Class 11 Mathematics Chapter 6 Permutations And Combinations

Key Concept: Basic Counting

c) 12

[Solution Description] According to the Fundamental Principle of Counting, if there are $m$ ways to do one thing and $n$ ways to do another, then there are $m \times n$ ways of doing both. Here, $m = 3$ appetizers and $n = 4$ main courses. So, the total number of meal combinations is $3 \times 4 = 12$.

Key Concept: Permutations with Conditions

c) 336

[Solution Description] The problem involves selecting and arranging 3 students out of 8, which can be solved using the permutation formula $nPr = n(n-1)(n-2)…(n-r+1)$. Here, $n = 8$ and $r = 3$.

$8P3 = 8 \times 7 \times 6 = 336$

Thus, there are 336 possible ways to arrange the selected students.

Key Concept: Formula Application

c) 5040

[Solution Description]

To solve this problem, we use the permutation formula $nPr = \frac{n!}{(n-r)!}$. Here, $n = 10$ and $r = 4$.

\[
\begin{aligned}
10P4 &= \frac{10!}{(10-4)!} \\
&= \frac{10 \times 9 \times 8 \times 7 \times 6!}{6!} \\
&= 10 \times 9 \times 8 \times 7 \\
&= 5040
\end{aligned}
\]

Therefore, there are 5040 different ways.

Key Concept: Basic Definition

c) 10

[Solution Description]

The formula for combinations is given by:

$nCr = \frac{n!}{r!(n-r)!}$

Here, $n = 5$ and $r = 3$. Substituting these values into the formula gives:

$5C3 = \frac{5!}{3!(5-3)!}$

Simplifying further:

$= \frac{5 \times 4 \times 3!}{3! \times 2 \times 1}$

Canceling $3!$ in both the numerator and denominator, we have:

$= \frac{5 \times 4}{2 \times 1}$

Calculating this gives:

$= \frac{20}{2} = 10$

Therefore, there are 10 combinations.

Key Concept: Mixed Permutation

d) 125

[Solution Description] The word “STUDY” has 5 different letters. If repetition is allowed, then each of the 3 positions in the word can be filled by any of these 5 letters. Thus, the number of possible 3-letter words that can be formed is given by $5^3$.

Calculating:

$5^3 = 5 \times 5 \times 5 = 125$

Therefore, 125 different 3-letter words can be formed.

Key Concept: Counting Pairs

c) 12

[Solution Description]

Using the fundamental principle of counting, if there are $m = 4$ ways to choose a starter and $n = 3$ ways to choose a main course, then the total number of meal combinations is calculated as follows:

$m \times n = 4 \times 3 = 12$

Therefore, there are 12 different meal combinations available.

Key Concept: Basic Arrangement

c) 120

[Solution Description] The number of arrangements of $n$ distinct items is given by the factorial of $n$, denoted as $n!$. Here, we have 5 students, so the number of different arrangements is $5!$. Calculating this gives $5 \times 4 \times 3 \times 2 \times 1 = 120$.

Key Concept: Factorial and Real-World Application, Factorial in Combinatorial Identities

b) 120

[Solution Description]

To find the number of combinations for one team, use the formula $^6C_4 = \frac{6!}{4! \times (6-4)!}$. Calculate it step by step:

$^6C_4 = \frac{6!}{4! \times 2!}$

$= \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{(4 \times 3 \times 2 \times 1) \times (2 \times 1)}$

$= \frac{720}{48} = 15$

Since there are 8 different teams, multiply the combinations for one team by the number of teams:

$8 \times 15 = 120$

Therefore, there are 120 different combinations.

Key Concept: Conceptual Understanding

c) The order of selection does not matter in combinations.

[Solution Description] In combinations, the order of selection does not matter. This differentiates combinations from permutations, where order does matter.

Key Concept: Factorial Understanding

d) 120

[Solution Description] Factorial notation is represented as $n!$, where $n! = n \times (n-1) \times (n-2) \times \ldots \times 1$.

Calculate $5!$:

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

Therefore, the value of 5! is 120.

