CBSE Class 11 Maths Important Questions - Chapter 6 Permutations and Combinations

1 Mark Questions

1. Evaluate $^5P_3$.

Solution:

Using the formula for permutations:

\[^nP_r = \dfrac{n!}{(n-r)!}\]

\[^5P_3 = \dfrac{5!}{(5-3)!} = \dfrac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = 5\times 4 \times 3 = 60\]

2. What is the value of $0!$?

Solution:

By definition, $0! = 1$.

3.Find the number of arrangements of the word "BOOK".

Solution:

The word "BOOK" has 4 letters, where O is repeated twice.

The formula for arrangements with repetition is:

\[\text{Arrangements} = \dfrac{n!}{p!}\]

Here, $n = 4$, $p = 2$ (repetition of O):

\[\text{Arrangements} = \dfrac{4!}{2!} = \dfrac{4 \times 3 \times 2}{2} = 12\]

4. How many 3-digit numbers can be formed using the digits $1, 2, 3, 4, 5$, if no digit is repeated?

Solution:

The first digit has 5 options, the second has 4, and the third has 3:

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

6. How many ways can 5 people be seated in a row?

Solution:

The number of arrangements of $n$ items is $n!$.

Here, $n = 5$:

\[5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\]

7. How many 3-digit numbers can be formed by using the digits 1 to 9 if no. digit is repeated

9 x 8 x 7= 504

2 Marks Questions

1. Prove that $^nC_r + ^nC_{r-1} = ^{n+1}C_r$.

Solution:

Using the combination formula:

\[^nC_r = \dfrac{n!}{r!(n-r)!}, \quad ^nC_{r-1} = \dfrac{n!}{(r-1)!(n-r+1)!}\]

Adding these:

\[^nC_r + ^nC_{r-1} = \dfrac{n!}{r!(n-r)!} + \dfrac{n!}{(r-1)!(n-r+1)!}\]

Taking LCM:

\[^nC_r + ^nC_{r-1} = \dfrac{n! \cdot (n-r+1) + n! \cdot r}{r!(n-r+1)(n-r)!}\]

Simplify numerator:

\[= \dfrac{n!(n+1)}{r!(n-r+1)!} = ^{n+1}C_r\]

2. Find the number of permutations of the letters in the word "MISSISSIPPI".

Solution:

The total number of letters is 11, with repetitions:

I appears 4 times.

S appears 4 times.

P appears 2 times.

The formula for arrangements with repetition:

\[\text{Arrangements} = \dfrac{n!}{p_1! \times p_2! \times \dots}\]

Substituting:

\[\text{Arrangements} = \dfrac{11!}{4! \times 4! \times 2!} = \dfrac{39916800}{24 \times 24 \times 2} = 34650\]

3. How many ways can 4 boys and 3 girls sit in a row such that all girls sit together?

Solution:

Treat the 3 girls as a single block. Now there are 5 blocks to arrange:

\[\text{Arrangements of 5 blocks: } 5! = 120\]

Within the block of 3 girls:

\[\text{Arrangements: } 3! = 6\]

Total arrangements:

\[5! \times 3! = 120 \times 6 = 720\]

4. How many words can be formed using all the letters of the word "EQUATION" such that all vowels are together?

Solution:

Treat all vowels (E, U, A, I, O) as a single block. This gives 4 blocks (1 vowel block + 3 consonants).

Arrangements of blocks:

\[4! = 24\]

Arrangements of vowels within the block:

\[5! = 120\]

Total arrangements:

\[4! \times 5! = 24 \times 120 = 2880\]

3 Marks Questions

1. A committee of 5 is to be formed from 6 men and 4 women. How many ways can it be formed if at least 3 women are included?

Solution:

Case 1: 3 women and 2 men:

\[\binom{4}{3} \times \binom{6}{2} = 4 \times 15 = 60\]

Case 2: 4 women and 1 man:

\[\binom{4}{4} \times \binom{6}{1} = 1 \times 6 = 6\]

Total ways:

\[60 + 6 = 66\]

2. Prove that the product of $n$ consecutive integers is divisible by $n!$.

Solution:

Let the $n$ consecutive integers be $x, x+1, \dots, x+n-1$. The product:

\[P = x(x+1)(x+2)\dots(x+n-1)\]

Each integer from $1$ to $n$ appears in the factors of $P$, ensuring divisibility by $n!$.

