Question

a coach must choose five starters from a team of 14 players.

how many different ways can the coach choose the starters?

a 2,002
b 240,240
c 362,880
d 120

Explanation:

🆕 New Concept Discovered: Combinations
Choosing groups where order does not matter.

Step 1: Identify if order matters

We need to choose a group of \( 5 \) starters from a total pool of \( 14 \) players.

Because the problem asks for the number of different ways to choose the starters (and does not assign specific positions or order to the chosen players), the order in which we select them does not matter. This means we use the combination formula rather than a permutation.

Step 2: Apply the combination formula

The formula for combinations is:

$$ C(n, r) = \frac{n!}{r!(n-r)!} $$

Where:

• \( n = 14 \) (the total number of players to choose from)
• \( r = 5 \) (the number of players we need to choose)

Substitute these values into the formula:

$$ C(14, 5) = \frac{14!}{5!(14-5)!} = \frac{14!}{5! \cdot 9!} $$

Step 3: Simplify and calculate

Expand the factorials to simplify the calculation:

$$ C(14, 5) = \frac{14 \times 13 \times 12 \times 11 \times 10 \times 9!}{5 \times 4 \times 3 \times 2 \times 1 \times 9!} $$

Cancel out \( 9! \) from the numerator and denominator:

$$ C(14, 5) = \frac{14 \times 13 \times 12 \times 11 \times 10}{5 \times 4 \times 3 \times 2 \times 1} $$

Now, simplify the remaining terms:

• \( 5 \times 2 = 10 \), which cancels out the \( 10 \) in the numerator.
• \( 4 \times 3 = 12 \), which cancels out the \( 12 \) in the numerator.

This leaves us with:

$$ C(14, 5) = 14 \times 13 \times 11 $$

Calculate the final product:

$$ 14 \times 13 = 182 $$

$$ 182 \times 11 = 2,002 $$

Answer:

A 2,002