One urn contains two black balls (labelled $B_1$ and $B_2$) and one white ball. A second urn contains one black ball and two white balls (labelled $W_1$ and $W_2$). Suppose the following experiment is performed. One of the two urns is chosen at random. Next a ball is randomly chosen from the urn. Then a second ball is chosen at random from the same urn without replacing the first ball. (a) Write the sample space showing all possible outcomes. (b) What is the probability that two black balls are chosen? (c) What is the probability that two balls of opposite colour are chosen?

One urn contains two black balls (labelled B₁ and B₂) and one white ball. A second urn contains one black ball

NCERT Exemplar · Q12

B_1

(a) Write the sample space showing all possible outcomes.

(b) What is the probability that two black balls are chosen?

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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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The sample space has $12$ equally likely outcomes; $P(\text{two black})=\dfrac{1}{6}$ and $P(\text{opposite colours})=\dfrac{2}{3}$ .

Setting up

An urn is chosen with probability $\tfrac12$ , then two balls are drawn in order without replacement. Each urn holds $3$ balls, giving $3\times2=6$ ordered draws per urn, so $12$ outcomes in all, each with probability $\tfrac12\cdot\tfrac16=\tfrac{1}{12}$ .

(a) Sample space

Urn 1 $\{B_1,B_2,W\}$ :

(B_1,B_2),\ (B_1,W),\ (B_2,B_1),\ (B_2,W),\ (W,B_1),\ (W,B_2).

Urn 2 $\{B,W_1,W_2\}$ :

(B,W_1),\ (B,W_2),\ (W_1,B),\ (W_1,W_2),\ (W_2,B),\ (W_2,W_1).

Together these are the $12$ equally likely outcomes of $S$ .

(b) Two black balls

Only Urn 1 has two black balls. The favourable outcomes are $(B_1,B_2)$ and $(B_2,B_1)$ — that is $2$ outcomes:

$P(\text{two black})=\frac{2}{12}=\frac{1}{6}.$ …

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

Showing the 12 most recent of 18 on this concept.

KEAM 2026Set eng-2026-04174 marksMCQ
$S$
›Reveal solutionSolution

By symmetry each place-value column sums to $2\times(3+5+7)=30$ , giving $30\times111=3330$ . …

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KEAM 2026Set eng-2026-04174 marksMCQ
$Q. There are 3 boys and 4 girls in a group. The number of ways they can sit in a row so that between any two boys there is a girl and between any two girls there is a boy, is (A) 88 (B) 96 (C) 124 (D) 144 (E) 288$
›Reveal solutionSolution

The seating must strictly alternate; with 4 girls and 3 boys the pattern is fixed as $GBGBGBG$ , giving $4!\,3!=144$ .

The conditions "between any two boys there is a girl" and "between any two girls there is a boy" together force a strictly alternating seating. With $3$ boys and $4$ girls in $7$ seats, the only alternating arrangement is

$G\,B\,G\,B\,G\,B\,G,$ …
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KEAM 2026Set eng-2026-04174 marksMCQ
$Q. The number of arrangements of the letters of the word BANANA so that the arrangement starts and ends with the same letter, is (A) 12 (B) 16 (C) 18 (D) 20 (E) 24$
›Reveal solutionSolution
Only A or N can occupy both ends; counting each case gives $12+4=16$ .

BANANA has letters $A(\times3), N(\times2), B(\times1)$ . For an arrangement to start and end with the same letter, that letter must appear at least twice, so it is $A$ or $N$ .

Both ends $A$ : two $A$ 's used at the ends; the middle four positions hold $\{A,N,N,B\}$ , arrangeable in $\dfrac{4!}{2!}=12$ ways. …
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KEAM 2026Set eng-2026-04204 marksMCQ
$n_{P_5} = 6720$
›Reveal solutionSolution

$^nP_5=\frac{n!}{(n-5)!}$ and $^{n-1}P_4=\frac{(n-1)!}{(n-5)!}$ , so their ratio is $n$ . Thus $n=6720/840=8$ , giving $^nP_3=8\cdot7\cdot6=336$ .

Use the permutation formulas:
$^nP_5=\frac{n!}{(n-5)!},\qquad {}^{n-1}P_4=\frac{(n-1)!}{(n-5)!}.$
Dividing: …
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KEAM 2025Set eng-2025-04234 marksMCQ
${}^{11}P_r = 7920$
›Reveal solutionSolution
11109*8 = 7920, so r = 4.

