Before you downvote this for being a duplicate, kindly take cognisance of the face that I don't have enough reputation to comment on the germane answer.I'll attempt to pose my enquiry as a question In regard to the ingenious solution here, my queries are :

1) Does the function model capture the actual phenomena completely? Let's say after 1st pick up,the colours are $1\ 1\ 3\ 4$ (for $4$ balls coloured $1$ to $4$),but $f_2$ could be such that it maps $1\to1, 2\to2, 3\to2, 4\to4$ giving the result of $f_1f_2 = 1\ 1\ 2\ 4$,but obviously this configuration is not possible as colour 2 no longer exists to propagate further.

2)Even if the above model is apt, why is the Expectation, as it is? If we have $k$ distinct colours won't the number of ways in which k or k-1 distinct colours be obtained depend upon the multiplicity of each of the $k$ colours present? Example (for $5$ balls) if the configuration is $1,2,3,4,4$ number of ways to get $3$ distinct colours, as per my understanding of the solution is $6\cdot2$ (i.e $\binom{4}{2}2$) however shouldn't it be $9\cdot2(1-2,1-3,1-4,1-4,2-3,2-4,2-4,3-4,3-4)$.Is ONLY the number of distinct colours without the corresponding multiplicity sufficient to calculate the next configuration?


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