Using the digits 1,2 and 3, how many two digit numbers can be formed if repetition is not allowed

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The $7$ digits $1, 2, 3, 5, 6, 8, 9$ are to be used to make $5$ digit numbers with each being used not more than once in a number. How many numbers can be made which are more than 60 000 AND even? I got something like $3\cdot 3\cdot 5P_3 = 540$. But I think it's wrong because there are numbers that we can choose which are both even and $\geq 6$.

For a number to be divisible by $6$, it must be divisible by both $2$ and $3$. If it is divisible by $2$, it must be even, so the units digit must be $2$ or $4$. If it is divisible by $3$, the sum of its digits must be divisible by $3$.

The only one-digit positive integer that is divisible by $6$ is $6$ itself, so the number must have at least two digits.

Two-digit numbers: If the units digit is $2$, the tens digit must have remainder $1$ when divided by $3$. Hence, the tens digit must be $1$ or $4$.

If the units digit is $4$, the tens digit have remainder $2$ when divided by $3$. Hence, the tens digit must be $2$ or $5$.

Therefore, there are four two-digit numbers divisible by $6$ that can be formed using the digits $1, 2, 3, 4, 5$ without repetition. They are $12$, $24$, $42$, $54$.

Three-digit numbers: If the units digit is $2$, the sum of the hundreds digit and tens digit must have remainder $1$ when divided by $3$. Since the sum of the hundreds digit and tens digit must be at least $1 + 3 = 4$ and at most $5 + 4 = 9$, the only possibilities are that the sum of the hundreds digit and tens digit is $4$ or $7$. Since digits cannot be repeated, the only way to obtain $4$ is to use the digits $1$ and $3$ in either order, and the only way to obtain $7$ is to use the digits $3$ and $4$ in either order. Hence, there are four three-digit numbers divisible by $6$ that can be formed with the digits $1, 2, 3, 4, 5$ that have units digit $2$. They are $132$, $312$, $342$, and $432$.

If the units digit is $4$, then the sum of the hundreds digit and tens digit must have remainder $2$ when divided by $3$. Since the sum of the hundreds digit and tens digit must be at least $1 + 2 = 3$ and at most $3 + 5 = 8$, the sum of the hundreds digit and tens digit must be $5$ or $8$. Since digits cannot be repeated, the only way to obtain $5$ is to use the digits $2$ and $3$ in either order, and the only way to obtain $8$ is to use the digits $3$ and $5$ in either order. Hence, there are also four three-digit numbers divisible by $6$ that can be formed with the digits $1, 2, 3, 4, 5$ that have units digit $4$. They are $234$, $324$, $354$, $534$.

Therefore, there are a total of eight three-digit numbers divisible by $6$ that can be formed from the digits $1, 2, 3, 4, 5$ without repetition.

Four-digit numbers: If the units digit is $2$, then the sum of the thousands digit, hundreds digit, and tens digit must have remainder $1$ when divided by $3$. Since the sum of the thousands digit, hundreds digit, and tens digit must be at least $1 + 3 + 4 = 8$ and at most $3 + 4 + 5 = 12$, the sum of the thousands digit, hundreds digit, and tens digit must be $10$. Since digits cannot be repeated, the only way to obtain a sum of $10$ is to use the digits $1$, $4$, and $5$ in some order. There are $3! = 6$ such orders. Hence, there are six four-digit numbers divisible by $6$ with units digit $2$ that can be formed from the digits $1, 2, 3, 4, 5$ without repetition. They are $1452$, $1542$, $4152$, $4512$, $5142$, and $5412$.

If the units digit is $4$, the remainder of the sum of the thousands digit, hundreds digit, and tens digit must be $2$ when divided by $3$. Since the sum of the thousands digit, hundreds digit, and tens digit must be at least $1 + 2 + 3 = 6$ and at most $2 + 3 + 5 = 10$, the sum of the thousands digit, hundreds digit, and tens digit must be $8$. Since digits cannot be repeated, the only way to obtain a sum of $8$ is to use the digits $1$, $2$, and $5$ in some order. Since there are $3! = 6$ such orders, there are also six four-digit numbers that can be formed from the digits $1, 2, 3, 4, 5$ without repetition. They are $1254$, $1524$, $2154$, $2514$, $5124$, and $5214$.

Hence, there are a total of $12$ four-digit numbers divisible by $6$ that can be formed from the digits $1, 2, 3, 4, 5$ without repetition.

Five-digit numbers: The sum of the five digits $1, 2, 3, 4, 5$ is $15$, which is divisible by $3$. Hence, any five digit number formed from these digits without repetition that has units digit $2$ or $4$ is divisible by $6$. There are two ways of filling the units digit and $4!$ ways of filling the remaining digits. Hence, there are $2 \cdot 4! = 48$ five-digit numbers that can be formed with the digits $1, 2, 3, 4, 5$ without repetition.

In total, there are $4 + 8 + 12 + 48 = 72$ numbers divisible by $6$ that can be formed from the digits $1, 2, 3, 4, 5$ without repetition.

If repetition is allowed and leading zeros are allowed, then you have 000 to 999, which is 1,000 possibilities (I will not list them here). If no leading zeros are allowed, then you have 100 through 999, so there are 900 possibilities. If there are other restrictions, such as no repetition, then I'll refer you to the related link at MathsIsFun.com

31. How many two digit numbers can be generated using the digits 1,2,3,4 without repeating any digit?
A. 10	B. 12
C. 4	D. 16

Answer: Option B

Explanation:

We have four digits 1,2,3,4. The first digit can be any of these four digits.

Now we have already chosen the first digit. Since we cannot repeat the digits, we are left with 3 digits now. The second digit can be any of these three digits.

Since the first digit can be chosen in 4 ways and second digit can be chosen in 3 ways, both the digits can be chosen in $4×3=12$ ways. [Reference: Multiplication Theorem]

i.e., 12 two digit numbers can be formed.

but it is specified that, without repeating digits! will it not change the ans?

The digits cannot be repeated. That is why first digit has 4 options and second digit has only 3 options. Therefore 4*3=12 is the answer.

Suppose digits can be repeated. Then first digit will have 4 options. Similarly second digit also will have 4 options. In that case, answer will be 4*4=16

The digits cannot be repeated. That is why first digit has 4 options and second digit has only 3 options. Therefore 4*3=12 is the answer.<p>Suppose digits can be repeated. Then first digit will have 4 options. Similarly second digit also will have 4 options. In that case, answer will be 4*4=16</p>

This is 4p2, right?

yes.

(use Q&A for new questions)