# The total number of numbers of not more than 20 digits that are formed by using the digits 0,1,2,3 and 4 is

$

{\text{A}}{\text{. }}{{\text{5}}^{20}} \\

{\text{B}}{\text{. }}{{\text{5}}^{20}} - 1 \\

{\text{C}}{\text{. }}{{\text{5}}^{20}} + 1 \\

{\text{D}}{\text{. None of these}} \\

$

Last updated date: 25th Mar 2023

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Answer

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307.5k+ views

Hint: Not more than twenty digit numbers can be formed from 5 different digits by filling each 20 places by 5 different digits one by one.

Complete step-by-step answer:

Given the digits 0,1,2,3 and 4 to form a number. The condition of forming the number is that it shouldn't have more than 20 digits. We need the numbers of such numbers that are satisfying the given condition.

Let’s take an imaginary number with 20 digit’s places . We have 5 numbers to fill each of the 20 digit’s places i.e. one’s places can be filled by any of the digits 0,1,2,3 or 4. Similarly , all other digit’s places will have 5 digits available to choose.

So, for each digit’s place we have 5 cases possible . If we have to make not more than two digit numbers and each digit places can have 5 cases possible i.e. can be filled by 5 different numbers. Then total numbers possible would be 5(5). Similarly, for not more than 20 digit’s places we will have

$5 \times 5 \times 5 \times ...20{\text{ times}}$. So , the total number of numbers possible is ${5^{20}}$.

Answer is option (A).

Note: In these types of questions, the key concept is to assume a number of maximum digits. As zero is given as an option to fill the digit’s places, the smaller digit number will be covered easily , by taking all the cases.

Complete step-by-step answer:

Given the digits 0,1,2,3 and 4 to form a number. The condition of forming the number is that it shouldn't have more than 20 digits. We need the numbers of such numbers that are satisfying the given condition.

Let’s take an imaginary number with 20 digit’s places . We have 5 numbers to fill each of the 20 digit’s places i.e. one’s places can be filled by any of the digits 0,1,2,3 or 4. Similarly , all other digit’s places will have 5 digits available to choose.

So, for each digit’s place we have 5 cases possible . If we have to make not more than two digit numbers and each digit places can have 5 cases possible i.e. can be filled by 5 different numbers. Then total numbers possible would be 5(5). Similarly, for not more than 20 digit’s places we will have

$5 \times 5 \times 5 \times ...20{\text{ times}}$. So , the total number of numbers possible is ${5^{20}}$.

Answer is option (A).

Note: In these types of questions, the key concept is to assume a number of maximum digits. As zero is given as an option to fill the digit’s places, the smaller digit number will be covered easily , by taking all the cases.

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