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I have a maths question that asks for the amount of 4 digit numbers that can be made out of 2,3,4 and 0 (0 cannot be used as the first digit), each number can only be used once. I don't understand it so could someone please tell me in the simplest way what is the answer and if there is a formula that I can use for other questions similar to this?

Thanks!

Ashley342
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  • you have 3 choices for the 1st digit, because 0 is not available. And you have 3 choices for the 2nd digit, because you can't repeat the digit you have already used. You have 2 choices for the 3rd digit, and the choices for the previous 3 digits determines the last digit. $3\cdot3\cdot2 = 18$ – Doug M Apr 04 '17 at 00:48

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The simplest way (I think) is to temporarily ignore the condition that zero cannot be the first digit. Find out how many numbers you can get without that condition, and then subtract the quantity of numbers that violate the condition.

The number of possible ways to list four of the digits $0,2,3,4,$ using each digit only once, is simply the number of ways you can permute four objects: $4! = 24.$

But among those $24$ sequences of digits are some that start with zero. The ones that start with zero are all three-digit numbers made of the other three digits $2,3,4.$ There are $3!=6$ ways to arrange the digits $2,3,4,$ so we must subtract these $6$ ineligible numbers from the earlier count of $24,$ with the result $24-6=18.$

David K
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You can use some tools from Combinatorics. For your specific question, here's the underlying reasoning (bolded numbers are the digits):

  1. How many options do you have for your first digit? You can use 2,3 or 4, so you have 3 options.
  2. How many options you have left for your second digit? It can't be the same as the first, but now you can use the 0. So if you chose, say, 2, you can use 3, 4 or 0. Thus you can choose from 3 digits.
  3. How many options you have left for the third digit? Well, you can use any you didn't use before, so there are 2 numbers to choose from
  4. Well, you only have one digit left, so only one option here.

Now, working backwards: for every third digit you have $1$ choice, so a total of $2$. For every second digit, you have 2 choices for third and fourth digits, so that's $3\times2=6$ options. And for each first digit, you have 6 options of second, third and fourth digits, so you have $3\times6=18$ options in total.

While doing exercises, you can list how many options you begin with, then narrow them down one step at a time with the assumptions you have. Then you usually multiply the number of options for every step (but beware of some traps). With enough practice, you can skip some steps or do those calculations mentally.

AspiringMathematician
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