This page provides comprehensive Class 11 Maths Exemplar Chapter 6 Exercise 6.2 Solutions. Detailed step-by-step solutions for Class 11 Maths NCERT Exemplar Chapter 6 Permutations and Combinations Exercise 6.2. Free PDF download and interactive practice.
Exercise 6.2
Short Answer Type Questions
Q1
Eight chairs are numbered 1 to 8. Two women and 3 men wish to occupy one chair each. First the women choose the chairs from amongst the chairs 1 to 4 and then men select from the remaining chairs. Find the total number of possible arrangements.
Women choose 2 chairs from 1 to 4: ${}^4P_2 = 4 \times 3 = 12$ ways. Remaining chairs = $8 - 2 = 6$. Men choose 3 chairs from remaining 6: ${}^6P_3 = 6 \times 5 \times 4 = 120$ ways. Total arrangements = $12 \times 120 = 1440$.
$1440$
Q2
If the letters of the word RACHIT are arranged in all possible ways as listed in dictionary. Then what is the rank of the word RACHIT?
Alphabetical order: A, C, H, I, R, T. Words starting with A: $5! = 120$ Words starting with C: $5! = 120$ Words starting with H: $5! = 120$ Words starting with I: $5! = 120$ Words starting with R: - RA... (Next is C): RAC... - RACH... (Next is I): RACHI... - RACHIT is the first word starting with RACHI. Rank = $4 \times 120 + 1 = 481$.
$481$
Q3
A candidate is required to answer 7 questions out of 12 questions, which are divided into two groups, each containing 6 questions. He is not permitted to attempt more than 5 questions from either group. Find the number of different ways of doing questions.
Out of 18 points in a plane, no three are in the same line except five points which are collinear. Find the number of triangles that can be formed joining the points.
We wish to select 6 persons from 8, but if the person A is chosen, then B must be chosen. In how many ways can selections be made?
Case 1: A is chosen. Then B must be chosen. We need to select $6-2=4$ more from remaining $8-2=6$. Ways = ${}^6C_4 = 15$. Case 2: A is not chosen. Then B can be chosen or not. We need to select 6 from remaining $8-1=7$ (excluding A). Ways = ${}^7C_6 = 7$. Total ways = $15 + 7 = 22$.
$22$
Q6
How many committee of five persons with a chairperson can be selected from 12 persons.
We need to select a committee of 5, where one is a chairperson. Method: Select 5 members first: ${}^{12}C_5 = 792$. From these 5, select 1 chairperson: ${}^5C_1 = 5$. Total ways = $792 \times 5 = 3960$.
$3960$
Q7
How many automobile license plates can be made if each plate contains two different letters followed by three different digits?
A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected from the lot.
Select 2 Black from 5: ${}^5C_2 = 10$. Select 3 Red from 6: ${}^6C_3 = 20$. Total ways = $10 \times 20 = 200$.
$200$
Q9
Find the number of permutations of n distinct things taken r together, in which 3 particular things must occur together.
Select $r$ things such that 3 particular things are included. We need to select $r-3$ from remaining $n-3$. Ways = ${}^{n-3}C_{r-3}$. Arrange these $r$ things such that the 3 particular things are together. Treat the 3 things as 1 unit. Total units = $(r-3) + 1 = r-2$. Arrangements = $(r-2)!$. The 3 things can be arranged among themselves in $3!$ ways. Total permutations = ${}^{n-3}C_{r-3} \times (r-2)! \times 3!$.
$3! (r-2)! {}^{n-3}C_{r-3}$
Q10
Find the number of different words that can be formed from the letters of the word ‘TRIANGLE’ so that no vowels are together.
Word: TRIANGLE. Vowels: A, E, I (3). Consonants: T, R, N, G, L (5). Arrange consonants first: $5! = 120$. This creates 6 gaps: _ C _ C _ C _ C _ C _. Place 3 vowels in 6 gaps: ${}^6P_3 = 6 \times 5 \times 4 = 120$. Total words = $120 \times 120 = 14400$.
$14400$
Q11
Find the number of positive integers greater than 6000 and less than 7000 which are divisible by 5, provided that no digit is to be repeated.
