NCERT Exemplar Solutions Class 11 Maths Permutations and Combinations — True or False and Matching Q51–64

NCERT Exemplar Problems, Class 11 Mathematics — Permutations and Combinations, True or False and Matching questions 51 to 64. All 14 questions solved.

Nine true-or-false statements (51 to 59), each needing a justification, then five matching questions (60 to 64) in which every part is a short count. For a true-or-false statement the safest method is to do the count yourself and compare, not to judge whether the formula “looks right”. The statements are mostly one word away from being correct.

One answer in the printed key is wrong, Question 61 part (c). It is explained where it occurs.

Key insight. To test a statement, count the thing it describes from first principles and see whether the stated formula gives the same number. Several of these statements swap $5$ and $3$, or $20$ and $22$, which only a fresh count exposes.

Question 51

State whether the statement is True or False. Also give justification.

There are 12 points in a plane of which 5 points are collinear, then the number of lines obtained by joining these points in pairs is ${}^{12}C_2 – {}^5C_2$.

Solution. The statement is false. The ${}^5C_2 = 10$ pairs among the collinear points do not give 10 lines but a single line, so after subtracting all ten we have to add that one line back. The correct count is ${}^{12}C_2 – {}^5C_2 + 1 = 66 – 10 + 1 = 57$, not $56$.

Question 52

Three letters can be posted in five letterboxes in $3^5$ ways.

Solution. The statement is false. Each of the three letters is independent and can go in any of the 5 boxes, so there are $5 \times 5 \times 5 = 5^3 = 125$ ways. The statement has the base and the exponent the wrong way round ($3^5 = 243$).

Question 53

In the permutations of $n$ things, $r$ taken together, the number of permutations in which $m$ particular things occur together is ${}^{n-m}P_{r-m} \times {}^rP_m$.

Solution. The statement is false. Choose the other $r – m$ things from the remaining $n – m$ (unordered, ${}^{n-m}C_{r-m}$), join the $m$ particular things into one block, arrange the $r – m + 1$ objects, and order the things inside the block: the correct number is ${}^{n-m}C_{r-m}\,(r-m+1)!\,m!$. The stated formula does not equal it. A small test shows this: with $n = 5$, $r = 3$ and $m = 2$ (two particular things together in a line of three drawn from five), listing the arrangements gives 12, while the stated formula gives ${}^3P_1 \times {}^3P_2 = 3 \times 6 = 18$.

Question 54

In a steamer there are stalls for 12 animals, and there are horses, cows and calves (not less than 12 each) ready to be shipped. They can be loaded in $3^{12}$ ways.

Solution. The statement is true. There are 12 stalls, and each stall can take a horse, a cow or a calf, so it has 3 choices. Because there are at least 12 animals of each kind, the supply never runs out, so the choices for different stalls are independent and the total is $3^{12}$.

Question 55

If some or all of $n$ objects are taken at a time, the number of combinations is $2^n – 1$.

Solution. The statement is true. Each object is either taken or left, giving $2^n$ choices for the whole set. One of them leaves everything out, which is not “some or all”, so we subtract it: ${}^nC_1 + {}^nC_2 + \cdots + {}^nC_n = 2^n – 1$.

Question 56

There will be only 24 selections containing at least one red ball out of a bag containing 4 red and 5 black balls. It is being given that the balls of the same colour are identical.

Solution. The statement is true. Because balls of one colour are identical, a selection is described only by how many red and how many black balls it contains. Red can be $1, 2, 3$ or $4$ (at least one): 4 choices. Black can be $0, 1, \ldots, 5$: 6 choices. So there are $4 \times 6 = 24$ selections.

Question 57

Eighteen guests are to be seated, half on each side of a long table. Four particular guests desire to sit on one particular side and three others on other side of the table. The number of ways in which the seating arrangements can be made is $\dfrac{11!}{5!\,6!}(9!)(9!)$.

