Probability, counting and permutation I am trying to answer this question:
The letters of the word TROGLODYTE are arranged at random. Find the probability that there are at least 7 letters separating “G” and “Y”.
I assumed the events: A: having 7 letters separating “G” and “Y”
B: having 8 letters separating “G” and “Y”
P(A) =? For this we need to select 7 letters and treat G and those 7
letters and Y as 1 identity and then this 9 letter group and the rest 1
identity can be arranged in 2! ways, G and Y can interchange
their positions , this will be done in 2 ways and selection of 7
letters that could lie between G and Y could be done in ${C_8}^7$ ways, hence total number of ways this arrangement could be
done are $2!\times2!C_8^7$. Hence $$P(A)=\frac{2!\times2!C_8^7}{\frac{10!}{2!2!}}$$
and P(B) in the same manner.
Is this correct?
A: A simpler way to calculate it is as follows. The two spaces filled by G and Y can be any of $\binom {10}2=45$ possibilities, with equal probability. There are $3$ of these possibilities which have at least $7$ spaces separating them (first and last, second and last, first and second last). So the answer is $3/45=1/15$.
(Here $\binom{10}2$ is an alternative notation for your $C_{10}^2$.)
A: The total number of distributions ie the sample space is just $\frac{10! }{2! 2! }$. If G and Y are fixed at any 2 places then you just need to distribute
1) 7 items in between them and 1 not between them ie you need to distribute a total of 8 items. The total ways to distribute is then $2\times 2\times \frac{8! }{2! 2! }$. The first 2 in the above is due the fact that G and Y can be swapped giving 2 possible ways and the second 2 is due to the fact the G and Y along with the group of 7 seven letters in between them can be either in front of the 10th letter or in the end of the 10th letter. This will happen for both cases when G is occuring before Y or the opposite hence $2\times2 = 4$ cases. 
2) there are 8 items in between G and Y and the total number of arrangements is $2\times 1\times \frac{8! }{2! 2! }$
Hence the total probability is
$\frac{ 2\times 2\times \frac{8!}{2! 2!}}{\frac{10! }{2!2!}} + \frac{ 2\times 1\times \frac{8!}{2! 2!}}{\frac{10! }{2! 2! }} = \frac{1}{15}$
Note $ 8! =8 \times \; ^8C_7$
