SOLUTION: a) How many 4-digit numbers can be formed from the set A = {0,1,2,3,4,5,6}if there is no repetition? b) How many of the numbers in part a) are odd? c) How many of the numbers

Here is a sample choice:

3206

The number that goes where the 3 is can be any of 1, 2, 3, 4, 5, or 6.
0 cannot come first, so we can choose it any of 6 ways.

The number that goes where the 2 is can be 0 or any one of the
5 that was not chosen first.  That's 6 ways.

The number that goes where the 0 can be any of the 5 that weren't
chosen first or second. That's 5 ways.

The number that goes where the 6 is can be any one of the 4 that
weren't chosen 1st, 2nd or 3rd.  That's 4 ways.

The answer is  or  ways. 

Here is a sample choic:

4361

Let's choose the 4th or last digit, where the 1 is, 
first.  It can only be 1, 3, or 5, since the number 
must be odd. That's 3 ways.

Next choose the first digit.  It cannot be 0 or the
digit that was chosen as the 4th digit.  That's
5 ways.

Next choose the second digit.  It can be 0 or any
digit that was not chosen as the 4th digit.  That's
5 ways.

Finally choose the third digit.  It can be any one of
the 4 remaining digits.  

The answer is  or  ways.

There are two ways to solve this.  

First way.

A. The number of ways the 3 can come first.

Sample choice 3256:

There are 6 choices for the digit where the 2 is.
There are 5 choices for the digit where the 5 is.
There are 4 choices for the digit where the 6 is.

That's  or  ways.

B. The number of ways the 3 can come 2nd.

Sample choice 4306:

There are 5 choices for the digit where the 4 is.
There are 5 choices for the digit where the 0 is.
There are 4 choices for the digit where the 6 is.

That's  or  ways.

C. The number of ways the 3 can come 3rd.

This is also  for we could just swap the
2nd and 3rd digits of B.

D. The number of ways the 3 can come 4th.

This is also  for we could just swap the
2nd and 4th digits of B.

That's a total of  or  ways.

Second way.

Find the number which do not contain 3 and then subtract
from the total 720.

Now we are choosing from this set:

{0,1,2,4,5,6}

Here is a sample choice:

2046

The number that goes where the 2 is can be any of 1, 2, 4, 5, or 6.
0 cannot come first, so we can choose it any of 5 ways.

The number that goes where the 0 is can be 0 or any one of the
5 that was not chosen first.  That's 5 ways.

The number that goes where the 4 is can be any of the 4 that weren't
chosen first or second. That's 4 ways.

The number that goes where the 6 is can be any one of the 3 that
weren't chosen 1st, 2nd or 3rd.  That's 3 ways.

The answer is  or  ways. 

So we subtract these which don't contain 3 from the 720 and
get  or .  This is the same answer
we got when doing it the other way.

In order to be divisible by 5, the last digit can only be 0 or 5.

A.  Cases when 0 comes last.

Here is a sample choice:

5240 

There are 6 choices for the digit where the 5 is.
There are 5 choices for the digit where the 2 is.
There are 4 choices for the digit where the 4 is.

That's  or  ways.

B.  Cases when 5 comes last.

Here is a sample choice:

3625

There are 5 choices for the digit where the 3 is,
as it cannot be 0.
There are 5 choices for the digit where the 6 is.
There are 4 choices for the digit where the 2 is.

That's  or  ways.

The total from A and B is

 or  ways.

Edwin