CP #16

BTG Challenge Question:

How many unique 12-letter "words" can be created using 8 A's and 4 B's such that no 2 B's are adjacent? (For example, the word "ABAABABAAABA" is permissible, but the word "AABBABABAAAA" is not permissible, since it has 2 adjacent B's)

Answer:

we have 8 A's and 4 B's and we need to figure out how many ARRANGEMENTS are possible such that no two B's are together.

First let us arrange 8 A's. As all are identical we have just one way to arrange them.

Here my advice is to actually write 8 A's on your note pad. Now see my friend we have nine places where 4 B's can go. just count.

_ A _ A _ A _ A _ A _ A _ A _ A _

these 4 places can be chosen in 126 ways. apply combination formula ie 9C4.

Now as all 4 B's are identical we don't need to worry about arranging those 4 B's ( after placing them in four of the nine places.) So, here is your answer friend. i would have further counted for the permutations if the 4 B's were not identical. But as they are identical the answer is 126.