Question 9

Mathematics Permutation and Combination Easy

Number of three digit numbers that can be formed using 0, 1, 2, 3 and 5 where these digits are allowed to repeat any number of times, is equal to

(A) 125
(B) 140
(C) 100
(D) 115
✓ Correct Answer: Option C Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

We need the number of three-digit numbers formed from the digits \(\{0,1,2,3,5\}\) with repetition allowed. A three-digit number cannot begin with \(0\). Thus the first (hundreds) digit has \(4\) choices \(\{1,2,3,5\}\), and each of the tens and units digits has \(5\) choices. Therefore the total number is \[ 4\times 5\times 5=100. \] \[ \boxed{100} \]

Practice Similar Permutation and Combination Questions

Reinforce your conceptual grasp with related questions from NIMCET 2025:

Q15 Number of permutations of the letters of the word BANGLORE such that the...
Solve →
Q31 The number of 3-digit integers that are multiple of 6 which can be formed...
Solve →
Q1 The slope of the normal line to the curve \(x = t 2 + 3t - 8\) and \(y =...
Solve →
Q2 The value of the integral \(∈t 0 π/2 (1+2cos x ) (2+cos x ) 2 dx\) lies in...
Solve →
Q3 Which one of the following is NOT a correct statement?
Solve →
Full-Length Test Series

Prepare for NIMCET Under Real Exam Conditions

Take timed full-length mock tests with authentic question distributions, negative marking (+12/-3 in Math), and instant All India Rank calculation.

Start Free Mock Test Series →