# String of Letters

• Sep 10th 2006, 10:59 AM
fifthrapiers
String of Letters
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?
• Sep 10th 2006, 12:37 PM
CaptainBlack
Quote:

Originally Posted by fifthrapiers
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?

If no B is followed by a U the B's must be the last 4 chars in the string, so
there is just one such string.

RonL
• Sep 10th 2006, 12:59 PM
Plato
Quote:

Originally Posted by fifthrapiers
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?
• Sep 10th 2006, 08:18 PM
fifthrapiers
Quote:

Originally Posted by Plato
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?
• Sep 10th 2006, 08:23 PM
fifthrapiers
2380 for #4? Trying to create a program for it but not sure how successful it'll be.
• Sep 11th 2006, 02:52 AM
malaygoel
Quote:

Originally Posted by fifthrapiers
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.

you missed
TTTTAATAAT
TTTTATAAAT
TTTTATAATA
TTTATATAAT
TTTAATATAT
TTTATAATAT

SO 16
you have
A-A-A-A-
Thesefour blanks eithercontain T or not . hance there are $\displaystyle 2^4$ways.
Keep Smiling
Malay
• Sep 11th 2006, 02:55 AM
malaygoel
Quote:

Originally Posted by fifthrapiers
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?

for d
you have thefolloing elements to arrange
BU,BU,BU,BU,13U
THSE CAN BE DONE IN
$\displaystyle \frac{17!}{13!4!}$

KeepSmiling
Malay
• Sep 11th 2006, 09:07 PM
fifthrapiers
Yup! I got that answer 17!/(13!4!) by the program I created (2380 ways).

Thanks for showing the easier way of finding it.