# How many 3-digit numbers can be formed using the digits 2, 3, 4 and 5 as often as desired?

The method I tried:

5 x 4 x 3 = 60 different numbers

I solved it this way because if a number needs to be formed with a certain number of digits (with no restrictions - which I think 'as often as desired' means), you assign a scale to the given constituent digits and then pick the same number of digits as the new number needs, going from highest to lowest. Then multiply these digits by each other to give the amount of different numbers that could be created.

However my answer of 60 possible numbers conflicts with the textbook's answer of 64.

The next part of the question was finding how many 3-digit numbers can be formed using 2, 3, 4 and 5 using at most one each. I was able to get this question, by changing 2, 3, 4 and 5 to 1, 2, 3 and 4; then multiplying 4 by 3 by 2 to give 24 possibilities.

• I really understand your method at all; I have no idea what you mean by assigning a scale, etc. There are four choices for the first digit, four for the second, four for the third; this gives a total of $4 \times 4 \times 4 = 64$. – user296602 Jul 29 '16 at 1:54
• The answer of $5\times 4\times 3=60$ is correct for the (almost unrelated) question "How many 3-digit numbers can be formed from the digits $\color{red}{1},2,3,4,5$ $\color{red}{\text{where digits may not be repeated}}$?" – JMoravitz Jul 29 '16 at 1:57
• Note that the choice of digits is completely irrelevant to the problem (except for zero, as we do not call a three-digit string of numbers which starts with zero a three digit number). The digits could have been 2,3,4,5. they could have been 2,5,8,9, whatever. There is no reason to "change 2,3,4,5 into 1,2,3,4" – JMoravitz Jul 29 '16 at 1:59
• @JMoravitz Thank you, you explained this very clearly. – Mad Banners Jul 29 '16 at 2:07

In each of the $3$ digit positions there are $4$ possibilities, so the number of possible combinations (reusing digits) is $4\cdot4\cdot4=4^3$.
If you cannot reuse digits, then the number of possibilities for the first digit is $4$; the number of possibilities for the second digit after choosing the first digit is $3$ since you've already used one; the number of possibilities for the third digit would be $2$. This gives you $4 \cdot 3 \cdot 2$.