3. How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, and 7, if no digit is repeated?

Ans: Using the numbers, 1, 2, 3, 4, 6 and 7, 3-digit numbers can be formed.

The unit's place can be filled by any of the digits 2, 4 or 6. Hence, there are 3 ways.

Since, it is given that, digits cannot be repeated, units place is already occupied, and the hundreds and tens place can be occupied by the remaining 5 digits.

Thus, the number of ways of filling hundreds and tens places = 

$^5{{\text{P}}_{2{\mkern 1mu} }}{\text{ }} = {\text{ }}{\mkern 1mu} \dfrac{{5!}}{{(5{\mkern 1mu} {\text{ }} - {\text{ }}{\mkern 1mu} 2)!}}{\text{ }} = {\text{ }}\dfrac{{5!}}{{3!}} = {\text{ }}\dfrac{{5 \times 4 \times 3!}}{{3!}}{\text{ }} = {\text{ }}20$

Therefore, by multiplication principle, number of 3-digit numbers = $3 \times 20{\mkern 1mu}= {\mkern 1mu} 60$

4. Evaluate $\dfrac{{{\text{n!}}}}{{{\text{(n}}{\mkern 1mu} {\text{-}}{\mkern 1mu} {\text{r)!}}}},$when 

(i) n = 6, r = 2 (ii) n = 9, r = 5

Ans: $ {\text{(i)}}{\mkern 1mu} {\text{When}}{\mkern 1mu} {\text{n=6,}}{\mkern 1mu} {\text{r=2:}}$

$\dfrac{{{\text{n!}}}}{{{\text{(n}}{\mkern 1mu} {\text{-}}{\mkern 1mu} {\text{r)!}}}}{\mkern 1mu} {\text{=}}{\mkern 1mu} \dfrac{{{\text{6!}}}}{{{\text{(6}}{\mkern 1mu} {\text{-}}{\mkern 1mu} {\text{2)!}}}}{\text{ }}$

${\text{ =}}\dfrac{{{\text{6!}}}}{{{\text{4!}}}}{\text{=}}\dfrac{{{\text{6}} \times {\text{5}} \times {\text{4}}!}}{{{\text{4!}}}}{\text{=30}}$

${\text{(ii)}}{\mkern 1mu} {\text{When}}{\mkern 1mu} {\text{n=9,}}{\mkern 1mu} {\text{r=5:}}$

$\dfrac{{{\text{n!}}}}{{{\text{(n}}{\mkern 1mu} {\text{-}}{\mkern 1mu} {\text{r)!}}}}{\mkern 1mu} {\text{=}}{\mkern 1mu} \dfrac{{{\text{9!}}}}{{{\text{(9}}{\mkern 1mu} {\text{-}}{\mkern 1mu} {\text{5)!}}}}{\mkern 1mu} $

${\text{=}}{\mkern 1mu} \dfrac{{{\text{9!}}}}{{{\text{4!}}}}{\mkern 1mu} {\text{=}}{\mkern 1mu} \dfrac{{{\text{9}} \times {\text{8}} \times {\text{7}} \times {\text{6}} \times {\text{5}} \times {\text{4}}!}}{{{\text{4!}}}}$

${\text{ =}}{\mkern 1mu} {\text{9}} \times {\text{8}} \times {\text{7}} \times {\text{6}} \times {\text{5}}{\mkern 1mu} {\text{=}}{\mkern 1mu} {\text{15120}}$

4 Marks Questions

1. How many numbers greater than 5000 can be formed using the digits $1, 2, 3, 4, 5$ if no digit is repeated?

Solution:

Case 1: Numbers starting with 5:

\[4 \times 3 \times 2 = 24\]

Case 2: Numbers starting with 4:

\[3 \times 3 \times 2 = 18\]

Total:

\[24 + 18 = 42\]

2. In how many ways can 3 boys and 3 girls be arranged in a row such that no two girls sit together?

Solution:

Arrange the boys first:

\[3! = 6\]

Place the girls in the gaps:

\[3! = 6\]

Total:

\[6 \times 6 = 36\]

3. Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?

