There will be only $24$ selections containing at least one red ball out of a bag containing $4$ red and $5$ black balls. It is being given that the balls of the same colour are identical.

There will be only 24 selections containing at least one red ball out of a bag containing 4 red and 5 black ba

NCERT Exemplar · Q56

24

Uttar Pradesh UpmspShort· 1mImportance★★★★★est

94% · 122/130 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 →

When selecting from identical items, the number of ways to choose from $n$ items is $n+1$ . For "at least one red ball", we subtract selections with zero red balls from the total. The calculation yields $24$ selections, so the given statement is True.

When dealing with selections from a group of identical items, the problem simplifies significantly. Instead of worrying about which specific item is chosen, we only care about how many items of that type are chosen. This is a fundamental concept in combinatorics, especially when items within a category are indistinguishable.

Consider a set of $n$ identical items. You can choose:

$0$ items
$1$ item
$2$ items
...
$n$ items

This gives a total of $n+1$ distinct ways to make a selection from these $n$ identical items.

For problems involving multiple types of identical items, the total number of selections is found by multiplying the number of ways to select items of each type.

The condition "at least one red ball" is a classic indicator for using the complementary counting principle. It's often easier to calculate the total number of possibilities and then subtract the "unwanted" possibilities (in this case, selections with zero red balls).

Let's apply these ideas to the problem.

Determine the number of ways to select red balls.

We have $4$ identical red balls.

The number of ways to select red balls is $4+1 = 5$ ways. These correspond to selecting $0, 1, 2, 3,$ or $4$ red balls.
Determine the number of ways to select black balls.

We have $5$ identical black balls.

The number of ways to select black balls is $5+1 = 6$ ways. These correspond to selecting $0, 1, 2, 3, 4,$ or $5$ black balls.
Calculate the total number of possible selections.

To find the total number of distinct selections from the bag, we multiply the number of ways to select red balls by the number of ways to select black balls. This includes the case where we select zero red balls and zero black balls (an empty selection).

Total selections = (Ways to select red balls) $\times$ (Ways to select black balls)

Total selections = $5 \times 6 = 30$ .
Identify selections that contain zero red balls.

The condition is "at least one red ball". This means we need to exclude all selections where no red balls are chosen.

If we select zero red balls, there is only $1$ way to do this (choose none).

For the black balls, we can still select any number from $0$ to $5$ , which is $5+1 = 6$ ways. …

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

←Previous questionQ55 — If some or all of n objects are taken at a time, the number of combina…→Next questionQ57 — Eighteen guests are to be seated, half on each side of a long table. F…

Q.There will be only 242424 selections containing at least one red ball out of a bag containing 444 red and 555 black balls. It is being given that the balls of the same colour are identical.

🎓Unlock everything free for 14 days

Q.There will be only $24$ selections containing at least one red ball out of a bag containing $4$ red and $5$ black balls. It is being given that the balls of the same colour are identical.

Unlock everything free for 14 days