying the above equation, we can write is as:
Total possibilities = 64
Case (3) : The number of 2 digit numbers possible are given by:
Any 2 digit number is less than 2000, because of the number of digits.
As repetition allowed, the number of possibilities for the first digit is 4.
As repetition allows, the number of possibilities for the second digit are 4.
As they all can be done together, we will use product rule:
Total possibilities $ =4\times 4=16 $
Case (4): The number of 1 digit numbers possible are possible are given by:
Any 1 digit number is less than 2000. The possibilities for 1 digit, is 4 numbers given:
By adding all cases we can say the equation that:
Total numbers = case (1)+ case (2)+ case (3)+ case (4).
By substituting their values, we get the equation in form:
Total numbers =64+64+16+4 = 148
Therefore 148 numbers are possible.

Note: Generally students forget repetition cases. While taking 4 digits they forget and take cases even for the first digit being 2. But for numbers to be less than 2000 the first digit must only be 1. Generally students forget and use sum rules instead of product rules . But look at definitions carefully to select one of them and apply them carefully. Don’t misplace terms.

There are a total of #1880# such numbers.

Explanation:

Assuming we are allowed to use each digit at most once to form decimal strings then, we can enumerate the possibilities as follows:

There are #6*5*4*3*2*1 = 720# suitable #6# digit numbers.
There are #6*5*4*3*2 = 720# suitable #5# digit numbers.
There are #6*5*4*3 = 360# suitable #4# digit numbers.
There are #4*5*4 = 80# suitable #3# digit numbers.

That makes a total of:

#720+720+360+80 = 1880#

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Updated On: 27-06-2022

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