Solution. The statement is true. Seat the seven fussy guests first: the four go on the given side and the three on the other. That leaves 11 guests, and each side needs 9 in all, so 5 more must go on the first side and 6 on the second. Choosing which 5 of the 11 go on the first side gives ${}^{11}C_5 = \dfrac{11!}{5!\,6!}$ ways. Now each side has 9 people, who can be seated along their side in $9!$ ways, and the two sides are independent: $(9!)(9!)$.

Question 58

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. He can choose the seven questions in 650 ways.

Solution. The statement is false. This is Question 3 of this exercise (short answer set): the splits $(5,2), (4,3), (3,4), (2,5)$ give $2\,({}^6C_5\,{}^6C_2 + {}^6C_4\,{}^6C_3) = 2(90 + 300) = 780$. Equivalently, ${}^{12}C_7 – 12 = 792 – 12 = 780$. So 650 is wrong.

Question 59

To fill 12 vacancies there are 25 candidates of which 5 are from scheduled castes. If 3 of the vacancies are reserved for scheduled caste candidates while the rest are open to all, the number of ways in which the selection can be made is ${}^5C_3 \times {}^{20}C_9$.

Solution. The statement is false. Select the 3 scheduled-caste candidates for the reserved vacancies from the 5: ${}^5C_3 = 10$. The other $12 – 3 = 9$ vacancies are open to all, so they are filled from everyone who is left, which is $25 – 3 = 22$ candidates (this includes the 2 scheduled-caste candidates not already chosen). So the second factor should be ${}^{22}C_9$, not ${}^{20}C_9$: the correct count is ${}^5C_3 \times {}^{22}C_9$.

Question 60

Match each item given under the column $C_1$ to its correct answer given under the column $C_2$.

There are 3 books on Mathematics, 4 on Physics and 5 on English. How many different collections can be made such that each collection consists of:

Solution.

• (a) One book from each subject: $3 \times 4 \times 5 = 60$, which is (ii).
• (b) At least one of each subject means a non-empty selection from each subject, and a subject with $k$ books gives $2^k – 1$ non-empty selections: $(2^3 – 1)(2^4 – 1)(2^5 – 1) = 7 \times 15 \times 31 = 3255$, which is (iii).
• (c) At least one English book: the English books give $2^5 – 1 = 31$ non-empty selections, and the other $3 + 4 = 7$ books can be taken in any way, $2^7 = 128$: $31 \times 128 = 3968$, which is (i).

Question 61

Five boys and five girls form a line. Find the number of ways of making the seating arrangement under the following condition:

Solution.

• (a) Alternating lines start with a boy or with a girl, 2 patterns. In each pattern the boys are arranged in $5!$ ways and the girls in $5!$ ways: $2 \times 5! \times 5! = 28800$. This number is written both as (iii) $(5!)^2 + (5!)^2$ and as (iv) $2!\,5!\,5!$, which are the same value, so either fits; the key’s choice of (iii) is fine.
• (b) No two girls together: line up the five boys first, $5!$ ways, creating 6 gaps. Put the five girls into five different gaps, in order, ${}^6P_5 = 6!$ ways. Total $5! \times 6!$, which is (i).
• (c) All the girls together: treat the five girls as one block. Together with the five boys that makes 6 objects, arranged in $6!$ ways, and the girls can be ordered inside the block in $5!$ ways: $6! \times 5!$, which is again (i). Both (b) and (c) have the same value, since the number of lines with no two girls adjacent happens to equal the number with the girls all in one block for five of each.
• (d) All the girls never together means the lines other than those in (c): $10! – 5!\,6!$, which is (ii).

The printed key says (c) $\leftrightarrow$ (iv), which is wrong. The expression $2!\,5!\,5!$ is the count for alternating lines (part a), not for lines with all five girls in a block; its value is $28800$, while the block count is $86400$.

Question 62

There are 10 professors and 20 lecturers out of whom a committee of 2 professors and 3 lecturer is to be formed. Find:

Solution.

