Question

A swim team must be chosen from 12 candidates. What is the greatest number of different 3-person teams that can be chosen?

Answer 1

Answer: 220 combinations

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Explanation:

We have 12*11*10 = 1320 permutations possible. This is where order matters. However, order does not matter with swim teams as there are no positions or ranks. All that matters is the group overall (rather than any individual in the group).

Consider the set {A,B,C}. We have 3*2*1 = 6 ways to arrange this set of letters. When considering a permutation, there are 6 permutations but only 1 combination since order doesnt matter and ABC is the same as BAC. Therefore, we divide 1320 over 6 to get the final answer of 1320/6 = 220.

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You can use the combination formula with n = 12 and r = 3. Doing so will give the following:

$_n C _r = \frac{n!}{r!*(n-r)!}\\\\_{12} C _3 = \frac{12!}{3!*(12-3)!}\\\\_{12} C _3 = \frac{12!}{3!*9!}\\\\_{12} C _3 = \frac{12*11*10*9!}{3!*9!}\\\\_{12} C _3 = \frac{12*11*10}{3!}\\\\_{12} C _3 = \frac{12*11*10}{3*2*1}\\\\_{12} C _3 = \frac{1320}{6}\\\\_{12} C _3 = 220$

As you can see, we get the same result and note how 1320/6 is also present as well.

Answer 2

The greatest number of different 3-person teams that can be chosen is 220.

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,

The 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

$^mC_k = \dfrac{m!}{k! \times (m-k)!} ways.$

So, total number of choices in that case would be:

$^mP_k = k! \times ^mC_k = k! \times \dfrac{m!}{k! \times (m-k)!} = \dfrac{m!}{ (m-k)!}\\\\^mP_k = \dfrac{m!}{ (m-k)!}$

This is called permutation of k items chosen out of m items (all distinct).

Consider the set {A,B,C}.

3*2*1 = 6 ways to arrange this set of letters.

When considering a permutation, there are 6 permutations but only 1 combination since order doesnt matter and ABC is the same as BAC.

We can use the combination formula with n = 12 and r = 3.

$^mC_k = \dfrac{m!}{k! \times (m-k)!} ways.\\\\^{12}C_3= \dfrac{12!}{3! \times (12-3)!} ways.\\\\\\^{12}C_3= \dfrac{12!}{3! \times (9)!} ways.\\\\^{12}C_3= \dfrac{12 \times 11 \times 10 \times 9}{3! \times (9)!} ways.$

1320/6 = 220

Thus, the greatest number of different 3-person teams that can be chosen is 220.

Learn more about combinations and permutations here:

https://brainly.com/question/16107928

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