NCERT Exemplar Solutions Class 11 Maths Permutations and Combinations — Short and Long Answer Type Q1–25

NCERT Exemplar Problems, Class 11 Mathematics — Permutations and Combinations, Short Answer and Long Answer Type questions 1 to 25. All 25 questions solved.

Twenty-five counting problems, and almost all of them reduce to two questions: does the order matter, and is there a restriction to handle first? Order matters for arrangements, words and numbers, where you use ${}^nP_r = \dfrac{n!}{(n-r)!}$ or a factorial. Order does not matter for committees, teams and selections, where you use ${}^nC_r = \dfrac{n!}{r!\,(n-r)!}$. “At least one” is nearly always done as total minus none, and “no two together” by arranging the free objects first and slotting the restricted ones into the gaps.

All 22 answers that the printed key gives match the working below.

Key insight. Decide first whether you are selecting or arranging, then handle the restriction before the free choices. Choose-then-arrange gives ${}^nC_r \times r!$, which is the same thing as ${}^nP_r$.

Question 1

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.

Solution. The two women are different people, so which woman sits where matters: they take two of chairs 1 to 4 in order, in ${}^4P_2 = 12$ ways. That leaves $8 – 2 = 6$ chairs, and the three different men occupy three of them in order, in ${}^6P_3 = 120$ ways. The two choices are made one after the other, so multiply: $12 \times 120$.

Question 2

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?

Solution. The letters in alphabetical order are A, C, H, I, R, T. Count the words that come before RACHIT. Any word starting with A, C, H or I is earlier, and each such first letter leaves $5! = 120$ arrangements of the rest, giving $4 \times 120 = 480$ words. Now fix R and look at the second letter: A is the smallest letter left, so no word starting with R comes earlier through the second place. Fix RA; the next letter C is the smallest left (C, H, I, T), so none is earlier. Fix RAC; H is the smallest left, so none earlier; RACH then I and T are already in the right order. So exactly $480$ words precede it, and RACHIT itself is the next one.

Question 3

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.

Solution. Seven questions with at most 5 from each group of 6 means the split between the groups is $(5,2)$, $(4,3)$, $(3,4)$ or $(2,5)$. Add the four cases:

$$2\,[\,{}^6C_5\,{}^6C_2 + {}^6C_4\,{}^6C_3\,] = 2\,[\,6 \times 15 + 15 \times 20\,] = 2(90 + 300) = 780.$$

As a check, all selections are ${}^{12}C_7 = 792$, and the forbidden ones take all 6 from one group and 1 from the other: $2 \times 6 = 12$. So $792 – 12 = 780$.

Question 4

Out of 18 points in a plane, no three are in the same line except five points which are collinear. Find the number of lines that can be formed joining the point.

Solution. Any two points give a line, so ${}^{18}C_2$ counts every pair. But the five collinear points give ${}^5C_2 = 10$ pairs which all lie on one and the same line, so those ten pairs are really a single line. Subtract the ten and add back the one line: $153 – 10 + 1$.

Question 5

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?

Solution. Split on whether A is chosen. If A is not chosen, the 6 come freely from the other 7 persons: ${}^7C_6 = 7$. If A is chosen, B is forced in too, and the remaining 4 come from the other 6: ${}^6C_4 = 15$. The two cases cannot overlap, so add them.

Question 6

How many committee of five persons with a chairperson can be selected from 12 persons.

Solution. The chairperson is a distinguished member, so pick that person first in $12$ ways. The other four members are an unordered selection from the remaining 11: ${}^{11}C_4 = 330$. Hence $12 \times 330$.

Question 7

How many automobile license plates can be made if each plate contains two different letters followed by three different digits?

Solution. The position of each character matters, so this is a count of arrangements. Two different letters from 26: $26 \times 25 = 650$. Three different digits from 10 (a plate may start with 0): $10 \times 9 \times 8 = 720$. Multiply: $650 \times 720$.

Question 8

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.

Solution. The black and red choices are independent selections, each unordered: ${}^5C_2 \times {}^6C_3 = 10 \times 20$.

Question 9

Find the number of permutations of $n$ distinct things taken $r$ together, in which 3 particular things must occur together.

Solution. Treat it in three steps. First, all three particular things must be in the arrangement, so only $r – 3$ other things are chosen, from the remaining $n – 3$: ${}^{n-3}C_{r-3}$ ways. Second, glue the three particular things into one block. The block and the $r-3$ others form $r – 2$ objects, which can be arranged in $(r-2)!$ ways. Third, the three things can be ordered inside their block in $3!$ ways.

$${}^{n-3}C_{r-3}\,(r-2)!\;3!$$

I confirmed this by listing all arrangements for $(n,r) = (6,4), (7,5), (7,4), (8,5)$: the counts $36, 216, 48, 360$ agree with the formula each time.

