In how many ways can a football team of $11$ players be selected from $16$ players? How many of them will (i) include $2$ particular players? (ii) exclude $2$ particular players?

> [!TLDR] > This problem involves combinations, as the order of selecting players does not matter. We calculate the total ways to select $11$ players from $16$, then adjust the pool and selection size for cases where $2$ particular players are included or excluded. The results are $\mathbf{4368}$ total ways, $\mathbf{2002}$ ways including $2$ particular players, and $\mathbf{364}$ ways excluding $2$ particular players. When we talk about "selecting" items, where the order of selection does not matter, we are dealing with combinations. If the order *did* matter (e.g., selecting players for specific positions like goalkeeper, defender, etc.), we would use permutations. Since the question asks for the number of ways to "select" a team, it implies that the $11$ chosen players form a single team, regardless of the order in which they were picked. > [!FORMULA] > The number of ways to choose $k$ items from a set of $n$ distinct items, where the order of selection does not matter, is given by the combination formula: > $$C(n, k) = \binom{n}{k} = \frac{n!}{k!(n-k)!}$$ > Here, $n!$ (read as "n factorial") is the product of all positive integers up to $n$, i.e., $n! = n \times (n-1) \times \dots \times 2 \times 1$. Let's break down the problem into its three parts. ### Total Ways to Select 11 Players from 16 1. **Identify $n$ and $k$:** * The total number of available players is $n = 16$. * The number of players to be selected for the team is $k = 11$. 2. **Apply the Combination Formula:** We need to find $\binom{16}{11}$. Using the property $\binom{n}{k} = \binom{n}{n-k}$, it's often simpler to calculate $\binom{16}{16-11} = \binom{16}{5}$. $$ \binom{16}{11} = \binom{16}{5} = \frac{16!}{5!(16-5)!} = \frac{16!}{5!11!} $$ $$ = \frac{16 \times 15 \times 14 \times 13 \times 12 \times 11!}{5 \times 4 \times 3 \times 2 \times 1 \times 11!} $$ We can cancel out $11!$ from the numerator and denominator: $$ = \frac{16 \times 15 \times 14 \times

In how many ways can a football team of 11 players be selected from 16 players? How many of them will (i) incl

NCERT Exemplar · Q23

11

(i) include

2

particular players?

(ii) exclude

2

particular players?

Yanam BieapLong· 3mImportance★★★★★est

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Concept understanding — Permutations Without Repetition

Permutations Without Repetition – The Idea of Arranging Things

Imagine you have three different books on a shelf: a Physics book, a Chemistry book, and a Maths book. How many different ways can you arrange them in a row?

You could try listing them out:

Physics, Chemistry, Maths
Physics, Maths, Chemistry
Chemistry, Physics, Maths
Chemistry, Maths, Physics
Maths, Physics, Chemistry
Maths, Chemistry, Physics

That's 6 arrangements. Notice that each arrangement uses all three books exactly once — no book is repeated, and no book is left out. This is the core idea: permutations without repetition count the number of ways to arrange a set of distinct objects in order, using each object exactly once.

Why "Without Repetition"?

The phrase "without repetition" means that once you place an object in a position, you cannot use it again. In our book example, once you put the Physics book in the first slot, you cannot put it in the second or third slot. Each object appears exactly once in the arrangement.

This is different from "permutations with repetition" (like creating 3-letter codes from the letters A, B, C where you can reuse letters — e.g., AAA, AAB, etc.). Here, no repeats allowed.

The Counting Logic – Why Multiply?

Let's build the arrangement step by step for 3 distinct books:

First position: You have 3 choices (any of the 3 books).
Second position: After placing the first book, only 2 books remain — so 2 choices.
Third position: Only 1 book is left — so 1 choice.

Total arrangements = $3 \times 2 \times 1 = 6$ .

This product $3 \times 2 \times 1$ is called 3 factorial, written as $3!$ .

P(n) = n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1

For $n$ distinct objects, the number of permutations (arrangements in order) is $n!$ .

What If You Only Arrange Some of Them?

Suppose you have 5 different books, but you only want to arrange 3 of them on a shelf. How many ways?

