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?
Both #1 & #3 have the same answer. In #3 the last 8 letters are BUBUBUBU.
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?
Hmmm. I agree with #1 and #3 both being 1 string.
For #2, however, I got this as every possible combination:
TTTTTTAAAA
TTTTTAAAAT
TTTTTAAATA
TTTTTAATAA
TTTTTATAAA
TTTTAAATAT
TTTTAATATA
TTTTATATAA
TTTATATATA
TTATATATAT
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
UBUBUBUBU
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?
Thoughts?