The total number of numbers that can be formed by using the digits 5, 3 and 7 only, if no repetitions are allowed, is:

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NIMCET 2013 Official Paper
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  1. 39
  2. 105
  3. 15
  4. 27

Answer (Detailed Solution Below)

Option 3 : 15
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NIMCET 2020 Official Paper
120 Qs. 480 Marks 120 Mins

Detailed Solution

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Concept:

Basic Principle of Counting:

If there are m ways for happening of an event A, and corresponding to each possibility there are n ways for happening of event B, then the total number of different possibilities for happening of events A and B are:

  • Either event A alone OR event B alone = m + n.
  • Both event A AND event B together = m × n.

 

Calculation:

Since, the repetition is not allowed, we cannot form a four-digit or a higher digit number.

We have the following possible cases:

One-digit number:

There are three possibilities (3, 5, 7) for the single digit. ⇒ 3 possibilities.

Two-digit number:

There are three possibilities for one of the digits, and corresponding to each possibility, the other digit will have two possibilities. ⇒ 3 × 2 = 6 possibilities.

Three-digit number:

There are three possibilities for one of the digits, and then two possibilities for the second digit and then only a single possibility for the third digit. ⇒ 3 × 2 × 1 = 6 possibilities.

∴ The total number of possible numbers = 3 + 6 + 6 = 15.

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