Using 2, 2, 3, 3, 3 as digits, how many distinct numbers greater than 30000 can be formed?
Using 2, 2, 3, 3, 3 as digits, how many distinct numbers greater than 30000 can be formed?
3
6
9
12
Solution
For a 5-digit number using all of 2, 2, 3, 3, 3 to be greater than 30000, it must start with the digit 3. The remaining 4 digits are two 2's and two 3's, arranged in $\dfrac{4!}{2! \, 2!} = 6$ ways. Hence 6 such numbers.