# Calculating Possible Combinations of Bases in a DNA Strand of a Given Length

In my Biology class we were asked this question:

This DNA strand consists of eight pairs of nitrogenous bases. How many different sequences of eight bases can you make? Explain how you found your answer.

I guessed either 28 or 8!. Apparently, the answer is 8! ÷ 24. I asked my teacher, but she did not know the answer. Does anyone know why this would be?

• The answer your teacher gave you might be the answer to the question of how many sequences of 8 bases can be formed using only the bases shown in the diagram, each one can be used once. The factorial comes from the fact that once you pick a base there are n-1 options left and so on. 2^4 is 2*2*2*2 which accounts for there being four duplicate bases so that count only unique sequences. Commented Mar 15, 2019 at 0:47
• Don't guess! You need to learn to approach this sort of simple statistics logically as laid out by @Remi.b. The relevence of this sort of problem to biology is more in relation to the frequency of restriction sites, which are smaller and a better place to start. You can find lots of practice questions and an explanation of how to answer them on a self-teaching resource I put up for students at my own university. Give it a try. Commented Mar 15, 2019 at 18:32

## 2 Answers

At each base, you can have 4 different bases (A,T,C or G). Therefore for the first base there are 4 possibilities, namely

• A
• T
• C
• G

For the first two base pairs there are $$4^2 = 16$$ possible combinations (I'm no good with math but the premise below is in error. If there is an A then it can only be paired with a T if there is a T it can only pair with an A C with G and G with C. So many of the below listed options will not occur - changed to italics)

• AA
• AT
• *AC
• AG*
• TA
• *TT
• TC
• TG*
• *CA
• CT
• CC*
• CG
• *GA
• GT*
• GC
• GG

For the first three bases, there are $$4^3$$ possible combinations. For 8 base pairs, there are $$4^8 = 65536$$ possible combinations. $$2^8$$, $$8!$$ and $$8! + 2^4$$ are all wrong.

• I think the problem asks combinations with bases in the given sequence. There is 2 of each of A, T, G, and C. So the combinations come out to 8!/(2*2*2*2) which is the answer their teacher provided. Commented Oct 28, 2020 at 6:02

Unfortunately @Remi.B is wrong. His answer allows the choice of any base for each position. This includes arrangements such as AAAAAAAA which is not possible as there are only two A bases. User40950 is correct in his comment. Here is a little more expanded version of his comment.
If we label the bases as A1 A2 C1 C2 G1 G2 T1 T2 then there are 8! possible arrangements. Notice however that interchanging A1 and A2 while leaving everything else unchanged yields the same arrangement, hence we divide the answer by 2. This same argument applies to each of the other duplicate pairs, so we divide the answer by $$2^4$$. Thus the final answer is
$$8!/2^4$$