A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be s

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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 need to select a group of black balls and a group of red balls independently, then multiply the number of ways for each selection using combinations, resulting in $\boxed{200}$ ways.

This problem asks us to determine the number of ways to select a specific number of balls of different colours from a given lot. The key idea here is "selection," which implies that the order in which the balls are chosen does not matter. For example, picking ball A then ball B is the same as picking ball B then ball A. This immediately tells us we are dealing with combinations, not permutations.

We have two distinct tasks:

Selecting black balls.
Selecting red balls.

Since these two selections are independent of each other, the total number of ways to perform both tasks is the product of the number of ways to perform each individual task. This is a fundamental principle in combinatorics known as the Multiplication Principle.

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Let's apply this step-by-step:

Selecting black balls:
- We have $5$ black balls in total.
- We need to select $2$ black balls.
- The number of ways to do this is $C(5, 2)$ .

C(5, 2) = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{(2 \times 1)(3 \times 2 \times 1)} = \frac{5 \times 4}{2 \times 1} = 10

There are $10$ ways to select $2$ black balls from $5$.

2. Selecting red balls:

* We have $6$ red balls in total.

* We need to select $3$ red balls.

* The number of ways to do this is $C(6, 3)$ . …

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Q.A bag contains 555 black and 666 red balls. Determine the number of ways in which 222 black and 333 red balls can be selected from the lot.

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Q.A bag contains $5$ black and $6$ red balls. Determine the number of ways in which $2$ black and $3$ red balls can be selected from the lot.

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