I need help for the following question:
How many possible ways of stacking $12$ red, $12$ green, and $12$ blue poker chips so that no blue chip is touching one another?
I need help for the following question:
How many possible ways of stacking $12$ red, $12$ green, and $12$ blue poker chips so that no blue chip is touching one another?
First stack the $24$ non-blue chips; in how may ways can you do this? Counting the position below the bottom non-blue chip and the position above the top non-blue chip, there are now $25$ positions in the stack into which you could put a blue chip. Since you don’t want two blue chips to touch each other, each blue chip has to go into a different one of these $25$ possible positions. How many ways are there to pick $12$ of the $25$ positions to accommodate the blue chips? Now how should you combine the results of these two calculations?
Added: This is a good example of a problem that can fairly easily be solved in more than one way; such problems are common in elementary combinatorics. El'endia Starman’s approach is perhaps a little less straightforward than this one, but its basic idea is, as noted, very useful, so it’s worth understanding both: the more tools you have at your disposal, the better.
I immediately recognized this as a version of the classic so-called "Hotdog Problem" in combinatorics. In this case, you can set this up like so:
_B_B_B_B_B_B_B_B_B_B_B_B_
In every inner blank, there must be at least one chip while the outside blanks may have none. This means that we have $12+12-11=13$ chips left that we can freely allocate to $13$ blanks. The number of ways that we can do this is $\binom{25}{12}=5200300$ (the $23$ comes from $11$ blanks plus $13$ free chips, minus one because the last item is determined by the others). We can further arrange the red and green chips amongst themselves, leading to $\binom{24}{12}=2704156$. These two orderings are independent of each other, so we can multiply them to get the total number of possible orderings. Thus, the answer is:
$\binom{25}{12}\binom{24}{12}=14062422446800$
The general method I used can be applied to basically every problem of this type, so be on the lookout for them. :)
Hint : Consider the number of stackings in which the chips are allowed to touch each other first. Then subtract the number of stackings in which at least two blue chips touch, which is equivalent to consider 11 blue poker chips instead of 12.