Questions & Answers

Question

Answers

$

{\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}} \\

$

Answer
Verified

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.

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

So, for each digitâ€™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 digitâ€™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 digitâ€™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.

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

So, for each digitâ€™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 digitâ€™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 digitâ€™s places, the smaller digit number will be covered easily , by taking all the cases.

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