Question

A college has 10 basketball players. A 5-member team and a captain will be selected out of these 10 players. How many different selections can be made?

• A. 1260
• B. 1400
• C. 1250
• D. 1600

Hint:

When a problem involves choosing a group *and* assigning a role (like captain) within that group, remember that "choosing 5 and then 1 captain from the 5" is equivalent to "choosing 1 captain from 10 and then 4 members from the remaining 9". Pick the method that seems simpler to calculate for your exam.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem involves selecting a team and a captain from a group of players. This is a problem of combinations and permutations.

Step 2: Key Formula or Approach:
The selection can be broken down into two parts:
1. Choose the 5 members for the team from the 10 players. This is a combination problem: \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \).
2. Choose a captain from the 5 selected members.
Alternatively, we can first choose a captain from the 10 players, and then select the remaining 4 members from the remaining 9 players. This is equivalent.

Step 3: Detailed Explanation:
Method 1: Choose team first, then captain
1. Select 5 players for the team from 10 players:
Number of ways = \( \binom{10}{5} = \frac{10!}{5!(10-5)!} = \frac{10!}{5!5!} \)
\( = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} \)
\( = \frac{30240}{120} = 252 \)
2. Select 1 captain from the 5 chosen team members:
Number of ways = \( \binom{5}{1} = 5 \)
3. Total different selections:
Total = (Ways to choose team) \(\times\) (Ways to choose captain)
Total = $252 \times 5 = 1260$.
Method 2: Choose captain first, then remaining team members
1. Select 1 captain from 10 players:
Number of ways = \( \binom{10}{1} = 10 \)
2. Select the remaining 4 members from the remaining 9 players:
Number of ways = \( \binom{9}{4} = \frac{9!}{4!(9-4)!} = \frac{9!}{4!5!} \)
\( = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \)
\( = \frac{3024}{24} = 126 \)
3. Total different selections:
Total = (Ways to choose captain) \(\times\) (Ways to choose remaining members)
Total = $10 \times 126 = 1260$.
Both methods yield the same result.

Step 4: Final Answer:
1260 different selections can be made.

Answer 2

The problem needs the number of ways to pick a 5-member team plus a captain from 10 players. Since a captain must come from within the team, one clean way to check this is to see which option matches the structure of the count, choosing the team first (order doesn't matter) and then a captain from within it (order matters for that one role).

1. 1260: The number of ways to choose 5 players out of 10 is \( \binom{10}{5} = 252 \). Each such team can then have any one of its 5 members made captain, giving \( 5 \) choices per team. So the total is \( 252 \times 5 = 1260 \), which matches this option exactly.
2. 1400: For this to be correct, the count would need to be \( 252 \times 5.56 \), which is not a whole-number multiple of 252 for any sensible split of players and captains, so this value cannot arise from the given group sizes.
3. 1250: This is not a multiple of \( 252 \) or of \( \binom{9}{4} = 126 \) in a way consistent with picking 5 from 10 and then a captain from those 5, so it does not correspond to any valid factoring of the selection process.
4. 1600: Similarly, \( 1600 \) does not divide evenly by \( 252 \) or align with \( 10 \times 126 \), so it cannot represent this two-stage selection.

Only 1260 factors cleanly as (ways to choose the team) times (ways to choose the captain from that team), which is exactly the structure the question describes.

Therefore, the correct answer is 1260.