Combinatorics question about the four-letter sequence "GRIT"

AI Thread Summary
The discussion revolves around calculating the number of nine-letter words that can be formed from the letters in "FULBRIGHT" while containing the sequence "GRIT." It is established that "GRIT" can be treated as a single unit, simplifying the problem to arranging this unit with the remaining letters. The calculation shows there are six positions for "GRIT" and five letters left, leading to 6 x 5! = 720 possible combinations. Participants confirm this approach, agreeing that treating "GRIT" as a single letter is valid. The solution is verified as correct, providing clarity on the combinatorial aspect of the problem.
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Homework Statement
Given the letters in "FULBRIGHT" how many contain the four-letter sequence "GRIT".
Relevant Equations
6 x 5!
Question: "A total of 9! = 362880 different nine-letter ‘‘words’’ can be produced by rearranging the letters in FULBRIGHT. Of these, how many contain the four-letter sequence GRIT?"

Solution: There are six ways of getting the word "GRIT" with five letters left over giving 6 x 5! = 720 possibilities.

There is no answer in my book so just wanted to verify whether my solution is correct or not.
 
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Yes, you can consider GRIT as a single letter X. So ##6!##
 
Thank you I didn't think of it that way but that does make sense!
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
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