Maths : Basic Numeracy

Q 163 / 370

UPSC CSE Prelims 2021

Using 2, 2, 3, 3, 3, as digits, how many distinct numbers greater than 30000 can be formed?

EXPLANATION

Correct Option (2)

The given digits are 2, 2, 3, 3, 3, totaling five digits. The objective is to form 5-digit numbers greater than 30000 using all these digits.

For a 5-digit number to be greater than 30000, its first digit must be 3. If the first digit were 2, the resulting number would be less than 30000.

By fixing 3 as the first digit, the remaining digits available for arrangement are 2, 2, 3, 3. These are four digits, with the digit 2 appearing twice and the digit 3 appearing twice.

The number of distinct permutations of these four digits (n=4) with repetitions (p₁=2 for digit 2, p₂=2 for digit 3) is calculated using the formula:

Number of ways = n! / (p₁! p₂!)

Substituting the values:

Number of ways = 4! / (2!2!) = (4 × 3 × 2 × 1) / ((2 × 1) × (2 × 1)) = 24 / 4 = 6.

Therefore, 6 distinct numbers greater than 30000 can be formed using the given digits. These numbers are: 32233, 32323, 32332, 33223, 33232, 33322.

Incorrect Options:

Options 1 (3), 3 (9), and 4 (12) are incorrect as they do not correspond to the calculated number of distinct 5-digit numbers greater than 30000 that can be formed using the specified set of digits, based on the principles of permutations with repetitions.