Permutation problem. maybe?

This is a problem from my text. Of course my teacher expects us to know the answers without being taught how to do them, so i ask here.

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 so i use the permutation eqn n!/(n-r)! which is 7!/(7-7)!=5040 cause we have 7 numbers to choose from.

now

in how many passwords

a) are the first three digits even?

b) are the three even digits consecutive?

c) are the four odd digits consecutive?

d) are no tow odd digits consecutive?

Re: Permutation problem. maybe?

Hello, Aquameatwad!

Quote:

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.

Quote:

(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.

Quote:

b) are the three even digits consecutive?

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

. . but not necessarily in order.

So that 73**684**59 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.

Quote:

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.

Quote:

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.