How many 3-digit numbers greater than 300 can be formed using the digits 1, 2, 3, 4, and 5, when (a) repetition is not allowed and (b) repetition is allowed. Briefly explain the calculation process.

We can form numbers starting with 3, 4, or 5.

a)

If a number start with digit "3", then we can choose 2nd and 3rd digit by choosing two digits from $\{1,2,4,5\}$

Without repetition, this is:

$C^2_4=\frac{4!}{2!2!}=6$ ways

The same is for numbers starting with digit "4" or "5".

So, total number of ways:

$N=6+6+6=18$

b)

If a number start with digit "3", then we can choose 2nd and 3rd digit by

$5\cdot5=25$ ways

The same is for numbers starting with digit "4" or "5".

So, total number of ways:

$N=25+25+25=75$