How many two-digit numbers can be formed from the digits 3, 5,7 and 9 if repetition is not allowed
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 $3$. Show
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. Problem 1: How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that –(i) repetition of the digits is allowed?Solution:
(ii) repetition of the digits is not allowed?Solution:
Problem 2: How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?Solution:
Problem 3: How many 4-letter code can be formed using the first 10 letters of the English alphabet if no letter can be repeated?Solution:
Problem 4: How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?Solution:
Problem 5: A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?Solution:
Problem 6: Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?Solution:
How many 2Hence 12, 2-digit & 3-digit numbers can be formed by using the digits 3, 5, 6 without repeating any digit.
How many 3 digits number can be formed using digits 1 5 7 and 9?Solution : 357, 375, 537, 573, 735, 753. Therefore, '6' three-digit numbers can be formed.
Hence, the correct option is (d). Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams. How many 2∴ Required number of numbers = (1 x 5 x 4) = 20. Was this answer helpful?
How many 3 digits number can be formed using the digits 1 3 5 7 9 where we are allowed to repeat the digits?Therefore, a total of 100 3 digit numbers can be formed using the digits 0, 1, 3, 5, 7 when repetition is allowed.
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