CBSE Class 11 Maths Exemplar Chapter 7 Permutations and Combinations NCERT Solutions

The problem involves two stages: women choosing chairs and then men choosing chairs.

Stage 1: Women choosing chairs

There are 4 chairs available (numbered 1 to 4) for the 2 women. Since the order in which they choose matters (as they occupy distinct chairs), we use permutations. The number of ways the 2 women can choose chairs from 4 is given by the permutation formula $P(n, r) = \frac{n!}{(n-r)!}$.

Number of arrangements for women = $P(4, 2) = \frac{4!}{(4-2)!} = \frac{4!}{2!} = \frac{4 \times 3 \times 2!}{2!} = 4 \times 3 = 12$ ways.

Stage 2: Men choosing chairs

After the women have chosen their chairs, there are $8 - 2 = 6$ chairs remaining. The 3 men need to choose chairs from these 6 remaining chairs. Again, the order matters.

Number of arrangements for men = $P(6, 3) = \frac{6!}{(6-3)!} = \frac{6!}{3!} = \frac{6 \times 5 \times 4 \times 3!}{3!} = 6 \times 5 \times 4 = 120$ ways.

Total Arrangements

To find the total number of possible arrangements, we multiply the number of ways for each stage (by the multiplication principle).

Total possible arrangements = (Arrangements for women) $\times$ (Arrangements for men)

Total possible arrangements = $P(4, 2) \times P(6, 3) = 12 \times 120 = 1440$ ways.

Thus, the total number of possible arrangements is 1440.

To find the rank of the word 'RACHIT', we first list the letters of the word in alphabetical order: A, C, H, I, R, T. The word 'RACHIT' has 6 distinct letters.

We need to count how many words come before 'RACHIT' in the dictionary.

1. Words starting with 'A':

If the word starts with 'A', the remaining 5 letters (C, H, I, R, T) can be arranged in $5!$ ways.

Number of words = $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.

2. Words starting with 'C':

If the word starts with 'C', the remaining 5 letters (A, H, I, R, T) can be arranged in $5!$ ways.

Number of words = $5! = 120$.

3. Words starting with 'H':

If the word starts with 'H', the remaining 5 letters (A, C, I, R, T) can be arranged in $5!$ ways.

Number of words = $5! = 120$.

4. Words starting with 'I':

If the word starts with 'I', the remaining 5 letters (A, C, H, R, T) can be arranged in $5!$ ways.

Number of words = $5! = 120$.

5. Words starting with 'R':

Now, the word starts with 'R'. We look at the second letter. The letters remaining are A, C, H, I, T. The second letter in 'RACHIT' is 'A'. Since 'A' is the first letter alphabetically among the remaining letters, all words starting with 'RA' will come next.

So, we have counted all words starting with A, C, H, and I. The total count so far is $120 + 120 + 120 + 120 = 480$.

The word 'RACHIT' is the first word that starts with 'R' and has 'A' as the second letter. Therefore, its rank is the total count of preceding words plus 1.

Rank of 'RACHIT' = (Number of words starting with A, C, H, I) + 1

Rank = $4 \times 5! + 1 = 4 \times 120 + 1 = 480 + 1 = 481$.

The rank of the word 'RACHIT' is 481.

The total number of questions is 12, divided into two groups (Group I and Group II), each with 6 questions. The candidate must answer exactly 7 questions in total. There is a constraint: the candidate cannot attempt more than 5 questions from either group.

Let $n_1$ be the number of questions attempted from Group I and $n_2$ be the number of questions attempted from Group II. We must have $n_1 + n_2 = 7$. The constraint is $n_1 \le 5$ and $n_2 \le 5$. Since each group has 6 questions, this means $n_1$ can range from $6-5=1$ to 5, and similarly for $n_2$.

We need to find pairs $(n_1, n_2)$ such that $n_1 + n_2 = 7$, $1 \le n_1 \le 5$, and $1 \le n_2 \le 5$. The possible pairs are:

• Case 1: $n_1 = 5$, $n_2 = 2$. Number of ways = $C(6, 5) \times C(6, 2)$.
• Case 2: $n_1 = 4$, $n_2 = 3$. Number of ways = $C(6, 4) \times C(6, 3)$.
• Case 3: $n_1 = 3$, $n_2 = 4$. Number of ways = $C(6, 3) \times C(6, 4)$.
• Case 4: $n_1 = 2$, $n_2 = 5$. Number of ways = $C(6, 2) \times C(6, 5)$.

