RAS question
How many 3-digit numbers can be formed using the digits 1, 2, 3, 4, 5 without repetition?
Correct answer: (D) 60.
Using the digits 1, 2, 3, 4 and 5 without repetition, exactly 60 three-digit numbers can be formed.
Explanation
This is a permutation problem because the order of the selected digits changes the number: 123 and 321 are different numbers. NCERT defines a permutation as an arrangement of objects in a definite order, and gives the count of n different objects taken r at a time as nPr. Here, five distinct digits are available and three places have to be filled. The hundreds place has 5 choices, the tens place then has 4 choices, and the units place has 3 choices because repetition is not allowed. By the fundamental counting approach, the total is 5 x 4 x 3 = 60, the same as P(5,3) = 5!/(5-3)! = 60.
Why the other options are wrong
- (A) 80 does not follow the decreasing-choice pattern for three non-repeated positions, which must be 5 x 4 x 3.
- (B) 120 is 5!, the number of arrangements of all five digits, while only three-digit numbers are required.
- (C) 125 is 5^3, which would allow repetition in all three positions, but repetition is not allowed.
Concept
This tests permutations and the fundamental principle of counting. It recurs in RAS reasoning because digit-formation questions quickly check whether candidates distinguish ordered arrangements from simple selection or repeated choices.