X is any 5-digit number formed using the digits 1,2,3,4,5 without repetition

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.

A five-digit number is formed by the digits (0, 1, 2, 3, 4) without repetition. Find the probability that the number formed is divisible by 4.

1. \(\frac{{16}}{5}\)
2. \(\frac{{1}}{16}\)
3. \(\frac{{5}}{16}\)
4. \(\frac{{1}}{6}\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{{5}}{16}\)

Free

20 Questions 20 Marks 20 Mins

Total number of ways by which a 5-digit number is formed by the digits (0, 1, 2, 3, 4) without repetition

= 4 × 4! (Because digit ‘0’ cannot be placed at the first place of a 5-digit number)

The numbers formed by using these digits should have the last two digits as 12 or 24 or 32 or 20 or 04 or 40 so that they are divisible by 4.

For the last two digits as 20 or 04 or 40:

In each case, the five-digit number can be formed using the remaining 3 digits

Hence the number divisible by 4 in the above case can be formed in 3! × 3 = 18 ways.

For the last two digits as 12 or 24 or 32:

In each case, the five-digit number can be formed using the remaining 3 digits, but ‘0’ cannot exist in the first place.

So, the number divisible by 4 in the above case can be formed in 2 × 2 × 3 = 12 ways.

Hence total number of ways by which 5-digit number divisible by 4 = 12 + 18 = 30

Calculation:

Favourable outcomes = 30

Total number of outcomes = 4 × 4! = 96

Probability = (Favourable outcomes) / (Total number of outcomes)

\(= \frac{{30}}{{96}} = \frac{5}{{16}}\)

Latest UPPCL Assistant Engineer Updates

Last updated on Sep 22, 2022

The Uttar Pradesh Power Corporation Limited (UPPCL) released the final written result of UPPCL Assistant Engineer 2022 on 25th May 2022. The candidates who have secured qualifying marks in the written test will be called for a personal interview. PI is the last stage of this exam. Accumulated marks of written test and personal interview will be taken into consideration in making the final merit list. So candidates should download their UPPCL Assistant Engineer result and start preparing for personal interview the right way.

How many 5 digit numbers can be formed from 1 2 3 4 5 without repetition when the digit at the unit's place must be greater than that in the tens place?

How many 5 digit numbers can be formed using the digits 1 2 5 without repetition such that the number is divisible by the first digit from left?

How many 5 digit numbers can be formed using digits 1 2 3 4 and 5 if repetition of digits is not allowed?

What is the 5 digit number without repeating digits?