We know that $C(n, k) = C(n, n-k)$. So, $C(6, 5) = C(6, 1) = 6$ and $C(6, 2) = C(6, 4) = \frac{6 \times 5}{2 \times 1} = 15$. Also, $C(6, 3) = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.

Calculating the number of ways for each case:

• Case 1: $C(6, 5) \times C(6, 2) = 6 \times 15 = 90$.
• Case 2: $C(6, 4) \times C(6, 3) = 15 \times 20 = 300$.
• Case 3: $C(6, 3) \times C(6, 4) = 20 \times 15 = 300$.
• Case 4: $C(6, 2) \times C(6, 5) = 15 \times 6 = 90$.

The total number of different ways is the sum of the ways in all possible cases:

Total ways = $90 + 300 + 300 + 90 = 780$.

Alternatively, notice that Case 1 and Case 4 are identical, and Case 2 and Case 3 are identical. So, we can write the total as:

Total ways = $2 \times [C(6, 5) \times C(6, 2) + C(6, 4) \times C(6, 3)]$

Total ways = $2 \times [6 \times 15 + 15 \times 20] = 2 \times [90 + 300] = 2 \times 390 = 780$.

Therefore, the number of different ways of doing the questions is 780.

We are given a total of 18 points in a plane. Among these, 5 points are collinear, meaning they lie on a single straight line. The remaining $18 - 5 = 13$ points are such that no three of them are collinear.

To form a straight line, we need to choose any 2 points. The total number of ways to choose 2 points from 18 points is given by the combination formula $C(n, r) = \frac{n!}{r!(n-r)!}$.

Total possible lines if no points were collinear = $C(18, 2) = \frac{18!}{2!(18-2)!} = \frac{18!}{2!16!} = \frac{18 \times 17}{2 \times 1} = 9 \times 17 = 153$.

However, the 5 points that are collinear will form only one line instead of the $C(5, 2)$ lines they would form if they were not collinear.

Number of lines formed by the 5 collinear points if they were not collinear = $C(5, 2) = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4}{2 \times 1} = 10$.

Since these 5 points form only 1 line, we must subtract the $C(5, 2)$ lines and add back the 1 line they actually form.

Required number of straight lines = (Total lines from 18 points) - (Lines from 5 collinear points) + (1 line from the 5 collinear points)

Required number of lines = $C(18, 2) - C(5, 2) + 1$

Required number of lines = $153 - 10 + 1 = 144$.

Therefore, the number of straight lines that can be formed is 144.

We need to select 6 people from a group of 8 people. The condition is that if person A is selected, then person B must also be selected. We can analyze this problem by considering two main cases based on whether person A is selected or not.

Case 1: Person A is selected.

If person A is selected, then according to the condition, person B must also be selected. This means that 2 specific people (A and B) are already chosen. We need to select the remaining $6 - 2 = 4$ people from the remaining $8 - 2 = 6$ people.

The number of ways to select these 4 people from the remaining 6 is given by the combination formula $C(n, r)$.

Number of ways in Case 1 = $C(6, 4) = \frac{6!}{4!(6-4)!} = \frac{6!}{4!2!} = \frac{6 \times 5}{2 \times 1} = 15$.

Case 2: Person A is not selected.

If person A is not selected, the condition 'if A is chosen, then B must be chosen' does not impose any restriction on B. Person B can either be selected or not selected. We need to select 6 people from the remaining $8 - 1 = 7$ people (since A is not considered).

The number of ways to select 6 people from these 7 people is:

Number of ways in Case 2 = $C(7, 6) = \frac{7!}{6!(7-6)!} = \frac{7!}{6!1!} = \frac{7}{1} = 7$.

Total Number of Ways

The total number of ways to make the selection is the sum of the number of ways in Case 1 and Case 2, as these cases are mutually exclusive and cover all possibilities.

Total ways = (Ways when A is chosen) + (Ways when A is not chosen)

Total ways = $C(6, 4) + C(7, 6) = 15 + 7 = 22$.

Therefore, there are 22 different ways to make the selection.