A box contains $2$ white balls, $3$ black balls and $4$ red balls. The number of ways three balls be drawn from the box if at least one black ball is to be included in the draw is ______.

A box contains 2 white balls, 3 black balls and 4 red balls. The number of ways three balls be drawn from the

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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 one black ball, we use complementary counting: subtract the number of ways to draw no black balls from the total number of ways to draw three balls. The result is $\boxed{64}$ .

When a problem asks for "at least one" of a certain item, it means we could have one, two, three, or more of that item, up to the maximum possible. Calculating each of these individual cases and summing them up can be tedious. A more efficient approach is often to use the principle of complementary counting.

The idea behind complementary counting is simple:

Total number of ways = (Number of ways that satisfy a condition) + (Number of ways that do not satisfy the condition).

Rearranging this, we get:

Number of ways that satisfy a condition = Total number of ways - Number of ways that do not satisfy the condition.

In our problem, the condition is "at least one black ball". The opposite condition (the complement) is "no black balls". So, we can find the answer by:

(Ways with at least one black ball) = (Total ways to draw 3 balls) - (Ways to draw 3 balls with no black balls).

Let's break this down step-by-step.

Identify the total number of balls of each type.
- White balls ( $W$ ): $2$
- Black balls ( $B$ ): $3$
- Red balls ( $R$ ): $4$
- The total number of balls in the box is $2 + 3 + 4 = 9$ .
Calculate the total number of ways to draw three balls from the box without any restrictions.

We are selecting $3$ balls from a total of $9$ distinct balls. The order of drawing does not matter, so this is a combination problem.

The number of ways is given by $\binom{n}{k}$ , where $n$ is the total number of items and $k$ is the number of items to choose.

\text{Total ways} = \binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84

So, there are $84$ ways to draw three balls from the box.

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

If we draw no black balls, it means we must select all three balls from the white and red balls only.

The number of non-black balls is $2$ (white) $+ 4$ (red) $= 6$ balls.

We need to choose $3$ balls from these $6$ non-black balls.

\text{Ways with no black balls} = \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 5 \times 4 = 20

There are $20$ ways to draw three balls such that none of them are black.

4. Apply the complementary counting principle.

The number of ways to draw three balls with at least one black ball is the total number of ways minus the number of ways with no black balls. …

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Q.A box contains 222 white balls, 333 black balls and 444 red balls. The number of ways three balls be drawn from the box if at least one black ball is to be included in the draw is ______.

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Q.A box contains $2$ white balls, $3$ black balls and $4$ red balls. The number of ways three balls be drawn from the box if at least one black ball is to be included in the draw is ______.

Unlock everything free for 14 days