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I have these math problem:

Facts:

  • 4 numbers consisting of numbers 0 to 9
  • The first number ist not zero
  • A number may not occur more than once (No repetition)

How many numbers can I generate?

My Ideas: I have 8 possibilities for the first number.I have to calculate the other numbers with the Binomial-coefficient, but I don't exactly know how.

2 Answers 2

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As first place digit we don't have 0 at first place. So first digit can be choosen in 9 ways.

After placing one digit at first place we have 9 remaining digits. As second digit can be 0.

Similarly for 3rd place 8 options and for 4th we have 7 ways.

So total number of ways = $ 9 \times 9 \times 8 \times 7 = 4536$

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    Now it's clear, thanks! – 2017-01-22
  • 0
    Your welcome... – 2017-01-22
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You are correct in thinking that the first cannot be zero, or it would be three digit.

You have 9 possibilities for the first number, since you are excluding zero.

For the second, too, you have 9, since zero can be Included here.

For the third you have 8, since 2 numbers have already been used, and similarly 7 for the last. So the total number of ways is $$9×9×8×7=4536$$