Concept and Intuition

The permutation P(n,r) is a product of r descending factors starting at n. Matching 7920 to such a descending product identifies r.

Step-by-Step Solution

P(11, r) = 1110...*(11-r+1).

1110 = 110; 1109 = 990; 990*8 = 7920.

That used four factors, so r = 4. …
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KEAM 2025Set eng-2025-04234 marksMCQ
$A = \{0, 2, 4, 6, 8\}$
›Reveal solutionSolution
From 5! total arrangements we remove the 4! that start with 0, leaving 96.

Concept and Intuition

Using all five distinct digits gives 5! orderings, but a valid 5-digit number cannot begin with 0. Subtract the arrangements with 0 in the leading position.

Step-by-Step Solution

Total arrangements of {0,2,4,6,8} = 5! = 120.

Arrangements with 0 leading = 4! = 24. …
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KEAM 2025Set eng-2025-04264 marksMCQ
$\frac{25!}{(5!)^5}$
›Reveal solutionSolution

Groups are of equal size and unlabelled, so divide by $(5!)^{5}$ for internal order and by $5!$ for the interchange of the identical groups: $\frac{25!}{(5!)^{6}}$ . …

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KEAM 2025Set eng-2025-04274 marksMCQ
$Q. The number of integers greater than 7000 using 2,4,6,7,8 without repetition, is (A) 168 (B) 336 (C) 196 (D) 256 (E) 512$
›Reveal solutionSolution

Count 4-digit numbers starting with $7$ or $8$ ( $48$ ) plus all 5-digit arrangements ( $120$ ): total $168$ .

The available digits are $2,4,6,7,8$ (all distinct, no repetition). A number using these digits and greater than $7000$ is either a 4-digit or a 5-digit number.

4-digit numbers $>7000$ : the thousands digit must be $\ge7$ , so it is $7$ or $8$ ( $2$ choices). The remaining three places are filled from the other $4$ digits: …
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KEAM 2025Set eng-2025-04274 marksMCQ
$Q. Five digit number is formed using the digits 0,1,2,3,4 and 5 without repetitions. Number of five digit numbers which are divisible by 10 is (A) 360 (B) 240 (C) 120 (D) 480 (E) 520$
›Reveal solutionSolution

Fix the last digit as $0$ ; the other four positions use the digits $1$ – $5$ : $P(5,4)=120$ .

A 5-digit number formed from $\{0,1,2,3,4,5\}$ (no repetition) is divisible by $10$ iff its units digit is $0$ .

Fixing the units digit as $0$ , the remaining four positions are filled by choosing and arranging $4$ of the remaining $5$ digits $\{1,2,3,4,5\}$ : …
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KEAM 2025Set eng-2025-04284 marksMCQ
$n(A) = 8$
›Reveal solutionSolution

Subsets with exactly $2$ elements: $\binom{8}{2}=28$ ; with exactly $6$ : $\binom{8}{6}=28$ . Total $=56$ . …

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KEAM 2025Set eng-2025-04284 marksMCQ
$Q. Four digit numbers are formed using 0, 3, 4, 5, 9, 8 without repetitions. Then the number of such 4 digits numbers is (A) 270 (B) 300 (C) 320 (D) 400 (E) 450$
›Reveal solutionSolution

$6$ available digits $\{0,3,4,5,9,8\}$ , no repetition. Leading digit can't be $0$ ( $5$ ways); fill remaining $3$ places from the other $5$ digits ( $5\cdot4\cdot3$ ). Total $=300$ .

For a $4$ -digit number the first (thousands) place cannot be $0$ , so it has $5$ choices. …
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KEAM 2025Set eng-2025-04284 marksMCQ
$Q. A bag contains 5 red balls, 4 black balls, and 3 white balls. Then the number of ways of selecting three balls at random that contains at least one white ball is (A) 220 (B) 210 (C) 180 (D) 136 (E) 74$
›Reveal solutionSolution

At least one white $=$ (all selections) $-$ (no white) $=\binom{12}{3}-\binom{9}{3}=220-84=136$ .

There are $5+4+3=12$ balls. Choosing $3$ in all: $\binom{12}{3}=220$ . …
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Setting up

(a) Sample space

(b) Two black balls

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