Question

Suppose a sequence of 8 nucleotides contains 2 each of a, c, g, t. how many such sequences are there?

Answer 1

Using the concept of permutations with repetitions, we can calculate that there are 2,520 different sequences possible where each of the nucleotides A, C, G, and T appears exactly twice.

Step-by-step explanation:

To find out how many such sequences of nucleotides there are, we can use the combinatorics in mathematics, specifically the concept of permutations. We have a total of 8 nucleotides, composed of 2 adenine (A), 2 cytosine (C), 2 guanine (G), and 2 thymine (T) nucleotides. The question is equivalent to figuring out how many different ways we can arrange these nucleotides in a sequence.

The number of permutations of a set of n elements where there are repetitions is given by the formula:
Permutations = n! / (n1! * n2! * ... * nk!)

where n! (n factorial) is the product of all positive integers up to n, and ni! is the factorial of the count of the ith repeated element.

In this case, n = 8, and we have four groups of repeated elements (A, C, G, T). Each of these groups has a count of 2, so the formula becomes:
8! / (2! * 2! * 2! * 2!) = 40,320 / 16 = 2,520

So, there are 2,520 different sequences possible where each of the nucleotides A, C, G, and T appears exactly twice.

Answer 2

The answer is 2520. Suppose that n items are divided into two or more categories, the formula for the combinations of n items is n factorial over the product of the factorial of the number of item in each category. In our case, 8 nucleotides are divided into 4 categories a, c, g, t, and the number of items in each category is 2, so the answer is 8!/(2!*2!*2!*2!)=2520.