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Q: Where d represents a digit from 0 9 how many different ISBN numbers are possible if repetition of digits is allowed?

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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.

5040

1000

290

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

If repetition is allowed . . . . . 343 If repetition is not allowed . . . . . 210

-4

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

Yes.

If repetition is allowed [such as 122 or 555], then there are 5Â³ = 125 possibilities.

5 ^12

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

if the repetition is allowed the there is 6*6*6 possible ways = 216

Three numbers can be arranged in 27 different sequences if repetition is allowed, and in 6 different sequences if it's not.

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

9,999 if zeroes can be used than it would be 10,999

many i believe 8 or so??

Assuming 9 numbers chosen from 56, with no repetition allowed, there are 7575968400 possible combinations.

20 of them, if repetition is not allowed.

If no repetition, 6! ie 720; if repetition allowed, 66 ie 46656.

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

10

If repetition is allowed the formula would be 4x4x4 = 64 codes. If you must chose a different letter each time (no repetition) the formula would be 4x3x2 = 24 codes.

10^7 if the repetition of digits is allowed. 9*8*7*6*5*4*3 , if the repetition of digits is not allowed.

If order is important and repetition is allowed, then : 000 through 999 (1000 possible). If no leading zeros, then start at 100 through 999 (900 possible). See related link, if you need to limit to no repetition, or order doesn't matter (like if 123 is the same as 231).