There are 12 points in a plane of which 5 points are collinear, then the number of lines obtained by joining t

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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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When some points are collinear, they all lie on a single line instead of forming multiple lines; we subtract the "overcounted" pairs from the collinear set and add back the one line they actually form. The statement is incorrect — the correct count is ${}^{12}C_2 - {}^5C_2 + 1$ .

The question asks us to verify whether the given formula correctly counts the number of distinct lines formed by joining 12 points in pairs, where 5 of those points are collinear.

Why the naive approach fails

If all 12 points were in general position (no three collinear), every pair of points would determine a unique line. That would give us ${}^{12}C_2 = \frac{12 \cdot 11}{2} = 66$ distinct lines.

But we have a constraint: 5 points are collinear. These 5 points all lie on the same line. When we count ${}^{12}C_2$ , we're treating each of the ${}^5C_2 = 10$ pairs from these 5 collinear points as if they determine different lines. In reality, all 10 pairs lie on just one line.

The correct counting strategy

Start with the total pairs.

We have ${}^{12}C_2 = 66$ pairs of points.
Identify the overcounting.

Among the 5 collinear points, there are ${}^5C_2 = 10$ pairs. Each of these 10 pairs was counted as a separate line in our initial count, but they all determine the same line.
Subtract the excess and add back the truth.

We overcounted by including 10 lines where there should be only 1. So we subtract the 10 duplicate entries and add back the 1 actual line:

\text{Number of lines} = {}^{12}C_2 - {}^5C_2 + 1

Calculate the result. …

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Q.There are 121212 points in a plane of which 555 points are collinear, then the number of lines obtained by joining these points in pairs is 12C2−5C2{}^{12}C_{2} - {}^{5}C_{2}12C2​−5C2​.

Why the naive approach fails

The correct counting strategy

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Q.There are $12$ points in a plane of which $5$ points are collinear, then the number of lines obtained by joining these points in pairs is ${}^{12}C_{2} - {}^{5}C_{2}$ .

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