Answer to Question #330395 in Calculus for ash

Question

of 10 genes studied in yeast, 7 are found to form a network of metabolic interactions. If this set of 7 genes was chosen arbitrarily from the 10 that were studied, how many different sets would be possible? Show your calculation and the reasoning.a

Answer 1

Let n = 10, k = 7

We can choose the first gene in n ways, the second in n - 1 ways (because the first is already selected and there are n - 1 remaining) and we should divide this amount by two because there is no difference whether we select the second element before the first or after it (samples with such permutations will be equivalent), similarly we can select the third gene from the remaining n - 2 genes, and because there is no difference in what order we selected this gene (before previous two, between them or at the end; in total 3 permutations that produce equivalent samples), we should divide this value by 3. Continuing this chain until k genes are selected, we get the total amount of different samples is $n\cdot\frac{n-1}{2}\cdot\frac{n-2}{3}\cdot...\cdot\frac{n-k+1}{k}=\frac{n(n-1)...(n-k+1)}{k!}$

Let's substitute values n = 10 and k = 7:

$\frac{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4}{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7}=\frac{10\cdot9\cdot8}{1\cdot2\cdot3}=120$