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The choir club at a local school is running a field to a waterpark. There are 6 parks the club is considering. The director has asked the members to each pick three to narrow down the list of possibilities before making a final decision. How many different possibilities are there for the list of three?

A20
B150
C120
D3

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6C3 

6! / (3! (6 - 3)!) => 
6! / (3! 3!) => 
720 / (6 * 6) => 
720 / 36 => 
20

Answer:

The correct answer is A. 20

Step-by-step explanation:

Number of parks available for the club = 6

Capacity of the list of parks which is to be formed = 3

So, We need to select a list of 3 parks

Thus, we need to find the total number of different ways in which list of 3 parks can be formed from the total number of 6 parks.

\(\text{Number of different 3-parks list which can be chosen = }_3^6\txterm{C}\\\\\text{Number of different 3-parks list which can be chosen = } \frac{6!}{3!\times 3!}\\\\\text{Number of different 3 parks list which can be chosen = }\frac{6\times 5\times 4}{6}\\\\\text{Number of different 3 parks list which can be chosen = }20\)

Therefore, The correct answer is A. 20

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