Ans: From the digits 1,2,3,4 and 5, 4-digit numbers can be formed.

There are permutations of 5 different things taken 4 at a time.

Thus, the number of 4-digit numbers =

$^5{{\text{P}}_4}{\mkern 1mu} {\text{=}}{\mkern 1mu} \dfrac{{{\text{5!}}}}{{{\text{(5}}{\mkern 1mu} {\text{-}}{\mkern 1mu} {\text{4)!}}}}{\mkern 1mu} {\text{ }} = {\text{ }}{\mkern 1mu} \dfrac{{5!}}{{1!}} = 1 \times 2 \times 3 \times 4 \times 5{\mkern 1mu} {\text{ }} = {\text{ }}{\mkern 1mu} 120$

Out of 1, 2, 3, 4, and 5, we know that even numbers end either by 2 or 4.

Thus, the ways in which unit places can be filled is 2.

Since, repetition is not allowed, units place is already occupied by a digit and remaining vacant places can be filled by remaining 4 digits.

Thus, the number of ways in which remaining places can be filled = 

$^{\text{4}}{{\text{P}}_{\text{3}}}{\mkern 1mu} {\text{ }} = {\text{ }}{\mkern 1mu} \dfrac{{4!}}{{(4{\mkern 1mu} {\text{ }} - {\text{ }}{\mkern 1mu} 3)!}}{\mkern 1mu} {\text{ }} = {\text{ }}{\mkern 1mu} \dfrac{{4!}}{{1!}} = 4 \times 3 \times 2 \times 1{\mkern 1mu} {\text{ }} = {\text{ }}{\mkern 1mu} 24$

Therefore, by multiplication principle, number of even numbers = $24 \times 2{\mkern 1mu}= {\mkern 1mu} 48$

4. Find r if 

${\text{ (i)}}{{\text{ }}^5}{P_r} = {2^6}{P_{r - 1}}$

Ans: $ { \Rightarrow \dfrac{{5!}}{{(5 - r)!}} = 2 \times \dfrac{{6!}}{{(6 - r + 1)!}}} $

$ { \Rightarrow \dfrac{{5!}}{{(5 - r)!}} = \dfrac{{2 \times 6!}}{{(7 - r)!}}} $

${ \Rightarrow \dfrac{{5!}}{{(5 - r)!}} = \dfrac{{6 \times 5!}}{{(7 - r)(6 - r)(5 - r)!}}}$

${ \Rightarrow 1 = \dfrac{{2 \times 6}}{{(7 - r)(6 - r)}}}$

$ \Rightarrow (7 - r)(6 - r) = 12$

$ \Rightarrow 42 - 6r - 7r + r = 12$

$ \Rightarrow {r^2} - 13r + 30 = 0$

$ \Rightarrow {r^2} - 3r + 10r + 30 = 0$

$ \Rightarrow r(r - 3) - 10(r - 3) = 0$

$ \Rightarrow (r - 3)(r - 10) = 0$

$ \Rightarrow r = 3 = 0$ or $(r - 10) = 0$

$ \Rightarrow r = 3$ or $r = 10$

It is known that, $^n{P_r} = \dfrac{{n!}}{{(n - r)!}}$, where $0 \leqslant r \leqslant n$

