Hello, Aquameatwad!

1) Find the number of passwords that use each of the digits 3,4,5,6,7,8,9

So for passwords, i would think order matters . Yes!

so i use the permutation eqn n!/(n-r)! which is 7!/(7-7)! = 5040. . Right!

We are given: .3 even digits and 4 odd digits.

(a) In how many passwords are the first three digits even?

We want: .even - even - even - odd - odd - odd - odd

The 3 even digits can be arranged in ways.

The 4 odd digits can be arranged in ways.

Hence, there are: . passwords of the form EEEOOO.

b) are the three even digits consecutive?

I assume "consecutive" means the even digits are adjacent,

. . but not necessarily in order.

So that 7368459 is an acceptable password.

First, duct-tape the even digits together.

Then we have 5 "numbers" to arrange: .

. . There are possible arrangement.

But the 3 even digits can be permuted in ways.

Hence, there are: . such passwords.

c) are the four odd digits consecutive?

Duct-tape the odd digits together.

Then we have 4 "numbers" to arrange: .

. . There are possible arrangements.

But the 4 odd digits can be permutated in ways.

Hence, there are: . such passwords.

d) are no two odd digits consecutive?

There is one basic arrangement: .odd - even - odd - even - odd - even - odd

The 4 odd digits can be arranged in ways.

The 3 even digits can be aranged in ways.

Hence, there are: . such passwords.