6 character password must contain 3 digits, 2 letters and 1 special character
Assuming their are 7 special characters, REPETITION IS NOT ALLOWED, The password is NOT Case sensitive, how many ORDERED arrangements are there?
Is my Solution Correct?:
10c3 . 26c2 . 7c1 . 6! = 196 560 000 combinations
or Should it be 10p3 . 26p2 . 7p1 = 3276000 combinations? since you use perms to find the ordered arrangements.
As scoopfaze indicated in the comments, your first approach is correct. Your second approach is the answer to a different question:
We can fill the first three slots with three distinct decimal digits in $P(10, 3) = 10 \cdot 9 \cdot 8$ ways, the next two slots with distinct lowercase letters in $P(26, 2) = 26 \cdot 25$ ways, and the final slot with one the seven special characters in $P(7, 1) = 7$ ways. Therefore, the number of passwords with three distinct decimal digits, two distinct lowercase letters, and one of seven special characters that appear in that order is $$P(10, 3)P(26, 2)P(7, 1) = 10 \cdot 9 \cdot 8 \cdot 26 \cdot 25 \cdot 7$$ Notice that we could also write the answer to this question in the following way: $$\binom{10}{3}3!\binom{26}{2}2!\binom{7}{1}1!$$ Here is the rationale: Choose three distinct decimal digits, arrange them in the first three positions, choose two distinct lowercase letters, arrange them in the next two positions, choose one of the seven special characters, arrange it in the final position.
Notice that we are only allowing the digits to appear in the first three positions, the letters to appear in the next two positions, and the special character to appear in the last position. This is different from the question you wish to answer, where the digits, letters, and special characters may appear in any of the positions.
We can choose three distinct decimal digits, two distinct letters, and one special character, then arrange the six selected objects in $$\binom{10}{3}\binom{26}{2}\binom{7}{1}6!$$ ways.