<s>[INST] 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.
 [/INST] 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 $\frac{A}{B}=\frac{2\times4!\times4!}{8!}=\frac{2\times4\times3\times2\times4!}{8!}$
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