# Need help with this

• August 30th 2006, 05:58 PM
ashkash
Need help with this
The question is as follows:

You have 12 programs that need to be processed. How many ways can you order these 12 programs if

a. there are no restrictions?
b. you consider 4 programs higher in priority than the other 8 and want to process those 4 first?
c. you seperate the programs into 4 of top priority, 5 of lesser priority, and 3 of least priority and you want to process them so the top priority is processed first and the 3 programs of least priority are processed last?

So far this is what I have:
a. 12!
b. 12!/4!
c. 12!/(4!5!3!)

Can anyone please help me and let me know if this looks right or if I am doing anything wrong. thanks.
• August 30th 2006, 06:12 PM
JakeD
Quote:

Originally Posted by ashkash
The question is as follows:

You have 12 programs that need to be processed. How many ways can you order these 12 programs if

a. there are no restrictions?
b. you consider 4 programs higher in priority than the other 8 and want to process those 4 first?
c. you seperate the programs into 4 of top priority, 5 of lesser priority, and 3 of least priority and you want to process them so the top priority is processed first and the 3 programs of least priority are processed last?

So far this is what I have:
a. 12!
b. 12!/4!
c. 12!/(4!5!3!)

Can anyone please help me and let me know if this looks right or if I am doing anything wrong. thanks.

I'd say (a) correct, (b) 4!8! and (c) 4!5!3!.

For problems like this, it helps to try a smaller example to see if your formula works. Say it was 4=2+2 programs for (b). The number of ways to order them is

1234
1243
2134
2143

which is 2!2!, but not 4!/2!.
• August 30th 2006, 06:27 PM
Soroban
Hello, ashkash!

Quote:

You have 12 programs that need to be processed.
How many ways can you order these 12 programs if

a. there are no restrictions?

You are correct . . . There are $12!$ orderings.

Quote:

b. you consider 4 programs higher in priority than the other 8
and want to process those 4 first?

There are $4!$ orderings for the four high-priority programs
. . and $8!$ orderings for the remaining eight programs.

Therefore, there are: $(4!)(8!)$ possible orderings.

Quote:

c. you seperate the programs into 4 of top priority, 5 of lesser priority, and 3 of least priority
and you want to process them so the top priority is processed first
and the 3 programs of least priority are processed last?

There are $4!$ orderings of the four top-priority programs,
. . $5!$ orderings of the five lesser-priority programs,
. . and $3!$ orderings of the three least-priority programs.

Therefore, there are: $(4!)(5!)(3!)$ possible orderings.

• August 30th 2006, 06:38 PM
ashkash
got it, thanks for your help guys