Key Concept: Advanced Conditional Combinations, Complex Permutations

c) 20,196

[Solution Description] To satisfy the condition of having at least twice as many boys as girls, we have the following cases:

– Case 1: 4 Boys and 2 Girls

$\binom{12}{4} \times \binom{8}{2} = \frac{12!}{4!(12-4)!} \times \frac{8!}{2!(8-2)!}$

Calculating gives:

$\binom{12}{4} = 495, \quad \binom{8}{2} = 28$

Number of ways =

$495 \times 28 = 13,860$

– Case 2: 5 Boys and 1 Girl

$\binom{12}{5} \times \binom{8}{1} = \frac{12!}{5!(12-5)!} \times 8$

Calculating gives:

$\binom{12}{5} = 792, \quad \binom{8}{1} = 8$

Number of ways =

$792 \times 8 = 6,336$

Adding both cases:

$13,860 + 6,336 = 20,196$

Key Concept: Permutation Formula

c) 453600

[Solution Description] The word “BOOKKEEPER” has 10 letters where ‘O’ occurs twice, ‘K’ occurs twice, and ‘E’ occurs twice. Hence, the number of distinct permutations is given by the formula for permutations of multiset:

$\frac{10!}{2! \times 2! \times 2!} = \frac{3628800}{8} = 453600.$

Key Concept: Basic Counting

c) 60

[Solution Description] To solve this problem, use the permutation formula for arranging $$r$$ items from $n$ items without repetition:

$P(n, r) = \frac{n!}{(n-r)!}$

Here, $n = 5$ and $r = 3$. Substitute the values:

$P(5, 3) = \frac{5!}{(5-3)!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = 60$

Therefore, there are 60 ways to arrange 3 books from 5 books.

Key Concept: Simple Calculation

c) 20

[Solution Description]

We will use the combination formula $nCr = \frac{n!}{r!(n-r)!}$ to calculate this.

Here, $n = 6$ and $r = 3$.

$^6C_3 = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$

Therefore, there are 20 ways to select 3 items from 6.

Key Concept: Even Numbers

b) 50

[Solution Description]

To form a 3-digit even number, the units place must be filled with an even digit. From the given digits (1, 2, 3, 4, 5), the even digits are 2 and 4. Thus, there are 2 choices for the units place. For the hundreds and tens place, any of the 5 digits can be chosen since repetition is allowed.

Total combinations:

$5 \times 5 \times 2 = 50$

Key Concept: Mixed Permutations

c) 3360

[Solution Description] The word “ENGINEER” consists of 8 letters where ‘E’ appears 3 times and ‘N’ appears 2 times. To find the number of distinct permutations, we use the formula for permutations with indistinguishable objects:

$\frac{8!}{3! \times 2!}$

Calculating factorials:

\begin{align*}
8! &= 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320, \\
3! &= 3 \times 2 \times 1 = 6, \\
2! &= 2 \times 1 = 2.
\end{align*}

Therefore, the number of distinct permutations is

$\frac{40320}{6 \times 2} = \frac{40320}{12} = 3360.$

Key Concept: Simple Permutation Calculation

c) 12

[Solution Description] Using the formula for permutations $nPr = \frac{n!}{(n-r)!}$:

– Here, $n = 4$ and $r = 2$.

– Calculate $4P2 = \frac{4!}{(4-2)!} = \frac{4 \times 3 \times 2 \times 1}{2 \times 1} = \frac{24}{2} = 12$.

Therefore, there are 12 permutations.

Key Concept: Counting Pairs

c) 12

[Solution Description] The number of ways to choose an appetizer is 3. The number of ways to choose a main course is 4. Using the fundamental principle of counting, we multiply the two numbers together: $3 \times 4 = 12$. Therefore, there are 12 different meals.

Key Concept: Basic Definition

c) Selection of items without regard to order

[Solution Description]

A combination is defined as a selection of items where the order does not matter. This is contrasted with permutations, where the order would be important.

Key Concept: Simple Permutations

c) 120

[Solution Description]

To determine the number of ways to arrange 3 books from 6, use the permutation formula $^nP_r = \frac{n!}{(n-r)!}$.

Here, $n = 6$ and $r = 3$. Thus,

$^6P_3 = \frac{6!}{(6-3)!} = \frac{6 \times 5 \times 4 \times 3!}{3!} = 120$

Hence, there are 120 different ways to arrange 3 books out of 6.

