Matilda Webb
2022-05-08
Answered

Coach Ringdahl has 10 players on his varsity basketball team. How many different starting line-ups (5 players) can he choose?

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Carolyn Farmer

Answered 2022-05-09
Author has **16** answers

Step 1

This is a combinations question - we don't care about the order in which we select the 5 players in the starting line-up, only that they are chosen.

So we have: $C}_{10,5}=\frac{10!}{(5!)(10-5)!}=\frac{10!}{(5!)(5!)$

Step 2

And now let's evaluate this: $\frac{10!}{(5!)(5!)}=\frac{{\overline{)10}}\times {{\overline{)9}}}^{3}\times {{\overline{)8}}}^{2}\times 7\times 6\times \overline{)5}!}{(\overline{)5}!)({\overline{)5}}\times {\overline{)4}}\times {\overline{)3}}\times {\overline{)2}}\times 1)}=3\times 2\times 7\times 6=252$

This is a combinations question - we don't care about the order in which we select the 5 players in the starting line-up, only that they are chosen.

So we have: $C}_{10,5}=\frac{10!}{(5!)(10-5)!}=\frac{10!}{(5!)(5!)$

Step 2

And now let's evaluate this: $\frac{10!}{(5!)(5!)}=\frac{{\overline{)10}}\times {{\overline{)9}}}^{3}\times {{\overline{)8}}}^{2}\times 7\times 6\times \overline{)5}!}{(\overline{)5}!)({\overline{)5}}\times {\overline{)4}}\times {\overline{)3}}\times {\overline{)2}}\times 1)}=3\times 2\times 7\times 6=252$

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