A box contains two white, three black and four red balls. In how many ways can three balls be drawn from the b

NCERT Exemplar · Q14

Q. A box contains two white, three black and four red balls. In how many ways can three balls be drawn from the box, if atleast one black ball is to be included in the draw.

Yanam BieapShort· 3mImportance★★★★★est

62% · 80/130 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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 …

Start your 14-day free trial to unlock the full solution →

To find the number of ways to draw at least one black ball, we calculate the total number of ways to draw three balls and subtract the number of ways to draw no black balls. The result is 64 ways.

When a problem asks for "at least one" of a certain item, it often simplifies the calculation to consider the opposite scenario: "none" of that item. This is a fundamental principle in combinatorics, often called the Principle of Complementary Counting.

Here's why this approach is effective:

If you want to draw "at least one black ball," it means you could draw:

One black ball and two other balls (white or red)
Two black balls and one other ball (white or red)
Three black balls

Calculating each of these cases separately and then summing them up can be tedious and prone to errors.

Instead, consider the total number of ways to draw three balls without any restrictions. From this total, if we remove all the ways where no black balls are drawn, what remains must be all the ways where at least one black ball is drawn.

Let's apply this concept to the given problem.

Identify the total number of balls and their distribution.

We have:
- White balls: 2
- Black balls: 3
- Red balls: 4 The total number of balls in the box is $2 + 3 + 4 = 9$ . We need to draw 3 balls.
Calculate the total number of ways to draw 3 balls from the box without any restrictions.

This is a combination problem, as the order in which the balls are drawn does not matter. We use the combination formula $C(n, k) = \frac{n!}{k!(n-k)!}$ , where $n$ is the total number of items and $k$ is the number of items to choose.

Here, $n=9$ (total balls) and $k=3$ (balls to be drawn).

C(9, 3) = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} = \frac{9 \times 8 \times 7 \times 6!}{3 \times 2 \times 1 \times 6!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84

So, there are 84 total ways to draw 3 balls from the box.

3. Calculate the number of ways to draw 3 balls such that no black balls are included.

If no black balls are to be included, we must draw the three balls only from the white and red balls available.

The number of non-black balls is $2 \text{ (white)} + 4 \text{ (red)} = 6$ . …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ13 — There are 10 lamps in a hall. Each one of them can be switched on inde…→Next questionQ15 — If ⁿCr-1 = 36, ⁿCr = 84 and ⁿCr+1 = 126, then find rC₂.

Q.A box contains two white, three black and four red balls. In how many ways can three balls be drawn from the box, if atleast one black ball is to be included in the draw.

🎓Unlock everything free for 14 days

Unlock everything free for 14 days