In a football championship, 153 matches were played. Every two teams played one match with each other. The num

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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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The problem asks for the number of teams given the total matches played, where every two teams played one match. This is a combination problem ( $^nC_2$ ), leading to a quadratic equation whose positive solution gives the number of teams as $\boxed{18}$ .

When "every two teams played one match with each other," it means that each match is formed by selecting a unique pair of teams from the total pool of teams. The order in which we pick the two teams for a match does not matter; a match between Team A and Team B is the same as a match between Team B and Team A. This scenario is a classic application of combinations, where we are choosing a subset of items from a larger set without regard to their order.

Here, we need to find the total number of teams, let's call this $n$ . From these $n$ teams, we are forming groups of $2$ teams to play a match. The total number of such unique pairs (matches) is given as $153$ .

Here is the step-by-step solution:

Identify the mathematical concept: The problem describes a situation where we are selecting groups of 2 teams from a larger set of $n$ teams, and the order of selection does not matter. This is precisely the definition of a combination. If the order mattered (e.g., if Team A playing Team B was different from Team B playing Team A, perhaps home vs. away), it would be a permutation. But for a single match, order is irrelevant.
Define the variable: Let $n$ be the total number of teams participating in the championship.
Formulate the equation: The number of ways to choose 2 teams out of $n$ teams is given by the combination formula $^nC_2$ . We are told that this number of matches is $153$ .

Therefore, we can write the equation:

$^nC_2 = 153$
Expand the combination formula: The general formula for combinations is $^nC_r = \frac{n!}{r!(n-r)!}$ .

For our case, $r=2$ , so the formula simplifies to:

$^nC_2 = \frac{n!}{2!(n-2)!} = \frac{n \times (n-1) \times (n-2)!}{2 \times 1 \times (n-2)!} = \frac{n(n-1)}{2}$

Substituting this into our equation from step 3:

$\frac{n(n-1)}{2} = 153$
Solve the equation for $n$ :

Multiply both sides by $2$ :

$n(n-1) = 153 \times 2$

$n(n-1) = 306$

We need to find two consecutive integers whose product is $306$ . We can try estimating:

$10 \times 9 = 90$

$15 \times 14 = 210$ …

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Q.In a football championship, 153153153 matches were played. Every two teams played one match with each other. The number of teams, participating in the championship is ______.

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Q.In a football championship, $153$ matches were played. Every two teams played one match with each other. The number of teams, participating in the championship is ______.

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