${\therefore 0 \leqslant r \leqslant 5}$

$ {\therefore r = 3}$

${\text{ (ii)}}{{\text{ }}^5}{P_r}{ = ^6}{P_{r - 1}}$

${\text{Ans:}}$

${ \Rightarrow \dfrac{{5!}}{{(5 - r)!}} = \dfrac{{6!}}{{(6 - r + 1)!}}}$

${ \Rightarrow \dfrac{{5!}}{{(5 - r)!}} = \dfrac{{6 \times 5!}}{{(7 - r)!}}}$

${ \Rightarrow \dfrac{{5!}}{{(5 - r)!}} = \dfrac{6}{{(7 - r)(6 - r)(5 - r)!}}}$

${ \Rightarrow 1 = \dfrac{6}{{(7 - r)(6 - r)}}}$

${ \Rightarrow (7 - r)(6 - r) = 6}$

${ \Rightarrow 42 - 7r - 6r + {r^2} - 6 = 0}$

${ \Rightarrow {r^2} - 13r + 36 = 0}$

 $ { \Rightarrow {r^2} - 4r - 9r(r - 4) = 0}$

 $ { \Rightarrow (r - 4)(r - 9) = 0}$

 $ { \Rightarrow (r - 4)(r - 9) = 0}$

${ \Rightarrow (r - 4) = 0{\text{ or }}(r - 9) = 0}$

 $ { \Rightarrow r = 4{\text{ or }}r = 9}$

It is known that, $^n{P_r} = \dfrac{{n!}}{{(n - !)}}$, where $0 \leqslant r \leqslant n$

$\therefore 0 \leqslant r \leqslant n$

Hence, $r \ne 9$

$\therefore r = 4$

5. Determine n if

(i) $^{2n}{{C}_{3}}:{{\text{ }}^{n}}{{C}_{3}}=12:1$

(ii) $^{2n}{{C}_{3}}:{{\text{ }}^{n}}{{C}_{3}}=11:1$

Ans: $\frac{^{2n}{{C}_{3}}}{^{n}{{C}_{3}}}=\frac{12}{1}$

$\Rightarrow \dfrac{{{\text{(2n)!}}}}{{{\text{(2n}}\,{\text{ - }}\,{\text{3)!3!}}}}\,\times \dfrac{{{\text{3!(n - }}\,{\text{3)!}}}}{{{\text{n!}}}}\,{\text{ = }}\,\dfrac{{{\text{12}}}}{{\text{1}}}$

 $\Rightarrow \dfrac{{{\text{(2n)(2n}}\,{\text{ - }}\,{\text{1)(2n}}\,{\text{ - }}\,{\text{2)(2n}}\,{\text{ - }}\,{\text{3)!}}}}{{{\text{(2n}}\,{\text{ - }}\,{\text{3)!}}}}\,\times \,\dfrac{{{\text{(n}}\,{\text{ - }}\,{\text{3)!}}}}{{{\text{n(n}}\,{\text{ - }}\,{\text{1)(n}}\,{\text{ - }}\,{\text{2)(n}}\,{\text{ - }}\,{\text{3)!}}}}\,{\text{ = }}\,{\text{12}}$

$ \Rightarrow \dfrac{{{\text{2(2n}}\,{\text{ - }}\,{\text{1)(2n}}\,{\text{ - }}\,{\text{2)}}}}{{{\text{(n}}\,{\text{ - }}\,{\text{1)(n}}\,{\text{ - }}\,{\text{2)}}}}{\text{ = 12}}$

$ \Rightarrow \dfrac{{{\text{4(2n}}\,{\text{ - }}\,{\text{1)(n}}\,{\text{ - }}\,{\text{1)}}}}{{{\text{(n}}\,{\text{ - }}\,{\text{1)(n}}\,{\text{ - }}\,{\text{2)}}}}{\text{ = 12}}$

$ \Rightarrow \dfrac{{{\text{(2n}}\,{\text{ - }}\,{\text{1)}}}}{{{\text{(n}}\,{\text{ - }}\,{\text{2)}}}}{\text{ = 3}}$

$ \Rightarrow {\text{2n}}\,{\text{ - }}\,{\text{1}}\,{\text{ = }}\,{\text{3(n}}\,{\text{ - }}\,{\text{2)}}$

$ \Rightarrow {\text{2n}}\,{\text{ - }}\,{\text{1}}\,{\text{ = }}\,{\text{3n}}\,{\text{ - }}\,{\text{6}}$

$ \Rightarrow {\text{3n}}\,{\text{ - }}\,{\text{2n}}\,{\text{ = }}\,{\text{ - 1}}\,{\text{ + 6}}$

