A candy shop has $4$ different blue candies, $7$ different green candies, and 3 different red candies. How many different combinations can be arranged if no $2$ red candies can be next to each other?
My solution:
There's $3$ choose $2 = 3$ ways to pick $2$ red candies, and $14!$ ways to arrange all the candies in total. To subtract all of the cases where $2$ red candies are next to each other, I did $14!-3 \cdot 13!$ and I got a huge number. Can someone help me figure out how to do this problem?
Best Answer
The answer is definitely going to be a huge number, but not necessarily that one.
Here's a nice trick that helps solve problems where you need to keep things separated. We can arrange all of the blue and green candies in 11! ways. Now, think about the twelve possible spaces between those candies and on the ends of the line. For each of the three red candies, we need to pick one of those spaces to place it in, and we can't choose the same space for two different red candies because then they would be adjacent to each other.
That gives us a total of $12\cdot11\cdot10\cdot11!=5,269,0176,000$.