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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Learn more about permutations, combinations, and counting principles in probability and discrete mathematics.
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.