Key Concept: Real-World Application

c) 36

[Solution Description]

We need to calculate the number of ways to choose 2 players from 9 players for a handshake. Using the combination formula:

$nCr = \frac{n!}{r!(n-r)!}$

For $n = 9$ and $r = 2$, the calculation is as follows:

$9C2 = \frac{9!}{2!(9-2)!} = \frac{9 \times 8}{2 \times 1} = 36$

Hence, there will be 36 handshakes.

Key Concept: Complex Combination Problem, Multi-step Permutation Problem

b) 3220

[Solution Description] To solve this, we need to consider two cases:

Case 1: The committee has 3 women and 2 men.

$\text{Ways to choose 3 women} = \binom{8}{3} = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = 56$

$\text{Ways to choose 2 men} = \binom{10}{2} = \frac{10 \cdot 9}{2 \cdot 1} = 45$

Total ways for Case 1:

$56 \times 45 = 2520$

Case 2: The committee has 4 women and 1 man.

$\text{Ways to choose 4 women} = \binom{8}{4} = \frac{8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2 \cdot 1} = 70$

$\text{Ways to choose 1 man} = \binom{10}{1} = 10$

Total ways for Case 2:

$70 \times 10 = 700$

Finally, adding both cases gives:

$2520 + 700 = 3220$

Key Concept: Permutations with Fixed Positions

c) 9

[Solution Description] Since the first digit is fixed as 7, we only need to determine the possibilities for the remaining two positions. Each of those positions can contain any of the numbers 7, 8, or 9, allowing repetition. Thus, the number of 3-digit numbers is calculated by:

$3^2 = 9$

Key Concept: Basic Selection

c) 84

[Solution Description] To form a committee of 3 members, we select any 3 people out of a total of 9 individuals (5 men + 4 women). We use the combination formula:

$nCr = \frac{n!}{r!(n-r)!}$

Here, $n = 9$ and $r = 3$.

$^9C_3 = \frac{9!}{3! \cdot (9-3)!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$

So, there are 84 ways to form such a committee.

Key Concept: Non-Repetition Permutation

c) 60

[Solution Description] To form a 3-digit number using the digits 1, 2, 3, 4, and 5 without repetition, we calculate permutations of 5 different things taken 3 at a time: $nPr = \frac{5!}{(5-3)!} = \frac{5 \times 4 \times 3}{1} = 60$. Thus, there are 60 different 3-digit numbers possible.

Key Concept: Conditional Selection

c) 180

[Solution Description]

To solve this, we consider two possible cases:

– Case 1: The committee has exactly 2 women. We select 2 women out of 4 and 3 men out of 6:

$\text{Ways} = \binom{4}{2} \times \binom{6}{3} = 6 \times 20 = 120$

– Case 2: The committee has exactly 3 women. We select 3 women out of 4 and 2 men out of 6:

$\text{Ways} = \binom{4}{3} \times \binom{6}{2} = 4 \times 15 = 60$

Adding both cases together gives us the total number of ways:

$120 + 60 = 180$

Hence, there are 180 different ways to form such a committee.

Key Concept: Permutation with Conditions

b) 12

[Solution Description]

Consider the vowels A, A, and A as a single unit. Thus, the units to arrange are B, N, N, and AAA.

The number of distinct arrangements for these 4 units is $\frac{4!}{2!}$ because N repeats twice.

Additionally, the internal arrangement of the AAA is only 1 way since all three are identical.

Total arrangements = $\frac{4!}{2!} \times 1$

Calculating, $\frac{4!}{2!} = \frac{24}{2} = 12$.

Therefore, the number of different arrangements is 12.

Key Concept: Real-World Application

d) 720

[Solution Description] To choose a president, secretary, and treasurer in this order from 10 members, we apply the permutation formula for selecting and arranging 3 individuals out of 10.

Here, $n = 10$ and $r = 3$:

$nP_r = 10P_3 = 10 \times 9 \times 8 = 720$

Thus, there are 720 different ways to choose these three positions from the club members.

Key Concept: Non-Repetition Permutation

c) 120

[Solution Description] To form a 4-digit number, we need to choose 4 out of the 5 available digits. The number of permutations of 5 distinct digits taken 4 at a time, where no repetition is allowed, is given by:

$P(5, 4) = \frac{5!}{(5-4)!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{1!} = 120$

Thus, there are 120 such numbers.

