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

Question

Accepted Answer

To form a number greater than 30000 using the digits 2, 2, 3, 3, 3, the number must be a 5-digit number, and its first digit must be 3.

1.  The first digit is fixed as 3. (d1 = 3)
2.  We are left with the remaining 4 digits to arrange: 2, 2, 3, 3.
3.  We need to find the number of distinct permutations of these 4 digits. The formula for permutations with repetitions is n! / (n1! * n2! * ...), where n is the total number of items, and n1, n2, etc., are the counts of repeated items.
4.  Here, n = 4 (for the remaining 4 positions).
    The digit 2 appears 2 times (n1 = 2).
    The digit 3 appears 2 times (n2 = 2).
5.  Number of distinct arrangements = 4! / (2! * 2!)
    = (4 * 3 * 2 * 1) / ((2 * 1) * (2 * 1))
    = 24 / (2 * 2)
    = 24 / 4
    = 6

Therefore, 6 distinct numbers greater than 30000 can be formed.

The correct answer is B) 6.

Answer

3

Answer

6

Answer

9

Answer

12

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

Explanation

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