Question

The number of ways in which a team of 11 players can be formed out of 25 players, if 6 out of them are always to be included and 5 of them are always to be excluded, is ______.

• A. 2002
• B. $^{20}\text{C}_{11}$
• C. $^{20}\text{C}_6$
• D. $^{14}\text{C}_5$

Hint:

Inclusion/Exclusion trick: If $p$ are always included and $q$ are always excluded from $n$ items to select $r$, the formula is simply $^{(n - p - q)}\text{C}_{(r - p)}$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
When certain items are always included, we reduce both the total items and the required items. When items are excluded, we only reduce the total items available.

Step 2: Formula Application:
Total players remaining = $25 - 6 (\text{included}) - 5 (\text{excluded}) = 14$. Players still needed = $11 - 6 = 5$.

Step 3: Explanation:
Since 6 players are already in the team, we only need to choose the remaining 5 players from the 14 players who are neither already picked nor banned. Number of ways = $^{14}C_5 = \frac{14 \times 13 \times 12 \times 11 \times 10}{5 \times 4 \times 3 \times 2 \times 1} = 2002$.

Step 4: Final Answer:
The number of ways is 2002 (which is $^{14}C_5$ or $^{14}C_9$).