Key Concept: Advanced Permutation, Complex Factorial

c) 420

[Solution Description]

Consider all ‘I’s as one single object, so we have M, S, S, S, P, P plus this ‘III’ object. This gives us 7 objects to arrange. Out of these, the 3 ‘S’s and 2 ‘P’s are identical. The number of ways to arrange them is given by:

$\frac{7!}{3!2!} = \frac{5040}{12} = 420$

Therefore, there are 420 distinct permutations.

Key Concept: Permutations with Repetition Allowed

c) 16807

[Solution Description]

Here, we need to calculate the number of 5-letter arrangements using the letters from the word “NUMBERS”, allowing repetition.

Each position in the 5-letter word can be filled in 7 different ways (one for each letter in “NUMBERS”).

Thus, the total number of permutations is given by $7^5$:

$7^5 = 16807$

Therefore, 16807 distinct words can be formed.

Key Concept: Combination and Permutation Mix

c) 60

[Solution Description]

The word “ANGLE” has 5 unique letters: A, N, G, L, E.

Step 1: Choose 3 letters out of 5 using combination:

${5 \choose 3} = \frac{5 \times 4}{2 \times 1} = 10$

Step 2: Arrange these 3 letters using permutation ($3!$):

$3! = 3 \times 2 \times 1 = 6$

Multiply both results to get the total number of distinct words:

$10 \times 6 = 60$

Hence, 60 distinct words can be formed.

Key Concept: Permutations with Conditions

c) 12

[Solution Description]

To form a 3-digit number divisible by 5, the last digit must be 5. Therefore, we are left to choose two more digits from the remaining four (1, 2, 3, 4) for the first two positions.

Calculate permutations for the remaining two spots:

Number of ways to choose the first digit = 4 (since there are 4 choices: 1, 2, 3, or 4).

Number of ways to choose the second digit = 3 (after choosing the first digit).

So, total permutations = $4 \times 3 = 12$.

Thus, 12 different numbers can be formed.

Key Concept: Advanced Formula Application

d) 126

[Solution Description] Using the formula for combinations, $nCr = \frac{n!}{r!(n-r)!}$, we calculate:

Here, $n = 9$ and $r = 4$.

\[
{}^9C_4 = \frac{9!}{4!(9-4)!}
= \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}
= \frac{3024}{24} = 126.
\]

Therefore, there are 126 ways to choose the team.

Key Concept: Permutations with Exclusions

c) 900

[Solution Description]

To find the number of permutations where the two L’s are not together, we first calculate the total permutations without any restrictions and then subtract the permutations where the L’s are together.

Total permutations without restrictions:

There are 7 letters in BALLOON, with 2 L’s and 2 O’s, so the number of unrestricted permutations is:

$\frac{7!}{2! \times 2!} = \frac{5040}{4} = 1260.$

Treating the two L’s as a single unit (LL), we now have 6 units to arrange: B, A, LL, O, O, N. The number of permutations where L’s are together is:

$\frac{6!}{2!} = \frac{720}{2} = 360.$

Hence, the number of permutations where the L’s are not together is:

$1260 – 360 = 900.$

Key Concept: Basic Combinations

b) 15

[Solution Description]

To calculate the number of combinations, use the formula $nCr = \frac{n!}{r!(n-r)!}$. Here, $n=6$ and $r=4$.

Calculate $6C4$:

$6C4 = \frac{6!}{4! \times (6-4)!} = \frac{6 \times 5 \times 4!}{4! \times 2!} = \frac{6 \times 5}{2 \times 1} = 15$

Therefore, the number of combinations is 15.

Key Concept: Factorial in Permutations

b) 360

[Solution Description]

The formula for permutations is given by $nPr = \frac{n!}{(n-r)!}$. For $6P4$, substitute 6 for $n$ and 4 for $r$:

$6P4 = \frac{6!}{(6-4)!} = \frac{6!}{2!}$

Calculate $6! = 720$ and $2! = 2$, then:

$\frac{720}{2} = 360$

Therefore, $6P4 = 360$.

