In a string of letters B U U B U U U U B, the 1st B is followed by two U's, the
2nd B is followed by four U's, and the 3rd B is following by no U's. How many strings of 17 U's and 4 B's have every B followed by:
1.) No U?
2.) at most one U?
3.) exactly one U?
4.) at least one U?
In some ways #2 is the trickiest. The answer is the sum k=0 to 4 combin(4,k).
For #4, think of the 17 U’s in a line with spaces between them for one B.
How many spaces are there before of some U ( ^U^U…^U)?
We can put each B’s into any one of those different spaces (no two in the same space) .
How many ways?
For #2, however, I got this as every possible combination:
So 10. Not 15 as you suggested.
And for d, it's going to be some really, really big number. Still have no idea on this one. My thoughts:
for d, you basically have 5 places where you have n U's
Each one of those T's can be any number of T's from 1-13, except the first one
The left one has a min val of 0, and the right 4 have a min val of 1
The number of possible combinations in which the sum of the 5 numbers is 17 is the answer to #4?