Question:
The sum of all possible numbers that can be formed by using the digits \(2,3,5,7\) without repetition of digits is:
The sum of all possible numbers that can be formed by using the digits \(2,3,5,7\) without repetition of digits is:
Show Hint
For numbers formed using all digits without repetition, each digit appears equally often in every place value. Use:
\[
(n-1)!
\]
for the frequency of each digit in a particular place.
Updated On: Jun 26, 2026
- \(17\times \dfrac{10^{4}-1}{9}\)
- \(33\times 34\times 101\)
- \(6\times \dfrac{10^{3}-1}{9}\)
- \(33\times 35\times 1001\)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Determine the number of four-digit numbers formed.
Using the digits \(2,3,5,7\) without repetition, the total number of four-digit numbers formed is \[ 4!=24 \]
Step 2: Find the contribution of each digit in each place.
Each digit appears equally often in the units, tens, hundreds, and thousands places.
The number of times each digit appears in any particular place is \[ (4-1)!=3!=6 \] The sum of the digits is \[ 2+3+5+7=17 \] Hence, the contribution from each place is \[ 17\times 6 \]
Step 3: Compute the total sum.
The place value sum is \[ 1000+100+10+1=1111 \] Therefore, \[ \text{Required Sum} = 17\times 6\times 1111 \] \[ = 102\times 1111 \] \[ = 113322 \]
Step 4: Express the answer in the given form.
Since \[ 1111=11\times 101 \] we get \[ 17\times 6\times 1111 = 17\times 6\times 11\times 101 \] \[ = 1122\times 101 \] and \[ 1122=33\times 34 \] Thus, \[ \text{Required Sum} = 33\times 34\times 101 \]
Step 5: Final conclusion.
Hence, \[ \boxed{33\times 34\times 101} \] Therefore, the correct option is \[ \boxed{(2)\ 33\times 34\times 101} \]
Using the digits \(2,3,5,7\) without repetition, the total number of four-digit numbers formed is \[ 4!=24 \]
Step 2: Find the contribution of each digit in each place.
Each digit appears equally often in the units, tens, hundreds, and thousands places.
The number of times each digit appears in any particular place is \[ (4-1)!=3!=6 \] The sum of the digits is \[ 2+3+5+7=17 \] Hence, the contribution from each place is \[ 17\times 6 \]
Step 3: Compute the total sum.
The place value sum is \[ 1000+100+10+1=1111 \] Therefore, \[ \text{Required Sum} = 17\times 6\times 1111 \] \[ = 102\times 1111 \] \[ = 113322 \]
Step 4: Express the answer in the given form.
Since \[ 1111=11\times 101 \] we get \[ 17\times 6\times 1111 = 17\times 6\times 11\times 101 \] \[ = 1122\times 101 \] and \[ 1122=33\times 34 \] Thus, \[ \text{Required Sum} = 33\times 34\times 101 \]
Step 5: Final conclusion.
Hence, \[ \boxed{33\times 34\times 101} \] Therefore, the correct option is \[ \boxed{(2)\ 33\times 34\times 101} \]
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 permutations and combinations Questions
- There are 6 different novels and 3 different poetry books on a table. If 4 novels and 1 poetry book are to be selected and arranged in a row on a shelf such that the poetry book is always in the middle, then the number of such possible arrangements is:
- If a five-digit number divisible by 3 is to be formed using the numbers 0, 1, 2, 3, 4, and 5 without repetition, then the total number of ways this can be done is:
- Four digit numbers with all digits distinct are formed using the digits 1, 2, 3, 4, 5, 6, 7 in all possible ways. If \( p \) is the total number of numbers thus formed and \( q \) is the number of numbers greater than 3400 among them, then \( p : q = \):
- If all the letters of the word ‘SENSELESSNESS’ are arranged in all possible ways and an arrangement among them is chosen at random, then, the probability that all the E’s come together in that arrangement is:
- The number of arrangements of the letters of the word ARRANGEMENT in which two E’s do not occur adjacently 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