$ \Rightarrow {\text{n}}\,{\text{ = }}\,{\text{5}}$ 

${\text{(ii) }}\frac{^{2n}{{C}_{3}}}{^{n}{{C}_{3}}}=\frac{11}{1}$

 $\Rightarrow \dfrac{{{\text{(2n)!}}}}{{{\text{3!(2n}}\,{\text{ - }}\,{\text{3)!}}}} \times \dfrac{{{\text{3!(n}}\,{\text{ - }}\,{\text{3)!}}}}{{{\text{n!}}}}\,{\text{ = }}\,{\text{11}}$

$ \Rightarrow \dfrac{{{\text{(2n)(2n}}\,{\text{ - }}\,{\text{1)(2n}}\,{\text{ - }}\,{\text{2)(2n}}\,{\text{ - }}\,{\text{3)!}}}}{{{\text{(2n}}\,{\text{ - }}\,{\text{3)!}}}}\, \times \,\dfrac{{{\text{(n}}\,{\text{ - }}\,{\text{3)!}}}}{{{\text{n(n}}\,{\text{ - }}\,{\text{1)(n}}\,{\text{ - }}\,{\text{2)(n}}\,{\text{ - }}\,{\text{3)!}}}}$

 $\Rightarrow \dfrac{{{\text{2(2n}}\,{\text{ - }}\,{\text{1)(2n}}\,{\text{ - }}\,{\text{2)}}}}{{{\text{(n}}\,{\text{ - }}\,{\text{1)}}\left( {{\text{n}}\,{\text{ - }}\,{\text{2}}} \right)}}{\text{ = 1}}$

 $\Rightarrow \dfrac{{{\text{4(2n}}\,{\text{ - }}\,{\text{1)}}}}{{{\text{(n}}\,{\text{ - }}\,{\text{2)}}}}{\text{ = 11}}$

 $\Rightarrow {\text{4(2n}}\,{\text{ - }}\,{\text{1)}}\,{\text{ = }}\,{\text{11(n}}\,{\text{ - }}\,{\text{2)}}$

$ \Rightarrow {\text{8n}}\,{\text{ - }}\,{\text{4}}\,{\text{ = }}\,{\text{11n}}\,{\text{ - }}\,{\text{22}}$

$ \Rightarrow {\text{11n}}\,{\text{ - }}\,{\text{8n}}\,{\text{ = }}\,{\text{22}}\,{\text{ - }}\,{\text{4}}$

$ \Rightarrow {\text{3n}}\,{\text{ = }}\,{\text{18}}$

 $\Rightarrow {\text{n}}\,{\text{ = }}\,{\text{6}}$ 

Some More Important Questions from Permutations and Combinations:

1. How many ways can a committee of 3 be chosen from a group of 8 people?

Answer: The number of ways to select a committee of 3 from 8 individuals is denoted as $\binom{8}{3}$ or $C(8,3)$, calculated as $\dfrac{8!}{3! \cdot (8-3)!}$.

2.In how many ways can the letters of the word "APPLE" be arranged? 

Answer: The number of arrangements for the word "APPLE" is determined by the factorial of the number of distinct letters, which is $\dfrac{5!}{2! \cdot 2!}$ due to repeated letters.

3.How many 4-digit numbers can be formed using the digits 1, 2, 3, and 4 without repetition? 

Answer: There are 4 options for the first digit, 3 for the second, 2 for the third, and 1 for the fourth. So, the total number is $4 \times 3 \times 2 \times 1 = 24$.

4.In a deck of cards, how many ways can you pick 2 cards from a set of 52? 

Answer: This involves combinations, and the formula is $\binom{52}{2} = \dfrac{52!}{2! \cdot (52-2)!}$, resulting in $26 \times 51 = 1326$ ways.

5.How many different arrangements are possible for the letters of the word "MATHEMATICS"? 

Answer: The total arrangements for the word "MATHEMATICS" can be calculated as $\dfrac{11!}{2! \cdot 2! \cdot 2!}$, considering the repeated letters, resulting in a specific number of unique arrangements.

Tips to Solve Permutations and Combinations Chapter:

Here are seven tips to tackle Permutations and Combinations problems:

1. Clearly Define the Problem: Begin by precisely understanding what the problem is asking. Clearly define the elements you're working with and the specific arrangement or selection required.

