The number of ways in which a team of eleven players can be selected from 22 players always including 2 of the

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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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We fix the 2 compulsory players and remove the 4 excluded ones, leaving 16 available players from whom we must choose the remaining 9. The number of ways is ${}^{16}C_{9}$ , which matches option (C).

This is a classic selection with mandatory inclusion and exclusion problem. The key is to treat the "always included" and "always excluded" players as already decided, so they don't participate in the choosing process — they simply adjust the pool size and the number of slots left to fill.

Let’s break it down step by step.

Total players available initially: 22.
Players who must be excluded: 4. These 4 are completely out of consideration — they cannot be in the team at all. So we remove them from the pool.

Remaining players after exclusion: $22 - 4 = 18$ .
Players who must be included: 2. These 2 are already guaranteed a spot in the team. So we place them directly into the squad.

Now, the team needs 11 players total, and 2 are already chosen. So the number of remaining slots to fill is $11 - 2 = 9$ .
Available players to choose from: After removing the 4 excluded players, we had 18. But we also must account for the 2 who are already selected — they are not available for further selection (they're already in). So the pool of players from whom we can choose the remaining 9 is $18 - 2 = 16$ .
The selection problem reduces to: Choose 9 players from a set of 16. The order doesn't matter, so it's a combination: …

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Q.The number of ways in which a team of eleven players can be selected from 222222 players always including 222 of them and excluding 444 of them is (A) 16C11{}^{16}C_{11}16C11​ (B) 16C5{}^{16}C_{5}16C5​ (C) 16C9{}^{16}C_{9}16C9​ (D) 20C9{}^{20}C_{9}20C9​

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Q.The number of ways in which a team of eleven players can be selected from $22$ players always including $2$ of them and excluding $4$ of them is
(A) ${}^{16}C_{11}$
(B) ${}^{16}C_{5}$
(C) ${}^{16}C_{9}$
(D) ${}^{20}C_{9}$

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