5 digit numbers from 1 3 3 0 0 How many 5 digit number positive integers can be made from the following numbers: $1,3,3,0,0$?
First I calculate all possible combinations $$\frac{5!}{2!2!}=30$$
Then I need to remove all 5 digit numbers startting with $0$ and $00$. They are not 5 digit numbers positive integers.
Starting with $00$ $$\frac{3!}{2!}=3$$
Starting with $0$ $$\frac{4!}{2!}=12$$
$12-3=9$, else I would remove numbers starting with $00$ double.
$30-3-9=18$
Is $18$ the correct answer, or where did i thought wrong?
 A: Working from the other direction, the first digit is either $1$ or $3$, and then there are no restrictions on the remaining digits which can be permuted in $4!/2!/2!=6$ and $4!/2!=12$ ways respectively. Thus I arrive at $18$ ways, and you are correct.
A: Another way to puzzle through:  There are $3$ ways to arrange the digits $1,3,3$.  Given one of these arrangements, say $313$ there are $3$ places to put two $0$'s:
$$3X1X3X$$
Think of each $X$ as a basket and now you have two zeros to put in the three baskets.  There are $3$ ways to put both zeros in one basket.  Then there are three ways to put the zeros in different baskets.  So for each of the three arrangements of non-zero digits, there are $6$ ways to toss in the zeros.  So $18$ is the answer.
A: In the simplest of terms (as a check), without using permutations and combinations:  You draw from the available choices randomly to calculate all combinations.
The first number is one of 3-choices (1, 3, or 3).
The second number is one of 4-choices (all that remain).
The third is one of the 3-that remains.
Then one of 2.
And finally, the last choice.
There are 4 different ways to arrange the number 13300 (first 3 first, second 3 first, and the same with the zeros in combination.
$$3\cdot 4\cdot 3\cdot 2\cdot 1 = 72$$
Then, to account for the double $0$s and double $3$s, divide by 2 twice.
$$\frac{72}{2\cdot 2} = 18$$
A: I would just compute the total permutations , and multiply it by the probabilty $\frac{3}{5}$ of starting with a non-zero
$\dfrac{5!}{2!2!}\cdot\dfrac35 = 18$