Question 10

Find the number of different words that can be formed from the letters of the word ‘TRIANGLE’ so that no vowels are together.

Solution. TRIANGLE has 3 vowels (I, A, E) and 5 consonants (T, R, N, G, L), all different. To keep the vowels apart, arrange the consonants first, in $5! = 120$ ways. They leave $6$ gaps (one before, one after, and four between). Put the three vowels into three different gaps, in order: ${}^6P_3 = 120$ ways. No two vowels can then be adjacent, so $120 \times 120$.

Question 11

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.

Solution. The number is a four-digit number beginning with 6. Divisible by 5 means the last digit is 0 or 5.

• Last digit 0: the two middle places take two different digits from the 8 left (not 6, not 0): $8 \times 7 = 56$.
• Last digit 5: the middle places take two different digits from the 8 left (not 6, not 5, but 0 is allowed): $8 \times 7 = 56$.

$56 + 56 = 112$ (listing all multiples of 5 between 6001 and 6999 with four distinct digits gives the same).

Question 12

There are 10 persons named $P_1, P_2, P_3, \dots, P_{10}$. Out of 10 persons, 5 persons are to be arranged in a line such that in each arrangement $P_1$ must occur whereas $P_4$ and $P_5$ do not occur. Find the number of such possible arrangements.

Solution. $P_1$ is forced in and $P_4$, $P_5$ are forced out, so the other 4 members are chosen from the remaining $10 – 3 = 7$ persons: ${}^7C_4 = 35$ ways. The five chosen people are then lined up in $5! = 120$ ways. The printed key does not give an answer for this question; the hint’s ${}^7C_4 \times 5!$ is what we get.

Question 13

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.

Solution. Each lamp is either on or off, giving $2^{10}$ switch patterns. The hall is illuminated if at least one lamp is on, which excludes only the pattern with all lamps off. The printed key has no answer here.

Question 14

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.

Solution. “At least one” is easiest as total minus none. Three balls from all 9: ${}^9C_3 = 84$. Three balls with no black ball (from the 6 non-black): ${}^6C_3 = 20$. So $84 – 20$. The hint’s direct count, ${}^3C_1\,{}^6C_2 + {}^3C_2\,{}^6C_1 + {}^3C_3 = 45 + 18 + 1$, gives the same $64$.

Question 15

If ${}^nC_{r-1} = 36$, ${}^nC_r = 84$ and ${}^nC_{r+1} = 126$, then find ${}^rC_2$.

Solution. Consecutive binomial coefficients are linked by $\dfrac{{}^nC_r}{{}^nC_{r-1}} = \dfrac{n-r+1}{r}$ and $\dfrac{{}^nC_{r+1}}{{}^nC_r} = \dfrac{n-r}{r+1}$. This turns the three given values into two linear equations and removes the factorials.

$$\frac{n-r+1}{r} = \frac{84}{36} = \frac73 \;\Rightarrow\; 3n + 3 = 10r, \qquad \frac{n-r}{r+1} = \frac{126}{84} = \frac32 \;\Rightarrow\; 2n = 5r + 3.$$

From the second, $n = \dfrac{5r+3}{2}$; substituting, $\dfrac{3(5r+3)}{2} + 3 = 10r$, so $15r + 9 + 6 = 20r$ and $r = 3$, $n = 9$. Check: ${}^9C_2 = 36$, ${}^9C_3 = 84$, ${}^9C_4 = 126$. The question asks for ${}^rC_2 = {}^3C_2$ (the printed key stops at $r = 3$).

Question 16

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.

Solution. Only four- and five-digit numbers can exceed 7000. Every five-digit number is above 7000: $5! = 120$ of them. A four-digit number is above 7000 only if its first digit is 7, 8 or 9, which is 3 choices; the other three places then take digits from the remaining 4, in $4 \times 3 \times 2 = 24$ ways, so $3 \times 24 = 72$. Add: $72 + 120$.

Question 17

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?

Solution. Each pair of lines meets in exactly one point (no two are parallel), and no point is shared by three lines, so different pairs give different points. The number of points is the number of pairs: ${}^{20}C_2$.

Question 18

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?

Solution. There are 5 choices for the first two digits, and each uses two different digits. The next four places must use digits different from those two and from each other: $8 \times 7 \times 6 \times 5 = 1680$. Hence $5 \times 1680$.

Question 19

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.

Solution. Questions 1 and 2 are fixed, so only the other 2 answers are chosen, from the remaining 3 questions: ${}^3C_2$.

Question 20

A convex polygon has 44 diagonals. Find the number of its sides.

