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in Permutations by (46.3k points)
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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

1 Answer

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by (49.3k points)
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Best 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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