Number must be 4 digits, starting with 6. Digits available: 0-9. Divisible by 5 $\implies$ ends in 0 or 5. Case 1 (Ends in 0): 6 _ _ 0. Remaining digits for 2 spots: 8 (excluding 6, 0). Ways = ${}^8P_2 = 56$. Case 2 (Ends in 5): 6 _ _ 5. Remaining digits for 2 spots: 8 (excluding 6, 5). Ways = ${}^8P_2 = 56$. Total = $56 + 56 = 112$.
$112$
Q12
There are 10 persons named P1, P2, P3, ... P10. Out of 10 persons, 5 persons are to be arranged in a line such that in each arrangement P1 must occur whereas P4 and P5 do not occur. Find the number of such possible arrangements.
We need to select 5 persons. P1 is included. P4, P5 excluded. Need to select $5-1=4$ more persons from $10-1-2=7$ available persons. Selection ways = ${}^7C_4 = 35$. Arrangement of 5 persons = $5! = 120$. Total arrangements = $35 \times 120 = 4200$.
$4200$
Q13
There are 10 lamps in a hall. Each one of them can be switched on independently. Find the number of ways in which the hall can be illuminated.
Each lamp has 2 states (On/Off). Total combinations = $2^{10}$. Hall is illuminated if at least one lamp is On. Subtract the case where all are Off (1 case). Total ways = $2^{10} - 1 = 1024 - 1 = 1023$.
$1023$
Q14
A box contains two white, three black and four red balls. In how many ways can three balls be drawn from the box, if atleast one black ball is to be included in the draw.
Total balls = $2W + 3B + 4R = 9$. Select 3. Total ways = ${}^9C_3 = 84$. Ways with NO black ball (select from 2W+4R=6): ${}^6C_3 = 20$. Ways with at least one black = Total - None = $84 - 20 = 64$.
$64$
Q15
If ${}^nC_{r-1} = 36$, ${}^nC_r = 84$ and ${}^nC_{r+1} = 126$, then find ${}^rC_2$.
Find the number of integers greater than 7000 that can be formed with the digits 3, 5, 7, 8 and 9 where no digits are repeated.
Digits: 3, 5, 7, 8, 9 (5 digits). 4-digit numbers > 7000: Start with 7, 8, 9 (3 options). Remaining 3 spots from remaining 4 digits: ${}^4P_3 = 24$. Total = $3 \times 24 = 72$. 5-digit numbers: All are > 7000. Total = $5! = 120$. Total integers = $72 + 120 = 192$.
$192$
Q17
If 20 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, in how many points will they intersect each other?
Since no two are parallel and no three concurrent, every pair of lines intersects at a unique point. Number of intersection points = ${}^{20}C_2 = \frac{20 \times 19}{2} = 190$.
$190$
Q18
In a certain city, all telephone numbers have six digits, the first two digits always being 41 or 42 or 46 or 62 or 64. How many telephone numbers have all six digits distinct?
Prefixes: 41, 42, 46, 62, 64 (5 options). For each prefix (2 digits used), we need to fill remaining 4 places with distinct digits from remaining 8 digits. Ways = ${}^8P_4 = 8 \times 7 \times 6 \times 5 = 1680$. Total numbers = $5 \times 1680 = 8400$.
$8400$
Q19
In an examination, a student has to answer 4 questions out of 5 questions; questions 1 and 2 are however compulsory. Determine the number of ways in which the student can make the choice.
Total questions = 5. To answer = 4. Q1 and Q2 are compulsory. So 2 questions are already selected. Need to select $4-2=2$ questions from remaining $5-2=3$ questions. Ways = ${}^3C_2 = 3$.
$3$
Q20
A convex polygon has 44 diagonals. Find the number of its sides.
Number of diagonals = ${}^nC_2 - n = 44$. $\frac{n(n-1)}{2} - n = 44$. $n^2 - n - 2n = 88 \implies n^2 - 3n - 88 = 0$. $(n-11)(n+8) = 0$. Since $n > 0$, $n = 11$.