2. Distinguish Between Permutations and Combinations: Differentiate between situations that involve order (permutations) and those that don't (combinations). This clarity will guide your approach to problem-solving.

3. Utilize Factorials Wisely: Factorials (\(n!\)) play a crucial role. Understand when to use them, especially in scenarios with repeated elements, and simplify expressions to avoid unnecessary complexity.

4. Be Systematic in Counting: Develop a systematic counting approach, especially in complex problems. Break down the problem into stages, making it easier to manage and reduce the chance of overlooking possibilities.

5. Utilize Combination Formulas: Familiarize yourself with combination formulas, such as \(\binom{n}{r} = \dfrac{n!}{r! \cdot (n-r)!}\). Knowing when to apply these formulas will streamline your calculations.

6. Practice with Real-Life Examples: Relate problems to real-life scenarios whenever possible. This aids in better conceptual understanding and helps in visualizing the practical application of permutations and combinations.

7. Stay Organized and Neat: Keep your work organized and neat. Label variables clearly, use systematic notations, and double-check your calculations. A tidy approach can significantly reduce errors.

By applying these tips, you can enhance your problem-solving skills in Permutations and Combinations, making the chapter more manageable and enjoyable.

Important Formulas from Permutations and Combinations

Here are some important formulas from the Permutations and Combinations chapter for Class 11 students:

1. Permutations of n distinct objects: The number of ways to arrange n distinct objects is given by \(n!\) (n factorial).

2. Permutations of n objects taken r at a time: The number of ways to arrange n objects taken r at a time is given by \(P(n, r) = \dfrac{n!}{(n-r)!}\).

3. Combinations of n objects taken r at a time: The number of ways to choose r objects from n distinct objects without considering the order is given by \(\binom{n}{r} = \dfrac{n!}{r! \cdot (n-r)!}\).

4. Permutations with indistinguishable objects: For a group of n objects with some indistinguishable, the number of permutations is given by \(\dfrac{N!}{n_1! \cdot n_2! \cdot \ldots \cdot n_k!}\), where \(n_1, n_2, \ldots, n_k\) are the counts of indistinguishable objects.

5. Fundamental Counting Principle: If there are \(n_1\) ways to do the first task, \(n_2\) ways to do the second task, and so on, then the total number of ways to perform both tasks is \(n_1 \times n_2 \times \ldots\).

Benefits of Referring to Important Questions for Class 11 Maths Chapter 6 Permutations and Combinations

1. Comprehensive Coverage: The PDF includes a wide variety of important questions that cover all key concepts and types of problems in the chapter, ensuring thorough preparation.

2. Exam-Oriented Focus: The questions are carefully curated based on previous years’ exam trends and commonly tested topics, helping students focus on what matters most.

3. Step-by-Step Solutions: Detailed solutions provided for every question make it easier to understand problem-solving techniques and boost confidence in tackling similar problems.

4. Time Management: Practising these important questions improves problem-solving speed and accuracy, which is crucial for performing well in exams.

5. Concept Clarity: By working through these questions, students can strengthen their grasp of fundamental concepts and application-based problems.

6. Convenient Format: The downloadable PDF format makes it easy to study anywhere, anytime, without requiring an internet connection.

7. Aligned with Curriculum: The questions are designed to align perfectly with the NCERT syllabus, ensuring relevance to class and board exams.

Conclusion

Reviewing essential questions in Class 11 Chapter 6 - Permutations and Combinations ensures a complete understanding. The extra and important questions engage in a concept-focused discussion, covering all chapter topics. This method proves time-efficient during exam preparation, providing an effective way to revise and enhance understanding. Practice with these important questions streamlines preparation, building confidence for upcoming exams.

Important Study Materials for Class 11 Maths Chapter 6 Permutations and Combinations

CBSE Class 11 Maths Chapter-wise Important Questions

CBSE Class 11 Maths Chapter-wise Important Questions and Answers cover topics from all 14 chapters, helping students prepare thoroughly by focusing on key topics for easier revision.

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