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Q: How many 3-digit codes using the digits 0-9 are possible if repetition are allowed?

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5040

-4

24 three digit numbers if repetition of digits is not allowed. 4P3 = 24.If repetition of digits is allowed then we have:For 3 repetitions, 4 three digit numbers.For 2 repetitions, 36 three digit numbers.So we have a total of 64 three digit numbers if repetition of digits is allowed.

125 There are five choices for each of the three digits (since repetition is allowed). So there are 5*5*5=125 combinations.

Just six numbers... 345, 354, 435, 453, 534 & 543

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If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.

5040

5 ^12

290

There are 10 to the 10th power possibilities of ISBN numbers if d represents a digit from 0 to 9 and repetition of digits are allowed. That means there are 10,000,000,000 ISBN numbers possible.

-4

If repetition of digits is allowed, then 56 can.If repetition of digits is not allowed, then only 18 can.

64 if repetition is allowed.24 if repetition is not allowed.

Possible solutions - using your rules are:- 11,13,17,31,33,37,71,73 &77

64 if repetition is allowed, 6 otherwise.64 if repetition is allowed, 6 otherwise.64 if repetition is allowed, 6 otherwise.64 if repetition is allowed, 6 otherwise.

24 three digit numbers if repetition of digits is not allowed. 4P3 = 24.If repetition of digits is allowed then we have:For 3 repetitions, 4 three digit numbers.For 2 repetitions, 36 three digit numbers.So we have a total of 64 three digit numbers if repetition of digits is allowed.

if the repetition is not allowed then the arrangement can be 3*2*1 = 6 ways if repetition is allowed the ways can be 6*6*6= 216 ways

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