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

This problem asks for the number of ways to pick a 5-member team plus one captain from 10 basketball players. Instead of choosing the team first and the captain second, we can think of it as choosing the full group of 6 people who will be involved at all (5 team members plus 1 captain), and then deciding who among that group of 6 becomes captain.

Step 1: Choose the group of 6 players (out of 10) who will be involved in the selection.

\[ \binom{10}{6} = \frac{10!}{6!\,4!} = 210 \]

Step 2: Out of these 6 chosen players, designate one of them as the captain (the remaining 5 automatically form the team).

\[ \binom{6}{1} = 6 \]

Step 3: Multiply the two counts together, since each choice of the group of 6 can be paired with any of the 6 possible captain choices.

\[ 210 \times 6 = 1260 \]

Checking against the other options: 1400 and 1600 don't correspond to any consistent way of combining these binomial coefficients, they would only appear from an incorrect count for the group selection. 1250 is close to 1260 but isn't obtainable from any valid combination formula for this scenario.

Therefore, the correct answer is 1260.