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Sagot :
This is a combination problem where the order doesn't matter.
We compute the number of combinations with the following formula:
[tex]C^{n_{}}_r=\frac{n!}{r!(n-r)!}[/tex]Where n is the number of things to choose from, and we choose r of them. No repetition, the order doesn't matter.
In our problem, we have n = 20 and for r we can pick r = 16 or r =4 (with both options we get the same result), picking r = 16 we get:
[tex]C^{20_{}}_{16}=\frac{20!}{16!\cdot(20-16)!}=\frac{20!}{16!\cdot4!}=4845[/tex]Answer
We have 4845 ways possible to get 4 problems correct and 16 problems wrong from a 20 problem set.
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