Answer :
In this scenario, you are asked to determine how many ways there are to pick a starting five from a basketball team of twelve members.
First, let's understand whether we need to use permutations or combinations. The key point is whether the order in which you pick the players matters. In forming a starting lineup, the order in which you select the players does not matter; what matters is which players are selected. Therefore, we need to use combinations, not permutations.
Now, we need to calculate the number of combinations. The formula for combinations is given by:
[tex]\[ C(n, k) = \frac{n!}{k! \cdot (n - k)!} \][/tex]
where \( n \) is the total number of items (12 players) and \( k \) is the number of items to choose (5 players).
Plugging in the values:
[tex]\[ C(12, 5) = \frac{12!}{5! \cdot (12 - 5)!} = \frac{12!}{5! \cdot 7!} \][/tex]
Given the calculations for this scenario, the result is:
[tex]\[ C(12, 5) = 792 \][/tex]
Thus, the correct answer is:
b. Combination; [tex]${ }_{12} C_5=792$[/tex]
The best answer from the choices provided is B.
First, let's understand whether we need to use permutations or combinations. The key point is whether the order in which you pick the players matters. In forming a starting lineup, the order in which you select the players does not matter; what matters is which players are selected. Therefore, we need to use combinations, not permutations.
Now, we need to calculate the number of combinations. The formula for combinations is given by:
[tex]\[ C(n, k) = \frac{n!}{k! \cdot (n - k)!} \][/tex]
where \( n \) is the total number of items (12 players) and \( k \) is the number of items to choose (5 players).
Plugging in the values:
[tex]\[ C(12, 5) = \frac{12!}{5! \cdot (12 - 5)!} = \frac{12!}{5! \cdot 7!} \][/tex]
Given the calculations for this scenario, the result is:
[tex]\[ C(12, 5) = 792 \][/tex]
Thus, the correct answer is:
b. Combination; [tex]${ }_{12} C_5=792$[/tex]
The best answer from the choices provided is B.
