Answer: (a) \(60480\), (b) \(20160\), (c) \(16800\).

(a) A 6-digit number is formed by arranging \(6\) different digits chosen from the \(9\) digits \(1,2,3,4,5,6,7,8,9\).

So the number of possible numbers is

\({}^{9}P_6=9\times8\times7\times6\times5\times4=60480\).

(b) For the number to be greater than \(700000\), the first digit must be \(7\), \(8\), or \(9\).

There are \(3\) choices for the first digit.

After choosing the first digit, there are \(8\) digits left, and the remaining five positions can be filled in

\({}^{8}P_5=8\times7\times6\times5\times4\)

ways.

Therefore the number of possible numbers is

\(3\times{}^{8}P_5=3\times8\times7\times6\times5\times4=20160\).

(c) For the number to be greater than \(750000\), split into cases.

Case 1: The first digit is \(8\) or \(9\). There are \(2\) choices for the first digit, and then the remaining five positions can be filled in \({}^{8}P_5\) ways. This gives

\(2\times{}^{8}P_5=2\times8\times7\times6\times5\times4=13440\).

Case 2: The first digit is \(7\). Then the second digit must be \(5\), \(6\), \(8\), or \(9\) to make the number greater than \(750000\). This gives \(4\) choices for the second digit.

After choosing the first two digits, \(7\) digits remain, and the last four positions can be filled in

\({}^{7}P_4=7\times6\times5\times4\)

ways. This gives

\(4\times{}^{7}P_4=4\times7\times6\times5\times4=3360\).

Total:

\(13440+3360=16800\).

Answer: (a) \(665280\), (b) \(30240\), (c) \(26880\).

There are \(5\) letters, \(4\) numbers and \(3\) symbols, so there are

\(5+4+3=12\)

different characters altogether.

(a) With no restrictions, the password is an ordered arrangement of \(6\) different characters chosen from \(12\). Hence the number of passwords is

\({}^{12}P_6=12\times11\times10\times9\times8\times7=665280\).

(b) The first and last characters must be symbols. There are \(3\) choices for the first symbol and then \(2\) choices for the last symbol.

After these two positions have been filled, \(10\) characters remain, and the middle four positions can be filled in

\({}^{10}P_4=10\times9\times8\times7\)

ways. Therefore the number of passwords is

\(3\times2\times{}^{10}P_4=3\times2\times10\times9\times8\times7=30240\).

(c) The first two positions must be either symbol-number or number-symbol.

There are \(2\) possible orders, \(3\) choices for the symbol and \(4\) choices for the number, giving

\(2\times3\times4\)

ways to fill the first two positions.

The last two positions must be letters, so there are

\(5\times4\)

ways to fill them.

So far, one symbol, one number and two letters have been used, leaving \(8\) available characters. The two remaining middle positions can be filled in

\(8\times7\)

ways.

Thus the number of passwords is

\(2\times3\times4\times5\times4\times8\times7=26880\).