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Q: How many 3 digit numbers can be form from the numbers 1-9 if repetition is not allowed?

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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.

10

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

It is 10000.

The number of six digit numbers that you can make from ten different digits ifrepetitions of same digit on the six digit number is allowed is 1 000 000 numbers(including number 000 000).If no repetitions of the the same digit are allowed then you have:10P6 = 10!/(10-6)! = 151 200 different six digit numbers(six digit permutations form 10 different 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.

If you allow the first digit to be zero, then you can form (5 x 4 x 3 x 2 x 1) = 120 numbers. If the first digit can't be zero, then you can only form (4 x 4 x 3 x 2 x 1) = 96 numbers.

7*7 = 49 numbers.7*7 = 49 numbers.7*7 = 49 numbers.7*7 = 49 numbers.

7432

yes.

That would be the numbers in the form "32x" (where "x" can be any digit). In other words, ten numbers.

If you're allowed to use zero as the first digit, then that would be 10!/7! = 10 * 9 * 8 = 720. If you're not allowed to use zero as your first digit, then it would be 9 * 9 * 8 = 648.

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