First position: 5 choices
Second position: 4 choices
Third position: 3 choices

Total = $5 \times 4 \times 3 = 60$ .

This is a permutation of 5 objects taken 3 at a time, written as $P(5,3)$ or ${}^5P_3$ .

P(n, r) = \frac{n!}{(n-r)!} = n \times (n-1) \times \cdots \times (n-r+1)

Here $n$ is the total number of distinct objects, and $r$ is how many you are arranging. The formula works because:

Numerator $n!$ counts all arrangements of all $n$ objects.
Denominator $(n-r)!$ removes the arrangements of the $n-r$ objects you are not using.

Key Points to Remember

Important

Order matters — swapping two objects gives a different permutation.
No repetition — each object is used at most once.
For arranging all $n$ objects: $n!$
For arranging r out of $n$ objects: $\frac{n!}{(n-r)!}$

Common Mistake to Avoid …

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This problem involves combinations, as the order of selecting players does not matter. We calculate the total ways to select $11$ players from $16$ , then adjust the pool and selection size for cases where $2$ particular players are included or excluded. The results are $\mathbf{4368}$ total ways, $\mathbf{2002}$ ways including $2$ particular players, and $\mathbf{364}$ ways excluding $2$ particular players.

When we talk about "selecting" items, where the order of selection does not matter, we are dealing with combinations. If the order did matter (e.g., selecting players for specific positions like goalkeeper, defender, etc.), we would use permutations. Since the question asks for the number of ways to "select" a team, it implies that the $11$ chosen players form a single team, regardless of the order in which they were picked.

k

Let's break down the problem into its three parts.

Total Ways to Select 11 Players from 16

Identify $n$ and $k$ :
- The total number of available players is $n = 16$ .
- The number of players to be selected for the team is $k = 11$ .
Apply the Combination Formula:

We need to find $\binom{16}{11}$ . Using the property $\binom{n}{k} = \binom{n}{n-k}$ , it's often simpler to calculate $\binom{16}{16-11} = \binom{16}{5}$ .

\binom{16}{11} = \binom{16}{5} = \frac{16!}{5!(16-5)!} = \frac{16!}{5!11!}

= \frac{16 \times 15 \times 14 \times 13 \times 12 \times 11!}{5 \times 4 \times 3 \times 2 \times 1 \times 11!}

We can cancel out $11!$ from the numerator and denominator:

= \frac{16 \times 15 \times 14 \times 13 \times 12}{5 \times 4 \times 3 \times 2 \times 1}

Calculate the Value: Simplify the expression:
- $5 \times 3 = 15$ , so $\frac{15}{5 \times 3} = 1$ .
- $4 \times 2 = 8$ , so $\frac{16}{4 \times 2} = \frac{16}{8} = 2$ .

= 2 \times 1 \times 14 \times 13 \times 12

= 28 \times 156

= 4368

There are $4368$ ways to select a team of $11$ players from $16$.

(i) Ways to Include 2 Particular Players

Adjust the Selection:

If $2$ particular players must be included in the team, it means they are already chosen.
- The number of players we still need to select is $11 - 2 = 9$ .
Adjust the Pool of Available Players:

Since these $2$ particular players are already accounted for (they are in the team), they are also removed from the pool of players from whom we can choose.
- The remaining number of players to choose from is $16 - 2 = 14$ .
Apply the Combination Formula:

Now, we need to select $9$ players from the remaining $14$ players.

\binom{14}{9} = \binom{14}{14-9} = \binom{14}{5}

= \frac{14!}{5!9!} = \frac{14 \times 13 \times 12 \times 11 \times 10 \times 9!}{5 \times 4 \times 3 \times 2 \times 1 \times 9!}

= \frac{14 \times 13 \times 12 \times 11 \times 10}{5 \times 4 \times 3 \times 2 \times 1}

Calculate the Value: Simplify the expression:
- $5 \times 2 = 10$ , so $\frac{10}{5 \times 2} = 1$ . …

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Total Ways to Select 11 Players from 16

(i) Ways to Include 2 Particular Players

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