• (a) Two professors from 10 and three lecturers from 20: ${}^{10}C_2 \times {}^{20}C_3$, which is (iv).
• (b) A particular professor is in: only 1 more professor is chosen, from the other 9, and the lecturers are as before: ${}^9C_1 \times {}^{20}C_3$, which is (iii).
• (c) A particular lecturer is in: only 2 more lecturers, from the other 19: ${}^{10}C_2 \times {}^{19}C_2$, which is (ii).
• (d) A particular lecturer is out: all 3 lecturers come from the other 19: ${}^{10}C_2 \times {}^{19}C_3$, which is (i).

The numerical values are $51300$, $10260$, $7695$ and $43605$ respectively.

Question 63

Using the digits 1, 2, 3, 4, 5, 6, 7, a number of 4 different digits is formed. Find

Solution.

• (a) Four different digits from seven, arranged: ${}^7P_4 = 7 \times 6 \times 5 \times 4 = 840$, which is (i).
• (b) Divisible by 2 needs an even last digit: 3 choices (2, 4, 6). The other three places take different digits from the remaining 6: $6 \times 5 \times 4 = 120$. So $3 \times 120 = 360$, which is (iii).
• (c) Divisible by 25 needs the last two digits to be 00, 25, 50 or 75. The digit 0 is not among 1 to 7, so only 25 and 75: 2 endings. The first two places take different digits from the remaining 5: $5 \times 4 = 20$. So $2 \times 20 = 40$, which is (iv).
• (d) Divisible by 4 needs the last two digits to form a multiple of 4. The pairs from different digits among 1 to 7 are 12, 16, 24, 32, 36, 52, 56, 64, 72, 76: ten endings. The first two places take different digits from the other 5: $5 \times 4 = 20$. So $10 \times 20 = 200$, which is (ii). A full enumeration of the 840 numbers confirms $360$, $40$ and $200$.

Question 64

How many words (with or without dictionary meaning) can be made from the letters of the word MONDAY, assuming that no letter is repeated, if

Solution. MONDAY has 6 different letters, of which O and A are vowels.

• (a) Four of the six letters, in order: ${}^6P_4 = 6 \times 5 \times 4 \times 3 = 360$, which is (iii).
• (b) All six in order: $6! = 720$, which is (i).
• (c) The first letter is a vowel, 2 choices (O or A), and the other five letters are arranged in $5! = 120$ ways: $2 \times 120 = 240$, which is (ii).

Common mistakes

• Question 51 (collinear points): Leaving out the “$+1$”. Students subtract ${}^5C_2$ because those pairs are not new lines, but the five collinear points do still make one line.
• Question 52 (letters in letterboxes): Counting from the boxes’ side, giving $3^5$. The letters are the ones making a choice, each picking a box, so the base is the number of boxes and the exponent is the number of letters.
• Question 53 (permutations with $m$ things together): Treating the formula as obviously true. The $m$ particular things are glued into a block, so the count involves $(r-m+1)!$, not a permutation of all $r$ in the second factor.
• Question 59 (reserved vacancies): Taking the remaining pool as $20$, the non-scheduled-caste candidates. After reserving three seats for scheduled-caste candidates, the other 9 seats are open to everyone left, including the 2 scheduled-caste candidates not yet chosen, so the pool is 22.
• Question 61 (line of boys and girls): Matching 2!5!5! to “all girls together” because it contains 5! twice. That expression is the alternating count; a block of girls needs $6!\times5!$, as the block is one object among six.
• Question 63(d) (divisible by 4): Testing only the last digit being even. Divisibility by 4 depends on the last two digits together, so list the allowed two-digit endings first.

Practise next

• The other pages of this unit: Short and Long Answer Type, Q1–25 and Objective Type and Fill in the Blanks, Q26–50.
• NCERT textbook: Exercise 6.1 for the multiplication principle, Exercise 6.2 for arrangements with restrictions, and Exercise 6.3 for selections.
• The Miscellaneous Exercise on Chapter 6 for further mixed counting such as Questions 61 to 64.