A basketball team has 11 player on its roster. Only 5 player can be on the court at one time. How many different groups of 5 players can the team put on the floor?

Options

A. `5^(11)`
B. `._(11)C_(5)`
C. `._(11)P_(5)`
D. `11^5`

Answer

To solve the problem of how many different groups of 5 players can be selected from a roster of 11 players, we can use the concept of combinations. Here's a step-by-step solution:

### Step 1: Understand the Problem
We need to choose 5 players from a total of 11 players. The order in which we choose the players does not matter, which is why we use combinations instead of permutations.

### Step 2: Use the Combination Formula
The formula for combinations is given by:
\[ \binom{n}{r} = \frac{n!}{r!(n - r)!} \]
where:
- \( n \) is the total number of items (players in this case),
- \( r \) is the number of items to choose (players to be selected).

### Step 3: Identify Values of n and r
From the problem:
- \( n = 11 \) (total players),
- \( r = 5 \) (players to choose).

### Step 4: Substitute Values into the Formula
Now we substitute \( n \) and \( r \) into the combination formula:
\[ \binom{11}{5} = \frac{11!}{5!(11 - 5)!} = \frac{11!}{5! \cdot 6!} \]

### Step 5: Calculate Factorials
Now we calculate the factorials:
- \( 11! = 11 \times 10 \times 9 \times 8 \times 7 \times 6! \)
- \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)
- \( 6! \) will cancel out in the numerator and denominator.

Thus, we have:
\[ \binom{11}{5} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5!} = \frac{11 \times 10 \times 9 \times 8 \times 7}{120} \]

### Step 6: Perform the Multiplication
Now we calculate the numerator:
\[ 11 \times 10 = 110 \]
\[ 110 \times 9 = 990 \]
\[ 990 \times 8 = 7920 \]
\[ 7920 \times 7 = 55440 \]

### Step 7: Divide by 120
Now we divide the result by 120:
\[ \frac{55440}{120} = 462 \]

### Final Answer
Therefore, the number of different groups of 5 players that can be selected from 11 players is:
\[ \boxed{462} \]