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RAmen RAmen_ (:)_ ramen 30
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2015-04-19 06:39 1 Amen ramen AmenRAmen rmen.
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Now, compute the number of favorable sequences where the first 4 positions contain exactly two distinct nucleotide types, and each appears exactly twice.
So for each choice of two types $X, Y$ and each way to assign 2 of each to the first 4 positions:
For each such choice, the number of full sequences is:
But better to check if this matches expected form.
Alternatively, leave in factorial form for exactness:
Instead, use combinatorial interpretation:
Now compute numerically or simplify:
Simplify: divide numerator and denominator by 16:
But this is probability — must be less than 1 — so error in counting.
The mistake: when choosing the remaining $28$ positions, we must account for the fact that the multinomial count assumes the two types X and Y have exactly 2 each already placed, so the rest must be filled with the other nucleotides — but C and G are still fully present with 8 each, so yes, $8$ Cs and $8$ Gs go into the remaining 28 positions — valid.