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How Many Natural Numbers Not Exceeding 4321 Can Be Formed with the Digits 1, 2, 3 and 4, If the Digits Can Repeat? - Mathematics

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How many natural numbers not exceeding 4321 can be formed with the digits 1, 2, 3 and 4, if the digits can repeat?

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Solution

Case I: Four-digit number
Total number of ways in which the 4 digit number can be formed =`4xx4xx4xx4=256`

Now, the number of ways in which the 4-digit numbers greater than 4321 can be formed is as follows:
Suppose, the thousand's digit is 4 and hundred's digit is either 3 or 4.
∴ Number of ways =`2xx4xx4=32`

But 4311, 4312, 4313, 4314, 4321 (i.e. 5 numbers) are less than or equal to 4321.
∴ Remaining number of ways =`256-(32-5)=229`

Case II: Three-digit number
The hundred's digit can be filled in 4 ways.
Similarly, the ten's digit and the unit's digit can also be filled in 4 ways each. This is because the repetition of digits is allowed.
∴ Total number of three-digit number =`4xx4xx4=64`

Case III: Two-digit number
The ten's digit and the unit's digit can be filled in 4 ways each. This is because the repetition of  digits is allowed.
∴ Total number of two digit numbers `4xx4=16`

Case IV: One-digit number
Single digit number can only be four.
∴ Required numbers = 229 + 64 + 16 +4 = 313

Concept: Permutations
  Is there an error in this question or solution?

APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 16 Permutations
Exercise 16.2 | Q 33 | Page 16

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