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

The correct option is 6 (which corresponds to Option 6 in the provided choices).

To find how many distinct numbers greater than 30,000 can be formed using the digits 2, 2, 3, 3, and 3, let's analyze the requirements step-by-step.

Step 1: Determine the total number of digits and available multiset.

We are given 5 digits in total: three 3s and two 2s.

Any 5-digit number formed using all these digits will be in the ten-thousands range.

Step 2: Condition for the number to be greater than 30,000.

For a 5-digit number to be strictly greater than 30,000, its first digit (the ten-thousands place) must be greater than or equal to 3.

Since our available digits are only 2 and 3, the first digit must be 3.

Step 3: Arrange the remaining digits.

Once we fix the first digit as 3, we are left with 4 positions to fill using the remaining 4 digits:

Remaining digits: two 3s and two 2s (i.e., 3, 3, 2, 2).

The number of distinct arrangements of these 4 remaining digits with repetitions (two 3s and two 2s) is given by the multinomial coefficient formula:

$\text{Number of arrangements}=\frac{4!}{2!\×2!}$

Step 4: Calculate the final value.

$4!=24$

$2!\×2!=2\×2=4$

$\text{Total distinct numbers}=\frac{24}{4}=6$

Thus, exactly 6 distinct numbers greater than 30,000 can be formed.