Three balls are drawn from a bag containing 5 red, 4 white and 3 black balls. The number of ways in which this

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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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To find the number of ways to draw three balls with at least two red, we consider two mutually exclusive cases: exactly two red balls and exactly three red balls. We calculate the combinations for each case and sum them, resulting in $\boxed{80}$ ways.

Concept: Combinations (Selection Without Repetition)

This problem asks for the number of ways to select balls from a bag, where the order in which the balls are drawn does not matter. For example, drawing a red ball then a white ball is considered the same as drawing a white ball then a red ball if we are only concerned with the final set of balls. This type of selection is called a combination.

When we select $k$ items from a set of $n$ distinct items without regard to order and without replacement (meaning an item cannot be selected more than once), the number of ways is given by the combination formula:

n

In this problem, we have different colored balls. When we choose, say, 2 red balls from 5 red balls, we are selecting 2 distinct positions for red balls from the 5 available red balls. Similarly for other colors. The "without repetition" aspect means we are not picking the same physical ball multiple times.

Problem Breakdown

We have a bag containing:

$5$ red balls
$4$ white balls
$3$ black balls Total balls = $5 + 4 + 3 = 12$ balls.

We need to draw $3$ balls such that at least $2$ are red. This condition means we must consider two separate scenarios, as they are the only ways to satisfy the condition and are mutually exclusive (they cannot happen at the same time):

Exactly $2$ red balls are drawn.
Exactly $3$ red balls are drawn.

We will calculate the number of ways for each scenario and then add them together to get the total number of ways.

Step-by-step Solution

Identify the total number of balls of each type.
- Red balls: $R = 5$
- White balls: $W = 4$
- Black balls: $B = 3$
- Total non-red balls: $W + B = 4 + 3 = 7$
Calculate ways for Scenario 1: Exactly $2$ red balls.

If exactly $2$ balls are red, then the remaining $3 - 2 = 1$ ball must be chosen from the non-red balls.
- Number of ways to choose $2$ red balls from $5$ red balls:

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

*   Number of ways to choose $1$ non-red ball from the $7$ non-red balls (4 white + 3 black):

C(7, 1) = \binom{7}{1} = \frac{7!}{1!(7-1)!} = \frac{7}{1} = 7

*   The total number of ways for Scenario 1 is the product of these two combinations: …

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Q.Three balls are drawn from a bag containing 555 red, 444 white and 333 black balls. The number of ways in which this can be done if at least 222 are red is ______.

Concept: Combinations (Selection Without Repetition)

Problem Breakdown

Step-by-step Solution

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Q.Three balls are drawn from a bag containing $5$ red, $4$ white and $3$ black balls. The number of ways in which this can be done if at least $2$ are red is ______.

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