I've been stuck on this problem for quite some time now, I can't seem to find a video or such where it references permutations in a specific order of left to right. I have no idea how to set this problem up in terms of a formula.
You have $140$ distinct food products, $80$ cans of soup, $20$ cereals and $40$ cookies. Assume that all $140$ products are different. In how many ways can you put the $140$ products on a shelf with the cans of soup on the left, cereal in the middle and cookies on the right?
If possible, can these two answers be verified? They're simple I know, but I always second guess myself when it comes to any form of math.
How many ways can you put $140$ products on the shelf? $140!$
In how many ways can you give away 10 products? $\frac{150!}{(150-10)!} = \frac{150!}{140!}$
Thank you in advance for your help!
Well, we need to think of them as three separate permutations.
How many ways to arrange the soups? $80!$
For any given arrangement of soups how many ways to arrange the cereal? $20!$
For any given arrangement of soup and cereal how many ways to arrange cookies? $40!$
So, we multiply to get $80!20!40!$
In general, if the $n$ sub-arrangements are independent (which they are here - left, middle and right can't get mixed up), we have to multiply the permutations $P_1*P_2*P_3...P_{n-1}*P_n$
In this case $P_1*P_2*P_3=80!*20!*40!$