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RAS question

How many 3-digit numbers can be formed from digits 1, 2, 3, 4, 5 without repetition?

Correct answer: (C) 60.

The digits 1, 2, 3, 4 and 5 can form 60 different 3-digit numbers when no digit is repeated.

  1. (A)

    125

  2. (B)

    120

  3. (C)

    60

  4. (D)

    80

Explanation

This is a permutation question because each number is an ordered arrangement: 123 and 132 are different numbers. NCERT defines a permutation as an arrangement of objects in a definite order and gives the rule for n objects taken r at a time as nPr. Here, 5 different digits are available and 3 positions have to be filled without repetition. By the multiplication principle, the hundreds place has 5 choices, the tens place then has 4 choices, and the units place has 3 choices. Therefore, the total count is 5 x 4 x 3 = 60, the same as P(5,3) = 5!/(5-3)! = 60.

Why the other options are wrong

  • (A) 125 counts 5 x 5 x 5, which would allow the same digit to be used again in later positions.
  • (B) 120 is 5!, the number of arrangements of all five digits, but the question asks for only 3-digit numbers.
  • (D) 80 does not follow the without-repetition choice pattern; the three places have exactly 5, then 4, then 3 choices.

Concept

This tests permutations and the multiplication principle in basic counting. RAS reasoning papers use such questions because they check whether a candidate can translate a short condition, such as no repetition, into the correct choice count.

Source

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