Permutations And Combinations Exercise 6.1 NCERT Solutions Class 11 Math

Anand Classes provides complete NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations Exercise 6.1, designed to help students understand the fundamental counting principles and arrangement methods in mathematics. These step-by-step solutions are prepared as per the latest CBSE and NCERT syllabus, offering clear explanations and solved examples for each question. Students can use these solutions to strengthen their problem-solving skills and prepare effectively for board exams. Click the print button to download study material and notes.

NCERT Question.1 : How many 3-digit numbers can be formed from the digits 1, 2, 3, 4, and 5 assuming that
(i) Repetition of the digits is allowed
(ii) Repetition of the digits is not allowed

Solution:

(i) Repetition of the digits is allowed

Total number of digits = 5

Let the 3-digit number be $XYZ$.

Each of the three places can be filled in 5 ways (since repetition is allowed).

$$
\text{Total numbers} = 5 \times 5 \times 5 = 125
$$

Hence, $\boxed{125}$

(ii) Repetition of the digits is not allowed

Let the 3-digit number be $XYZ$.

• Number of choices for $X = 5$
• For $Y$, 4 digits remain
• For $Z$, 3 digits remain

$$
\text{Total numbers} = 5 \times 4 \times 3 = 60
$$

Hence, $\boxed{60}$

NCERT Question.2 : How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if digits can be repeated?

Solution:

Let the 3-digit number be $XYZ$.

Since the number is even, $Z$ (the unit digit) can be $2, 4,$ or $6$.
So, choices for $Z = 3$

Repetition is allowed, so each of $X$ and $Y$ can be filled in 6 ways.

$$
\text{Total numbers} = 6 \times 6 \times 3 = 108
$$

Hence, $\boxed{108}$

NCERT Question.3 : How many 4-letter codes can be formed using the first 10 letters of the English alphabet if no letter is repeated?

Solution:

Let the 4-letter code be $ABCD$.

Choices for each position:

• $A = 10$, $B = 9$, $C = 8$, $D = 7$

$$
\text{Total codes} = 10 \times 9 \times 8 \times 7 = 5040
$$

Hence, $\boxed{5040}$

NCERT Question.4 : How many 5-digit telephone numbers can be formed using the digits 0–9 if each number starts with 67 and no digit is repeated?

Solution:

Let the 5-digit number be $ABCDE$.

Since the number starts with 67,
$A = 6$ and $B = 7$ (fixed).

Remaining digits available for $C, D, E = 8$ (as 2 digits already used).

Choices:

• $C = 8$, $D = 7$, $E = 6$

$$
\text{Total numbers} = 1 \times 1 \times 8 \times 7 \times 6 = 336
$$

Hence, $\boxed{336}$

NCERT Question.5 : A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?

Solution:

For each toss, there are 2 possible outcomes — Head (H) or Tail (T).

Since the coin is tossed 3 times:

$$
\text{Total outcomes} = 2 \times 2 \times 2 = 8
$$

Hence, $\boxed{8}$

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NCERT Question.5 : Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?

Solution:

Total number of flags = 5

Each signal consists of 2 flags arranged one below the other, so the order matters.
Since the same flag cannot be repeated, repetition is not allowed.

• Number of choices for the upper flag = 5
• Number of choices for the lower flag = 4

$$
\text{Total signals} = 5 \times 4 = 20
$$

Hence, the total number of different signals that can be generated is
$$\boxed{20}$$

For more learning on permutations and arrangements, explore detailed lessons and NCERT solutions by Anand Classes for Class 11 Maths, JEE, NDA, and CUET preparation.