Q.In how many ways can the letters of the word ARRANGEMENTS be arranged? a) Find the chance that an arrangement chosen at random begins with the letters EE. b) Find the probability that the consonants are together.

ARRANGEMENTS has 12 letters: A(2), R(2), N(2), G(1), E(2), M(1), T(1), S(1). Total distinct arrangements $=\dfrac{12!}{2!2!2!2!}=\dfrac{479001600}{16}=29937600$.

a) Begins with EE: since there are only 2 E's, both must occupy positions 1-2 (only 1 way for that pair), and the remaining 10 letters (A,A,R,R,N,N,G,M,T,S) fill the rest in $\dfrac{10!}{2!2!2!}=\dfrac{3628800}{8}=453600$ ways. $P=453600/29937600=1/66$. (Cross-check: $P(\text{1st}=E)\times P(\text{2nd}=E\mid\text{1st}=E)=\frac{2}{12}\times\frac{1}{11}=\frac{1}{66}$.) …

Permutations of a multiset (divide by the factorial of each repeated letter's count); block method for 'consonants together'; direct se …

Forgetting to divide by $2!$ for EACH of the four repeated letters (A, R, N, E) in the to …