Answer
Verified
37.2k+ 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.
Recently Updated Pages
To get a maximum current in an external resistance class 1 physics JEE_Main
f a body travels with constant acceleration which of class 1 physics JEE_Main
If the beams of electrons and protons move parallel class 1 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Let f be a twice differentiable such that fleft x rightfleft class 11 maths JEE_Main
Find the points of intersection of the tangents at class 11 maths JEE_Main
Other Pages
The mole fraction of the solute in a 1 molal aqueous class 11 chemistry JEE_Main
A convex lens is dipped in a liquid whose refractive class 12 physics JEE_Main
In the given circuit the current through the 5mH inductor class 12 physics JEE_Main
Formula for number of images formed by two plane mirrors class 12 physics JEE_Main
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
Explain the construction and working of a GeigerMuller class 12 physics JEE_Main