Question:
How many 3-digit numbers can be formed using three distinct single digit prime numbers?
How many 3-digit numbers can be formed using three distinct single digit prime numbers?
Show Hint
Hint:
First find how many single digit prime numbers exist, then work out how many ways you can pick 3 of them and arrange those 3 in order.
First find how many single digit prime numbers exist, then work out how many ways you can pick 3 of them and arrange those 3 in order.
Updated On: Aug 17, 2026
- 64
- 24
- 12
- 4
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: List the single digit prime numbers.
The single digit numbers are 0 through 9, and the prime numbers among them are 2, 3, 5, and 7. So there are exactly 4 digits available to build the 3-digit number from.
Step 2: Understand what the question is asking for.
We need 3-digit numbers where all three digits are distinct and each digit is one of 2, 3, 5, 7. Since we must pick 3 different digits out of these 4 and then arrange them, this is a counting problem in two parts, first choose which 3 digits to use, then decide the order they appear in.
Step 3: Choose which 3 of the 4 primes to use.
The number of ways to choose 3 digits out of 4 available digits is given by the combination formula:
\[ \binom{4}{3} = \frac{4!}{3! \times 1!} = 4 \]
So there are 4 different sets of 3 digits that could be used, for example {2,3,5}, {2,3,7}, {2,5,7}, and {3,5,7}.
Step 4: Arrange each chosen set of 3 digits.
Once a set of 3 distinct digits is picked, they can be arranged in any order to form a 3-digit number. The number of ways to arrange 3 distinct digits is:
\[ 3! = 3 \times 2 \times 1 = 6 \]
Since none of 2, 3, 5, 7 is zero, every arrangement is a valid 3-digit number, there is no risk of a leading zero making the number only 2 digits long.
Step 5: Combine the choices and the arrangements.
Each of the 4 possible sets of digits can be arranged in 6 different orders, so the total count is:
\[ 4 \times 6 = 24 \]
Step 6: Rule out the other options.
Option (A), 64, would come from allowing digits to repeat, for example choosing freely 3 times from 4 digits ($4^3 = 64$), but the question asks for three distinct digits, so repetition is not allowed here. Option (C), 12, undercounts, it misses some valid orderings. Option (D), 4, only counts the number of digit sets, without accounting for the different orders each set can be arranged in.
Final Answer:
Choosing 3 out of 4 single digit primes and arranging them gives $4 \times 6 = 24$ distinct 3-digit numbers. \[ \boxed{24} \]
The single digit numbers are 0 through 9, and the prime numbers among them are 2, 3, 5, and 7. So there are exactly 4 digits available to build the 3-digit number from.
Step 2: Understand what the question is asking for.
We need 3-digit numbers where all three digits are distinct and each digit is one of 2, 3, 5, 7. Since we must pick 3 different digits out of these 4 and then arrange them, this is a counting problem in two parts, first choose which 3 digits to use, then decide the order they appear in.
Step 3: Choose which 3 of the 4 primes to use.
The number of ways to choose 3 digits out of 4 available digits is given by the combination formula:
\[ \binom{4}{3} = \frac{4!}{3! \times 1!} = 4 \]
So there are 4 different sets of 3 digits that could be used, for example {2,3,5}, {2,3,7}, {2,5,7}, and {3,5,7}.
Step 4: Arrange each chosen set of 3 digits.
Once a set of 3 distinct digits is picked, they can be arranged in any order to form a 3-digit number. The number of ways to arrange 3 distinct digits is:
\[ 3! = 3 \times 2 \times 1 = 6 \]
Since none of 2, 3, 5, 7 is zero, every arrangement is a valid 3-digit number, there is no risk of a leading zero making the number only 2 digits long.
Step 5: Combine the choices and the arrangements.
Each of the 4 possible sets of digits can be arranged in 6 different orders, so the total count is:
\[ 4 \times 6 = 24 \]
Step 6: Rule out the other options.
Option (A), 64, would come from allowing digits to repeat, for example choosing freely 3 times from 4 digits ($4^3 = 64$), but the question asks for three distinct digits, so repetition is not allowed here. Option (C), 12, undercounts, it misses some valid orderings. Option (D), 4, only counts the number of digit sets, without accounting for the different orders each set can be arranged in.
Final Answer:
Choosing 3 out of 4 single digit primes and arranging them gives $4 \times 6 = 24$ distinct 3-digit numbers. \[ \boxed{24} \]
Was this answer helpful?
0
0
Top GATE PH General Aptitude Questions
Is there any good show _________ television tonight? Select the most appropriate option to complete the above sentence.
As the police officer was found guilty of embezzlement, he was _________ dismissed from the service in accordance with the Service Rules. Select the most appropriate option to complete the above sentence.
- The sum of the following infinite series is: \[ 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + \dots \]
- A thin wire is used to construct all the edges of a cube of 1 m side by bending, cutting, and soldering the wire. If the wire is 12 m long, what is the minimum number of cuts required to construct the wire frame to form the cube?
The figures I, II, and III are parts of a sequence. Which one of the following options comes next in the sequence at IV?
View More Questions
Top GATE PH Questions
- If \( F_1(Q, q) = Qq \) is the generating function of a canonical transformation from \((p, q)\) to \((P, Q)\), then which one of the following relations is correct?
- When an unpolarized plane electromagnetic wave in a dielectric medium with refractive index \( n_1 \) is incident on a plane interface separating it from another dielectric medium with refractive index \( n_2 \) (where \( n_2 > n_1 \)), and the angle of incidence is \(\tan^{-1} \left( \frac{n_2}{n_1} \right) \), the following statement is true:
- The wavefunction of a particle in an infinite one-dimensional potential well at time \( t \) is
\[ \Psi(x, t) = \sqrt{\frac{2}{3}} e^{-iE_1 t/\hbar}\psi_1(x) + \frac{1}{\sqrt{6}} e^{i\pi/6} e^{-iE_2 t/\hbar} \psi_2(x) + \frac{1}{\sqrt{6}} e^{i\pi/4} e^{-iE_3 t/\hbar} \psi_3(x) \]where \(\psi_1\), \(\psi_2\), and \(\psi_3\) are the normalized ground state, the normalized first excited state, and the normalized second excited state, respectively. \(E_1\), \(E_2\), and \(E_3\) are the eigen-energies corresponding to \(\psi_1\), \(\psi_2\), and \(\psi_3\), respectively. The expectation value of energy of the particle in state \(\Psi(x,t)\) is - If a thermodynamical system is adiabatically isolated and experiences a change in volume under an externally applied constant pressure, then the thermodynamical potential minimized at equilibrium is the
- The mean distance between the two atoms of an HD molecule is \( r \), where H and D denote hydrogen and deuterium, respectively. The mass of the hydrogen atom is \( m_H \). The energy difference between the two lowest lying rotational states of HD in multiples of \(\frac{h^2}{m_H r^2}\) is:
View More Questions