How many DNA chains of length 6 have at least one of each base \(T\), \(C\), \(A\), and \(G\)? Answer this question under the following assumptions:

  1. Only the number of bases of a given type matter.

We can think of it as the number of 4-combinations of a 4-set without replacement (because we need at least one of each base) multiplied by a number of 2-combinations of a 4-set with replacement (because the remaining two bases can be of any type):

\[C(4, 4) \cdot C^{R}(4, 2) = \binom{4}{4} \cdot C(5, 2) = \binom{4}{4} \cdot \binom{5}{2} = 10\]

  1. Order matters.

We can break it down into two cases, both of which are DNA chains of length 6 with all four bases (\(A\), \(C\), \(G\), \(T\)) present:

  1. the remaining two bases are of the same type. In this case we have only 4 possibilities: \(AA\), \(CC\), \(GG\), \(TT\) for the total of:

\[4 \cdot \frac{6!}{3!} = 480\]

  1. the remaining two bases are of different type. In this case there are \(\binom{4}{2} = 6\) possibilities, namely: \(AC\), \(AG\), \(AT\), \(CG\), \(CT\), \(GT\), for the total of:

\[6 \cdot \frac{6!}{2!2!} = 1080\]

The final total is the sum of both cases:

\[4 \cdot \frac{6!}{3!} + 6 \cdot \frac{6!}{2!2!} = 1560\]