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How many natural numbers not exceeding 4321 can be formed with the digits 1, 2, 3, 4 if repetition is allowed? (a) 123 (b) 113 (c) 222 (d) 313 |
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Answer» (d) 313 As there are 4 digits 1, 2, 3, 4 and repetition of digits is allowed. Total number of 1-digit numbers = 4 Total number of 2-digit numbers = 4 × 4 = 16 Total number of 3-digit numbers = 4 × 4 × 4 = 64 Number of 4-digit numbers beginning with 1 = 4 × 4 × 4 = 64 (∵ The first place is occupied by 1) Number of 4-digit numbers beginning with 2 = 4 × 4 × 4 = 64 Number of 4-digit numbers beginning with 3 = 4 × 4 × 4 = 64 Number of 4-digit numbers beginning with 41 = 4 × 4 = 16 Number of 4-digit numbers beginning with 42 = 4 × 4 = 16 Number of 4-digit numbers beginning with 431 = 4 Number of 4-digit numbers beginning with 432 = 1 (4321 only) ∴ Total number of 4-digit numbers = 64 + 64 + 64 + 16 + 16 + 4 + 1 = 229 ∴ Total number of natural numbers not exceeding 4321 = 4 + 16 + 64 + 229 = 313. |
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