Solution. Join every pair of the $n$ vertices: ${}^nC_2$ segments. Of these, $n$ are sides of the polygon, so the diagonals number ${}^nC_2 – n$.

$$\frac{n(n-1)}{2} – n = 44 \;\Rightarrow\; n^2 – 3n – 88 = 0 \;\Rightarrow\; (n-11)(n+8) = 0.$$

A number of sides cannot be negative, so $n = 11$.

Question 21

18 mice were placed in two experimental groups and one control group, with all groups equally large. In how many ways can the mice be placed into three groups?

Solution. Each group holds $18 \div 3 = 6$ mice, and the three groups are distinguishable: two experimental groups and one control group (the book’s key treats them as labelled). Choose 6 mice for the first group, then 6 of the remaining 12 for the second; the last 6 form the third:

$${}^{18}C_6 \times {}^{12}C_6 \times {}^6C_6 = \frac{18!}{6!\,6!\,6!} = 17153136.$$

Question 22

A bag contains six white marbles and five red marbles. Find the number of ways in which four marbles can be drawn from the bag if (a) they can be of any colour (b) two must be white and two red and (c) they must all be of the same colour.

Solution. There are 11 marbles in all.

• (a) Any four of 11: ${}^{11}C_4 = 330$.
• (b) Two of 6 white and two of 5 red: ${}^6C_2 \times {}^5C_2 = 15 \times 10 = 150$.
• (c) All white or all red, two separate cases to add: ${}^6C_4 + {}^5C_4 = 15 + 5 = 20$.

Question 23

In how many ways can a football team of 11 players be selected from 16 players? How many of them will (i) include 2 particular players? (ii) exclude 2 particular players?

Solution. A team is an unordered selection of 11 from 16: ${}^{16}C_{11} = 4368$. If two particular players must be in, only the other 9 places are chosen from the remaining 14: ${}^{14}C_9 = 2002$. If two particular players must be out, all 11 places come from the remaining 14: ${}^{14}C_{11} = 364$. The printed key lists only (i) and (ii).

Question 24

A sports team of 11 students is to be constituted, choosing at least 5 from Class XI and atleast 5 from Class XII. If there are 20 students in each of these classes, in how many ways can the team be constituted?

Solution. Eleven players with at least 5 from each class forces the split to be exactly $5 + 6$ or $6 + 5$: a 4-and-7 split would break the minimum. Both cases give the same count, so double one of them:

$$2\,({}^{20}C_5 \times {}^{20}C_6) = 2 \times 15504 \times 38760 = 1201870080.$$

Question 25

A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has (i) no girls (ii) at least one boy and one girl (iii) at least three girls.

Solution. The group has 11 people in all.

• (i) No girls means all five from the 7 boys: ${}^7C_5 = 21$.
• (ii) Subtract the all-boy teams (21) and the all-girl teams (none, as there are only 4 girls, ${}^4C_5 = 0$) from all teams: ${}^{11}C_5 – 21 = 462 – 21 = 441$.
• (iii) At least three girls is 3 girls with 2 boys, or 4 girls with 1 boy: ${}^4C_3\,{}^7C_2 + {}^4C_4\,{}^7C_1 = 84 + 7 = 91$.

Common mistakes

• Question 2 (rank): Students count the words before RACHIT as $4 \times 5!$ and stop, giving 480 or 120 as the rank. The rank is one more than the number of words before it, since RACHIT itself occupies a place.
• Question 4 (lines): Writing ${}^{18}C_2 – {}^5C_2$ forgets that the five collinear points still make one line. Each pair counted from the collinear points collapses into that single line, which has to be put back.
• Question 7 (licence plates): Using ${}^{26}C_2$ and ${}^{10}C_3$ treats a plate like a committee. A plate such as AB123 differs from BA123, so the order matters and permutations are needed.
• Question 10 (no vowels together): Arranging all 8 letters and subtracting the vowel-together cases becomes messy. Arranging consonants first and filling gaps avoids the subtraction; the mistake is using ${}^5P_3$ for the gaps instead of ${}^6P_3$, forgetting the two end gaps.
• Question 11 (numbers divisible by 5): The two cases for the last digit do not give the same count if the 0 is mishandled. With last digit 5, the middle digits may include 0, so there are still 8 choices, not 7.
• Question 21 (equal groups): Dividing by $3!$ or $2!$ because the groups “look alike” is wrong here. The two experimental groups and the control group are named differently, so the groups are labelled.

Practise next

• Continue with the other pages of this unit: Objective Type and Fill in the Blanks, Q26–50 and True or False and Matching, Q51–64.
• NCERT textbook: Exercise 6.1 for the counting principle and factorials, Exercise 6.2 for arrangements, and Exercise 6.3 for selections.
• Miscellaneous Exercise on Chapter 6 for mixed restrictions of the kind in Questions 10 to 12 and 24 to 25.