# 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.
– Cell
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. Mar 15, 2019 at 18:32

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

• 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. Oct 28, 2020 at 6:02