Respuesta :
The number of different assignments that are possible for this case when the teacher wants to assign 4 of the 15 practice problems is 1,365 ways.
How many ways k things out of m different things (m ≥ k) can be chosen if order of the chosen things doesn't matter?
We can use combinations for this case,
Total number of distinguishable things is m.
Out of those m things, k things are to be chosen such that their order doesn't matter.
This can be done in total of
[tex]^mC_k = \dfrac{m!}{k! \times (m-k)!}[/tex] ways.
For this case, the teacher have to choose 4 problems out of 15 practice problems so that assignments are different.
An assignment would be different than other if the 4 questions in one assigment are not exactly same in the second assigment, and their order won't matter (assumingly).(no counter statement is given in the question's statement) against this assumption).
Those 15 problems are assumingly distinguishable (here we used general logic that each problem would be somewhat different if not fully, and thus, no two problems being indistinguishable).
Then, we can use the combinations to find the number of ways any 4 problems out of 15 practice problems available can be chosen, as shown below:
[tex]^{15}C_4= \dfrac{15!}{4! \times (15-4)!}\\\\\\^{15}C_4=\dfrac{15!}{4! \times 11!} = \dfrac{15 \times 14 \times 13 \times 12 \times 11 \times \cdots \times 2 \times 1}{(4 \times 3 \times 2 \times 1) \times ( 11 \times \cdots \times 2 \times 1)}\\\\\\^{15}C_4=\dfrac{15 \times 14 \times 13 \times 12 }{4 \times 3 \times 2 \times 1} = 1365[/tex]
Thus, the number of different assignments that are possible for this case when the teacher wants to assign 4 of the 15 practice problems is 1,365 ways.
Learn more about combinations here:
https://brainly.com/question/11958814
