discrete mathematics
How many $11$-digit numbers are there which contain all of the digits $0$–$9$ at least once?
Note that in this question a number cannot begin with a digit of $0$.
My instructor explained this as a case problem, but rushed through it without pausing for questions. The answer ended up being $99 \times (10!/2!)$, but I have no idea how to get there!
Note: I initially solved the problem using tedious case work (Method 2). I then figured out a more efficient method (Method 1) in which I considered whether or not the digit $3$ is used twice. I decided to leave Method 2 as a check on Method 1.
Method 1: We consider whether or not the digit $3$ is used twice.
The requirement that the number must be odd means the units digit is either $3$, $5$, or $7$.
Case 1: The digit $3$ is used at most once. Hence, we are selecting from the set $\{2, 3, 5, 6, 7\}$ without repetition.
One-digit numbers: There are three possible single-digit odd numbers, namely $3$, $5$, and $7$.
Two-digit numbers: There are three choices for the units digit. For each such choice, there are $4$ choices for the tens digit (any number other than the units digit). Therefore, there are $4 \cdot 3 = 12$ such numbers.
Three-digit numbers: The requirement that the number be less than $600$ means the hundreds digit must be less than $6$.
If the units digit is $7$, there are three choices for the hundreds digit ($2$, $3$, or $5$) and three choices for the tens digit (since we must exclude the units digit and the hundreds digit). Thus, there are $3 \cdot 3 = 9$ such choices.
If the units digit is $3$ or $5$, there are two choices for the hundreds digit (since we must exclude the units digit from the choices $2$, $3$, or $5$) and three choices for the tens digit (since we must exclude the units digit and the hundreds digit). Thus, there are $2 \cdot 3 \cdot 2 = 12$ such choices.
In all, there are $3 + 12 + 9 + 12 = 36$ odd numbers less than $600$ that can be formed if the digit $3$ is used at most once.
Case 2: The digit $3$ is used twice.
There must be at least two digits.
Two-digit numbers: There is just one, namely $33$.
Three-digit numbers: There are two three-digit odd numbers in which both the hundreds digit and tens digit are $3$'s, namely $335$ and $337$.
If both the the hundreds digit and units digit are $3$'s, there are four choices for the tens digit, namely $2$, $3$, $6$, or $7$.
If both the tens digit and units are $3$'s, the requirement that the number be less than $600$ means that the units digit is either $2$ or $5$.
Hence, there are $1 + 2 + 4 + 2 = 9$ odd numbers less than $600$ that can be formed if the digit $3$ is used twice.
Total: Combining the two mutually exclusive cases yields $36 + 9 = 45$ odd numbers less than $600$ that can be formed using the digits $2, 3, 3, 5, 6, 7$ without repetition.
Method 2: Tedious case work.
The requirement that the number must be odd means the units digit is either $3$, $5$, or $7$.
Case 1: Single-digit odd numbers.
There are three such numbers, namely $3$, $5$, and $7$.
Case 2: Two-digit odd numbers.
The units digit is $3$. Then we have five choices for the tens digit, namely $2$, $3$, $5$, $6$, or $7$.
The units digit is $5$. Then we have four choices for the tens digit, namely $2$, $3$, $6$, or $7$.
The units digit is $7$. Then we have four choices for the tens digit, namely $2$, $3$, $5$, or $6$.
Hence, it is possible to form $5 + 4 + 4 = 13$ odd two-digit numbers if we use the digits $2, 3, 3, 5, 6, 7$ at most once in each number.
Case 3: Three digit odd numbers less than $600$.
The units digit is $3$. Since the number must be less than $600$, the hundreds digit is either $2$, $3$, or $5$.
The units digit is $5$. Since the number must be less than $600$, the hundreds digit is either $2$ or $3$.
The units digit is $7$. Since the number must be less than $600$, the hundreds digit is either $2$, $3$, or $5$.
In all, the number of three-digit odd numbers less than $600$ that can be formed using the digits $2, 3, 3, 5, 6, 7$ is $$3 \cdot 4 + 3 + 4 + 2 \cdot 3 + 4 = 12 + 3 + 4 + 6 + 4 = 29$$
Total: Thus, we have a total of $3 + 13 + 29 = 45$ odd numbers less than $600$ that can be formed using the digits $2, 3, 3, 5, 6, 7$ without repetition.
You are wrong in assuming that even numbers can't be divisible by 3. As a counter example, 6 is divisible by 3.
The divisibility rule for 3 states that the if the sum of the digits of the number is divisible by 3, the number itself is divisible 3.
Out of the seven digits given, you need to find groups of 5 for which the sum is divisible by 3. For example, $(1, 2, 3, 4, 8)$ is one such group. For every such group you need to find the number of 5 digit numbers that can be formed and get the total.
Best Answer
My thinking: When you have 11 items in which 9 are unique and 2 are the same numeral repeated, then that can be arranged in $\frac{11!}{2!}$ ways. Then, there are 10 different choices for which numeral is repeated. So we have $\frac{10\times11!}{2!}$. Finally, one tenth of these arrangements should have 0 as its first digit and those don't count. So we need to multiply by $\frac{9}{10}$
$$\frac{9}{10}\times\frac{10\times11!}{2!}=99\times\frac{10!}{2!}$$