How many even numbers less than 600 can be made from the digits: 3,3,4,8,9 with each only being used once. I can't figure out what to do for the 3rd case where 3 digits are needed
[Math] How many even numbers less than 600 can be made from the digits: 3,3,4,8,9 with each only being used once.
permutations
Related Solutions
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 hundreds digit is $2$. We have four choices for the tens digit, namely $3$, $5$, $6$, or $7$.
- The hundreds digit is $3$. We have four choices for the tens digit, namely $2$, $5$, $6$, or $7$.
- The hundreds digit is $5$. We have four choices for the tens digit, namely $2$, $3$, $6$, or $7$.
The units digit is $5$. Since the number must be less than $600$, the hundreds digit is either $2$ or $3$.
- The hundreds digit is $2$. We have three choices for the tens digit, namely $3$, $6$, or $7$.
- The hundreds digit is $3$. We have four choices for the tens digit, namely $2$, $3$, $6$, or $7$.
The units digit is $7$. Since the number must be less than $600$, the hundreds digit is either $2$, $3$, or $5$.
- The hundreds digit is $2$. We have three choices for the tens digit, namely $3$, $5$, or $6$.
- The hundreds digit is $3$. We have four choices for the tens digit, namely $2$, $3$, $5$, or $6$.
- The hundreds digit is $5$. We have three choices for the tens digit, namely $2$, $3$, or $6$.
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.
We can think to this "scheme": you have five ordered empty "boxes" and each of them has to be filled with one digit from $0,\ldots,5$ (which are $6$ "objects") with no repetitions, in order to get a $5$-digits number. This means that the last two digits must form a number which is divisible by $4$; then the possible last two digits are as follows: $04$, $12$, $20$, $24$, $32$, $40$, $52$. Now, take for instance $04$; how many numbers (strings) of five distinct digits ending with $04$ are there? Since the last two digits are taken, we have to choose the first three among the remaining ones, that is we have to choose and order $3$ objects out of the $4$ left: this means we have $D_{4,3} = 4 \cdot 3 \cdot 2$ such strings. The same reasoning applies to $20$ and $40$, but not to $12$, $24$, $32$ and $52$: in these cases, in facts, for the remaining three digits $0$ is available but cannot be chosen as first digit, since we want $5$-digits numbers. Thus, in these four cases, the first digit can be filled in $3$ ways (and not $4$, because $0$ is also excluded), the second in $3$ ways and the third in $2$, giving $3\cdot 3\cdot 2$ possible numbers for each couple of final two non-zero digits. Hence, since for $04$, $20$ and $40$ (which are $3$ "cases") we apply the first reasoning, and for $12$, $24$, $32$ and $52$ ($4$ "cases") the second one, the total amount of numbers of five distinct digits (chosen among $0,\ldots,5$) that are divisible by $4$ is $$3 \cdot (4 \cdot 3 \cdot 2) + 4 \cdot (3 \cdot 3 \cdot 2) = 144.$$
Best Answer
We wish to find how many even numbers less than $600$ can be formed from the digits $3, 3, 4, 8, 9$ if each digit is used at most once.
Since the number is even, the units digit of each number must be $4$ or $8$.
One-digit numbers: The only possibilities are $4$ or $8$, giving us two possibilities in this case, as you found.
Two-digit numbers: If the units digit is $4$, then the tens digit can be $3$, $8$, or $9$. If the units digit is $8$, then the tens digit can be $3$, $4$, or $9$. Hence, there are six possibilities, as you found.
Three-digit even numbers: If the units digit of the even number less than $600$ is $4$, the hundreds digit must be $3$. This leaves us with three choices for the tens digit, namely $3$, $8$, or $9$. Hence, we can form three three-digit even numbers less than $600$ with units digit $4$ by using the digits $3, 3, 4, 8, 9$ at most once. They are $334$, $384$, $394$.
If the units digit of the three digit even number less than $600$ is $8$, we have two possibilities for the hundreds digit, namely $3$ or $4$. If the hundreds digit is $3$, we have three possibilities for the tens digit, namely $3$, $4$, or $9$. If the hundreds digit is $4$, we have two possibilities for the tens digit, namely $3$ or $9$. Thus, we can form five three-digit even numbers less than $600$ with units digit $8$. They are $338$, $348$, $398$, $438$, $498$.
Hence, there are a total of eight three-digit even numbers less than $600$ that can be formed with the digits $3, 3, 4, 8, 9$ if each digit is used at most once.
In all, we can form $2 + 6 + 8 = 16$ even numbers less than $600$ using the digits $3, 3, 4, 8, 9$ at most once.