Question:
The number of numbers between \(2000\) and \(5000\) that can be formed with the digits \(0,1,2,3,4\), repetition of digits not allowed, and are multiples of \(3\), is
The number of numbers between \(2000\) and \(5000\) that can be formed with the digits \(0,1,2,3,4\), repetition of digits not allowed, and are multiples of \(3\), is
Show Hint
For divisibility by \(3\), first check the sum of digits. In digit-arrangement questions, decide the digit set first, then apply restrictions on the first digit.
Updated On: Jun 26, 2026
- \(48\)
- \(30\)
- \(24\)
- \(32\)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Identify the type of numbers required.
Numbers between \(2000\) and \(5000\) must be \(4\)-digit numbers.
Using digits \[ 0,1,2,3,4 \] without repetition, the thousand's digit can only be \[ 2,3,\text{ or }4 \]
Step 2: Use the divisibility rule of \(3\).
A number is divisible by \(3\) if the sum of its digits is divisible by \(3\).
We have to choose \(4\) digits from \[ 0,1,2,3,4 \] The sum of all five digits is \[ 0+1+2+3+4=10 \]
Step 3: Decide which digit is excluded.
A \(4\)-digit number will use \(4\) digits, so one digit is excluded.
If the excluded digit is \(r\), then the sum of selected digits is \[ 10-r \] For divisibility by \(3\), \[ 10-r \equiv 0 \pmod{3} \] Since \[ 10\equiv 1 \pmod{3}, \] we need \[ 1-r\equiv 0 \pmod{3} \] So, \[ r\equiv 1 \pmod{3} \] From the given digits, the possible excluded digits are \[ 1,4 \]
Step 4: Case 1, exclude \(1\).
The selected digits are \[ 0,2,3,4 \] All arrangements are \[ 4! \] But the thousand's digit cannot be \(0\).
Arrangements starting with \(0\) are \[ 3! \] Therefore, valid numbers are \[ 4!-3!=24-6=18 \]
Step 5: Case 2, exclude \(4\).
The selected digits are \[ 0,1,2,3 \] All arrangements are \[ 4! \] Arrangements starting with \(0\) are \[ 3! \] So, valid \(4\)-digit numbers are \[ 4!-3!=24-6=18 \] But the number must be between \(2000\) and \(5000\).
Among digits \(0,1,2,3\), the thousand's digit can be only \[ 2 \text{ or }3 \] For each fixed thousand's digit, the remaining \(3\) digits can be arranged in \[ 3! \] ways.
Thus, valid numbers are \[ 2\times 3!=2\times 6=12 \]
Step 6: Add both cases.
Total number of required numbers is \[ 18+12=30 \]
Step 7: Final conclusion.
Therefore, \[ \boxed{30} \]
Numbers between \(2000\) and \(5000\) must be \(4\)-digit numbers.
Using digits \[ 0,1,2,3,4 \] without repetition, the thousand's digit can only be \[ 2,3,\text{ or }4 \]
Step 2: Use the divisibility rule of \(3\).
A number is divisible by \(3\) if the sum of its digits is divisible by \(3\).
We have to choose \(4\) digits from \[ 0,1,2,3,4 \] The sum of all five digits is \[ 0+1+2+3+4=10 \]
Step 3: Decide which digit is excluded.
A \(4\)-digit number will use \(4\) digits, so one digit is excluded.
If the excluded digit is \(r\), then the sum of selected digits is \[ 10-r \] For divisibility by \(3\), \[ 10-r \equiv 0 \pmod{3} \] Since \[ 10\equiv 1 \pmod{3}, \] we need \[ 1-r\equiv 0 \pmod{3} \] So, \[ r\equiv 1 \pmod{3} \] From the given digits, the possible excluded digits are \[ 1,4 \]
Step 4: Case 1, exclude \(1\).
The selected digits are \[ 0,2,3,4 \] All arrangements are \[ 4! \] But the thousand's digit cannot be \(0\).
Arrangements starting with \(0\) are \[ 3! \] Therefore, valid numbers are \[ 4!-3!=24-6=18 \]
Step 5: Case 2, exclude \(4\).
The selected digits are \[ 0,1,2,3 \] All arrangements are \[ 4! \] Arrangements starting with \(0\) are \[ 3! \] So, valid \(4\)-digit numbers are \[ 4!-3!=24-6=18 \] But the number must be between \(2000\) and \(5000\).
Among digits \(0,1,2,3\), the thousand's digit can be only \[ 2 \text{ or }3 \] For each fixed thousand's digit, the remaining \(3\) digits can be arranged in \[ 3! \] ways.
Thus, valid numbers are \[ 2\times 3!=2\times 6=12 \]
Step 6: Add both cases.
Total number of required numbers is \[ 18+12=30 \]
Step 7: Final conclusion.
Therefore, \[ \boxed{30} \]
Was this answer helpful?
0
0
Top AP EAPCET Mathematics Questions
- $D = \left\{x\in\mathbb{R}: f\left(x\right) =\sqrt{\frac{x - \left|x\right|}{x - \left[x\right]}} \text{is defined} \right\}$ and $C$ be the range of the real function $g(x) = \frac{2x}{4 + x^{2}}$. Then $D \cap C$
- $f\left(x\right) = \frac{x}{e^{x}-1} + \frac{x}{2} + 2 \cos^{3} \frac{x}{2} $ on $R-\left\{0\right\} $ is
- Consider the following lists.
List-I List-II (A) $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ (I) $[\frac{1}{3} , 1 ]$ (B) $(x)=|[x]|,x \in [R$ (II) Z (C) $h(x) = |x - [x]| , x \in [R$ (III) W (D) $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ (IV) [0, 1) (V) { -1, 1} - If $[x]$ is the greatest integer less than or equal to $x$ and $|x|$ is the modulus of $x$. then the system of three equations $2x + 3 | y | + 5[z] = 0, x + |y| - 2[z] = 4, x + |y| + |z| = 1$ has
- Investigate the values of $\lambda$ and $\mu$ for the system $x+2 y+3 z=6, x+3 y+5 z=9$, $2 x+5 y+\lambda z=\mu$ and match the values in List - I with the items in List - II.
List I List II (A) $\lambda=8, \mu \neq 15$ 1. Infinitely many solutions (B) $\lambda \neq 8, \mu \in R$ 2. No solution (C) $\lambda=8, \mu=15$ 3. Unique solution
View More Questions
Top AP EAPCET Combinatorics Questions
- The least value of \( n \) such that \( {}^{n-1}C_6 + {}^{n-1}C_7<{}^nC_8 \) is
- A box contains 5 white balls and 7 black balls. In how many ways can 3 balls be drawn from the box such that at least one black ball is included?}
- Given 5 different green toys, 4 different blue toys, and 3 different red toys, how many combinations of toys can be chosen taking at least one green and one blue toy?
- If \( ^{2n}C_3 : ^nC_3 = 12 : 1 \), then \( n = \) ?
- There are 4 oranges, 5 apples, and 7 mangoes in a fruit basket. The number of ways of selecting at least one fruit from among the fruits in the basket is:
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions