A dna sequence consists of a series of letters representing a dna strand that spells out the genetic code. there are 4 possible letters (a, c, g, and t), each representing a specific nucleotide base in the dna strand (adenine, cytosine, guanine, and thymine, respectively). how many distinguishable sequences can be found using two as, two cs, three gs, and one t

3360 ways to arrange 2 a's, 2 c's, 3g's, and a t This problem uses the factorial function, n! which is simply the product of all integers from 1 ... n. The beauty is that n! is the number of ways that n objects can be ordered. So we've been given a total of 8 letters. 8 letters can be arranged 8! ways, giving 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 = 40320 But we do have some duplicate letters. For the duplicates, we really don't care which order they're in. After all, you can't tell the difference between "aa" and "aa" (trust me, I swapped them). So divide the total number of ways you can arrange 8 elements by the the number of ways the identical values can be arranged. Giving 40320 / 2! / 2! / 3! / 1! = 40320 / 2 / 2 / 6 / 1 = 3360