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

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