A bag contains six white marbles and five red marbles. Find the number of ways in which four marbles can be dr

NCERT Exemplar · Q22

Q. A bag contains six white marbles and five red marbles. Find the number of ways in which four marbles can be drawn from the bag if

(a) they can be of any colour

(b) two must be white and two red and

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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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This is a combinations (selection without order) problem. For (a) total ways = $\binom{11}{4} = 330$ ;

(b) ways with exactly 2 white and 2 red = $\binom{6}{2} \times \binom{5}{2} = 150$ ;

The key idea here is that we are selecting marbles from a bag — the order in which we draw them does not matter. We only care about which marbles end up in our hand. That is the classic combinations problem, also called "permutations without repetition" when order is ignored.

When you have a total of $n$ distinct objects and you want to choose $r$ of them, the number of ways is given by the binomial coefficient:

\binom{n}{r} = \frac{n!}{r!(n-r)!}

This formula counts every distinct group of $r$ objects exactly once. No repetition is allowed (you cannot draw the same marble twice), and the order of drawing is irrelevant.

Let’s apply this to each part.

Part (a): Any colour — total selections

The bag has $6$ white + $5$ red = $11$ distinct marbles.
We want to choose any $4$ marbles from these $11$ , with no restrictions.
The number of ways is simply $\binom{11}{4}$ .

Compute it:

\binom{11}{4} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = \frac{7920}{24} = 330

So there are 330 ways to draw any four marbles.

Tip

A quick check: $\binom{11}{4} = \binom{11}{7}$ — choosing 4 to keep is the same as choosing 7 to leave behind. This symmetry can simplify calculations sometimes.

Part (b): Exactly two white and two red

We need to pick 2 white marbles from the 6 available, and 2 red marbles from the 5 available.
These two selections are independent — the choice of whites does not affect the choice of reds.
So we multiply the number of ways for each colour:

\text{Ways} = \binom{6}{2} \times \binom{5}{2}

Compute each:

\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15

\binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10

Multiply:

15 \times 10 = 150

Thus there are 150 ways to draw exactly two white and two red marbles.

Watch out

A common mistake is to treat this as $\binom{11}{4}$ and then try to subtract something. That won't work because "exactly two of each" is a specific composition, not a simple complement. Always break it into independent colour choices.

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2026Set ANNUAL1 markMCQ
$Q. The number of permutation of the letters of the world MANIPUR is$
(a) 5040
(b) 4050
(c) 4540
(d) 5450
›Reveal solutionSolution

MANIPUR has 7 distinct letters, so it can be arranged in 7! = 5040 ways.

The word MANIPUR is spelled M-A-N-I-P-U-R — 7 letters, and checking each one, none repeats.

…

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Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2026Set ANNUAL1 markMCQ
$\dfrac{5!}{2!}$
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A)
(b) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A)
(c) Assertion (A) is true but Reason (R) is false
(d) Assertion (A) is false, but Reason (R) is true
›Reveal solutionSolution

Both statements are individually true, but the Reason explains a different scenario (repetition allowed in selection), not the Assertion's scenario (arranging letters with a repeated letter).

Checking the Assertion: APPLE has 5 letters — A, P, P, L, E — with the letter P repeated twice. The number of distinct arrangements of n objects where one object repeats r times is $\dfrac{n!}{r!}$ . Here that's $\dfrac{5!}{2!}$ , so the Assertion is TRUE.

Checking the Reason: The number of permutations of n different objects taken r at a time WITH repetition allowed is indeed $n^r$ — this is a correct, standard formula. So the Reason is also TRUE.

…

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Q.A bag contains six white marbles and five red marbles. Find the number of ways in which four marbles can be drawn from the bag if

Part (a): Any colour — total selections

Part (b): Exactly two white and two red

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