Probability for a word to start with $\text{2,0,0,4}$ Let $A= \left\{ 2,2,4,4,0,0,0,0\right\}$. We arrange  those $8$ numbers randomly. What is the probability to get a sequence starting with $2,0,0,4$?
The answer is:$$\frac{2\cdot 4 \cdot 3 \cdot 2 \cdot 4!}{8!}$$
The denominator is simple, there are $8!$ permutations. Can you explain the numerator?
Thanks.
 A: You have two choices for the first digit as there are two $2$'s.  Then four choices for the next, as there are four $0$'s.  Having used one of the zeros, there are three left...
A: The letters that are left to be distributed are {2, 4, 0, 0}. These can be arranged in $A=\binom{4}{2}\times\binom{2}{1}=\frac{2\times4!}{2!\times2!}$ different ways.
The total amount of ways those 8 numbers can be arranged is not 8!, but $B=\binom{8}{4}\times\binom{4}{2}=\frac{8!\times4!}{4!\times4!\times2!\times2!}$. The chance is then $\frac{A}{B}=\frac{2\times4!\times4!}{8!}=\frac{2\times4\times3\times2\times4!}{8!}$
A: Be careful: the solution implies all elements in the alphabet are unique, e.g. you don't have 4 0s, but in fact $\{ 0_1, 0_2, 0_3, 0_4 \}$ and so on. Then yes, the numerator is the product of the number of ways to get on of two 2's, two of 4 0's $(4 \cdot 3)$, etc. 
If there's no distinction within each class, you just need to select the remaining letters $\{ 2,0,0,4 \}$ in 4 slots (because the first four are 'fixed'), and this number is $\binom{4}{1} \binom{3}{1} \binom{2}{2}$. The denominator in this would be $\binom{8}{6} \binom{6}{2} \binom{4}{4}$
