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Q: How many different two digit numbers can you form using the digits 1 2 3 5 6 7 and 9 with repetition?

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You have seven different digits (symbols) to choose from, so you can form seven different one digit numbers and 7×7=72=49 different two digit numbers.

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.

I take it that you want to make three digits numbers with 8,7,3, and 6 without repetition. The first digit cane be selected from among 4 digits, the second from 3 digits, the third digit from 2, hence the number of three digit numbers that can be formed without repetition is 4 x 3 x 2 = 24

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.

-123456787

Assuming that the first digit can't be zero: If repetition of digits is permitted: (5 x 6 x 6) = 180 numbers of 3 digits. If repetition of digits is not permitted: (5 x 5 x 4) = 100 numbers of 3 digits.

-4

30 without repetition (6P2) 66 with repetition (12C2)

As there are 5 different digits, allowing for repetition of a digit, the first digit can be any of the 5, and for each of these, the second can be any of the 5, and so for the third, making: 5 x 5 x 5 = 125 different 3-digit numbers.

20 of them, if repetition is not allowed.

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.

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

There are seven possible digits for the first digit and 6 digits for the second (minus one digit for the digit used as the first digit) and 5 options for the last digit (minus one again for the second digit) and then you just multiply them all together to get a total possible combination of 210 numbers that are possible.

There are 4 possible numbers if the digits are not repeated; 18 if they are. Those are 3-digit numbers, assuming that zero would not be a leading digit. If zero is allowed for a leading digit, then you can have 6 for the non repeated, and 27 if repetition.

10

290

5040 different 4 digit numbers can be formed with the digits 123456789. This is assuming that no digits are repeated with each combination.

You can select 9 numbers for the first digit, 8 numbers for the second digit, and 7numbers for the third digit; so 504 (e.g. 9*8*7) different three digit numbers can be written using the digits 1 through 9.

12689 14689 12489

9*8*7 = 504 of them.

3*2*1 = 6 of them.

It is 415968.

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

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