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:
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?