Key Concept: Greater Than a Number

c) 48

[Solution Description] We need to find numbers greater than 50000, which means the first digit must be either 5 or 6. Let’s consider each case:

Case 1: First digit is 5. The remaining numbers are 3, 6, 2, 1. The number of permutations is $4!$.

$4! = 24.$

Case 2: First digit is 6. The remaining numbers are 3, 5, 2, 1. The number of permutations is also $4!$.

$4! = 24.$

Total numbers = $24 + 24 = 48$.

So, the total number of numbers greater than 50000 is 48.

Key Concept: Simple Permutations

b) 12

[Solution Description] To find the number of permutations of 2 objects taken from 4 distinct objects without repetition, we use the formula for permutations of distinct objects:

$nP_r = n(n-1)(n-2)\ldots(n-r+1).$

For this case:

$4P_2 = 4 \times 3 = 12$

Hence, 12 different 2-letter words can be formed.

Key Concept: Factorial Understanding

d) 5040

[Solution Description] Factorial notation $n!$ is defined as $1 \times 2 \times 3 \times \ldots \times n$. For $7!$, the calculation is:

$7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040$

The value of $7!$ is 5040.

Key Concept: Real-World Application

c) 255

[Solution Description]

Begin by calculating the total number of unrestricted ways to select 4 people from 12 individuals:

$\binom{12}{4} = 495$

Next, find the number of ways to form a committee of 4 persons not including any complete couple:

Select 4 individuals from 12 such that no two are a couple.

Choose 4 couples and take one member from each taking note that choosing no pair from couples means selecting from just singles:

$\text{Choose 4 couples} = \binom{6}{4} = 15$

For each chosen couple, select one person:

$2^4 = 16$

So, the number of ways without any couple:

$15 \times 16 = 240$

Subtract these from the total:

$495 – 240 = 255$

Key Concept: Complex Permutations with Restrictions, Advanced Permutations with Identical Objects

b) 180

[Solution Description] The word ‘SCHOOL’ contains 6 letters where S, C, H, O, L are unique and one ‘O’ repeats. Since we are to arrange them in such a way that ‘O’ comes before ‘C’, consider each permutation of SCHOOL while fixing ‘O’ position before ‘C’.

Calculate total permutations without restriction: $\frac{6!}{2!} = 360$.

Since only half of these arrangements have ‘O’ before ‘C’, divide by 2 for restricted arrangements:

$\frac{360}{2} = 180$

Therefore, there are 180 distinct words where ‘O’ is before ‘C’.

Key Concept: Permutation with Constraints

c) 5040

[Solution Description] Consider both G’s as a single unit. Now we have E, N, I, N, E, E, R, I, N, GG to arrange. This gives us 9 units with N appearing 3 times, E appearing 3 times, and I appearing 2 times.

Using the permutation formula for non-distinct objects:

$\frac{9!}{3!3!2!} = \frac{362880}{72} = 5040$

So, there are 5040 such arrangements where both G’s come together.

Key Concept: Permutation Definition

c) An arrangement where order matters

[Solution Description]

A permutation is an arrangement of objects in a specific order. The concept focuses on the order in which items are arranged or sequenced.

Key Concept: Permutation with Multiple Conditions

c) 120

[Solution Description] The word “ARRANGE” has 7 letters with two vowels A, E. Since the arrangement must start and end with a vowel, we choose positions for these vowels first.

The number of ways to select vowels for the first and last position is $2 \times 1 = 2$ ways (choosing one for the first position and the other automatically goes to the last).

After placing A and E in specific positions, we are left with 5 letters: R, R, N, G, and the remaining vowel. These can be arranged among themselves considering the repetition of R using the formula:

$\frac{5!}{2!} = \frac{120}{2} = 60$

Therefore, the total arrangements are $2 \times 60 = 120$.

Key Concept: Vowel Grouping

c) 2160

[Solution Description]

The word ‘EXAMPLES’ contains 8 letters, with the vowels E, A, and E. Consider the group of vowels as one single unit or block. Thus, we have 6 units to arrange: (EAE), X, M, P, L, S. These 6 units can be arranged in $6!$ ways. However, within the vowel block, E, A, and E can be arranged among themselves in $\frac{3!}{2!}$ ways due to repetition of E. Therefore, total number of arrangements is:

$6! \times \frac{3!}{2!} = 720 \times 3 = 2160$