Using the digits 1, 2, 3, 5, and 6, without repetition, how many five-digit even numbers can be formed?
Nội dung chính Show
Concept: Basic Principle of Counting: If there are m ways for happening of an event A, and corresponding to each possibility there are n ways for happening of event B, then the total number of different possibilities for happening of events A and B are:
Calculation: For an even number, the units place must be 2 or 6, so it can be filled in 2 ways. The remaining four places can now be filled in 4, 3, 2 and 1 ways respectively. The total number of ways in which the number can be written = 4 × 3 × 2 × 1 × 2 = 48. Stay updated with the Mathematics questions & answers with Testbook. Know more about Permutations and Combinations and ace the concept of Fundamental Principles of Counting. 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. How many numbers can be formed using digits 1,2 3 4?Hence, the required number of numbers = (6×3)=18. How many digit numbers can you make using the digits 1,2 3 and 4 without repeating the digits?Using the digits 1, 2, 3 and 4 without repeating them, we can form 24 three digit numbers. The probability of getting numbers with the digit 4 in hundreds place and the digit 2 in tens place is _______ How many numbers can be formed from 12345 without repetition?Total Number of Numbers which can be formed by numbers 1,2,3,4,5 (without repeating digitsi) = 5*4*3*2*! = 5! = 120. How many 3 digit numbers can be formed using 12345 repetition?∴ Total number of 3-digit numbers = 3×4×5=60. How many 5 digit numbers can be formed in 12345 if repetition is not allowed?Total numbers formed using 1, 2, 3, 4, and 5 without repetition is 5! = 120.
How many 3As repetition is allowed, So the number of digits available for Y and Z will also be 5 (each). Thus, The total number of 3-digit numbers that can be formed = 5×5×5 = 125.
How many numbers can be formed by the digits 12345?4*3*2*1=24 possible numbers. How many 4 digit numbers can be formed using the numbers 1,2,3,4,5 with digits not repeated?
How many combinations are there of 5 numbers without repeating?How many combinations with 5 numbers without repetition are possible? There's one (1) possible combination without repetitions C(n,r) and 126 combinations with repetitions C'(n,r) of arranging a group of five numbers (i.e., the 1-5 number list).
|
