NCERT Exemplar Solutions Class 11 Maths Permutations and Combinations — Objective Type and Fill in the Blanks Q26–50

NCERT Exemplar Problems, Class 11 Mathematics — Permutations and Combinations, Objective Type and Fill in the Blanks questions 26 to 50. All 25 questions solved.

Fifteen multiple-choice questions (26 to 40) followed by ten fill-in-the-blank questions (41 to 50). They test the same few ideas in different clothes: ${}^nC_r = {}^nC_{n-r}$ (26, 42), the multiplication principle (27, 38, 46), numbers and words made from restricted digits or letters (28 to 31, 36, 39, 44), pair-counting through ${}^nC_2$ (32, 33, 34, 47) and selections with “at least” conditions (37, 40, 45, 49, 50).

All 25 printed answers match the working below. One printing slip is worth noticing: in Question 40 two of the four options are the same number.

Key insight. Most of these are one line once you name the object. Handshakes, matches and diagonals are pairs, so use ${}^nC_2$. Choices that are independent multiply, and a choice that is “chosen or not chosen” doubles, giving $2^k$.

Question 26

If ${}^nC_{12} = {}^nC_8$, then $n$ is equal to

Solution. Two combinations with the same $n$ are equal when the lower indices are equal or add up to $n$, because ${}^nC_r = {}^nC_{n-r}$. Here $12 \ne 8$, so $12 + 8 = n$.

Question 27

The number of possible outcomes when a coin is tossed 6 times is

Solution. Each toss has 2 outcomes and the tosses are independent, so the total is $2^6$. (Option (A) uses $6^2$ and option (C) uses $2 \times 6$, which are the wrong way round.)

Question 28

The number of different four digit numbers that can be formed with the digits 2, 3, 4, 7 and using each digit only once is

Solution. Four different digits fill four places in order, in $4! = 24$ ways. None of the digits is 0, so no number is lost by a leading zero.

Question 29

The sum of the digits in unit place of all the numbers formed with the help of 3, 4, 5 and 6 taken all at a time is

Solution. There are $4! = 24$ numbers. By symmetry each digit takes the units place equally often: fixing a digit there leaves $3! = 6$ arrangements of the others, so each of 3, 4, 5, 6 appears 6 times. The sum of the unit digits is $6\,(3 + 4 + 5 + 6) = 6 \times 18$.

Question 30

Total number of words formed by 2 vowels and 3 consonants taken from 4 vowels and 5 consonants is equal to

Solution. First select, then arrange. Choose 2 of 4 vowels and 3 of 5 consonants: ${}^4C_2 \times {}^5C_3 = 6 \times 10 = 60$. Each such set of five letters can be arranged into a word in $5! = 120$ ways. Option (A) stops after the selection.

Question 31

A five digit number divisible by 3 is to be formed using the numbers 0, 1, 2, 3, 4 and 5 without repetitions. The total number of ways this can be done is

Solution. A number is divisible by 3 when its digit sum is. The six digits add up to $15$, and a five-digit number leaves one digit out, so the left-out digit must keep the remaining sum a multiple of 3. Omitting 0 gives sum 15, and omitting 3 gives sum 12; omitting any other digit gives a sum that is not a multiple of 3.

• Digits 1, 2, 3, 4, 5: no zero, so $5! = 120$ numbers.
• Digits 0, 1, 2, 4, 5: zero cannot lead, so $4 \times 4! = 96$ numbers.

Total $120 + 96 = 216$.

Question 32

Every body in a room shakes hands with everybody else. The total number of hand shakes is 66. The total number of persons in the room is

Solution. A handshake is a pair of people, so ${}^nC_2 = 66$, that is $n(n-1) = 132 = 12 \times 11$. Hence $n = 12$.

Question 33

The number of triangles that are formed by choosing the vertices from a set of 12 points, seven of which lie on the same line is

Solution. Any three points make a triangle unless all three are collinear. Total choices ${}^{12}C_3 = 220$; the three-point choices taken from the seven collinear points, ${}^7C_3 = 35$, form no triangle. So $220 – 35$.

Question 34

The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines is

Solution. A parallelogram is bounded by two lines from the first set and two from the second. Choose them independently: ${}^4C_2 \times {}^3C_2 = 6 \times 3$.

Question 35

The number of ways in which a team of eleven players can be selected from 22 players always including 2 of them and excluding 4 of them is

Solution. Of the 22 players, 2 are certainly in and 4 are certainly out, so $22 – 2 – 4 = 16$ players remain to choose from. Only $11 – 2 = 9$ places are still open: ${}^{16}C_9$ (equal to ${}^{16}C_7$, which is not one of the options).

Question 36

The number of 5-digit telephone numbers having atleast one of their digits repeated is

Solution. A telephone number may begin with 0, so there are $10^5 = 100000$ five-digit numbers. Those with all digits different number ${}^{10}P_5 = 30240$. “At least one digit repeated” is the complement: $100000 – 30240$. Option (C) is the number with no repetition, and option (A) wrongly bars a leading zero.

Question 37

The number of ways in which we can choose a committee from four men and six women so that the committee includes at least two men and exactly twice as many women as men is

Solution. If there are $m$ men there must be $2m$ women, with $m \ge 2$, $m \le 4$ and $2m \le 6$. That allows $m = 2$ or $m = 3$ only, since $m = 4$ would need 8 women.

• 2 men and 4 women: ${}^4C_2 \times {}^6C_4 = 6 \times 15 = 90$.
• 3 men and 6 women: ${}^4C_3 \times {}^6C_6 = 4 \times 1 = 4$.

Total $90 + 4 = 94$.

Question 38

The total number of 9 digit numbers which have all different digits is

Solution. The first digit cannot be 0: 9 choices. The second digit can be any of the remaining 9 digits (now 0 is allowed), then 8, and so on down to 2 choices for the ninth digit. That is $9 \times (9 \times 8 \times \cdots \times 2) = 9 \times 9!$.

Question 39

The number of words which can be formed out of the letters of the word ARTICLE, so that vowels occupy the even place is

Solution. ARTICLE has 7 letters, so the even places are 2, 4 and 6, three of them. The three vowels A, I, E go into those places in $3! = 6$ ways. The four consonants R, T, C, L fill the other four places in $4! = 24$ ways. Multiply: $6 \times 24$.

Question 40

Given 5 different green dyes, four different blue dyes and three different red dyes, the number of combinations of dyes which can be chosen taking at least one green and blue dye is

Solution. Each dye is chosen or not chosen, so the 5 green dyes can be taken in $2^5$ ways, of which one (taking none) is not allowed: $2^5 – 1 = 31$. Blue dyes likewise: $2^4 – 1 = 15$. Red dyes have no condition, so any subset, including none, is allowed: $2^3 = 8$. Multiply: $31 \times 15 \times 8 = 3720$.

The book prints options (A) and (D) both as 3600. That is a misprint in the question as printed (one of them was presumably meant to be a different number); it does not affect the answer, because only (B) is 3720.

Question 41

If ${}^nP_r = 840$, ${}^nC_r = 35$, then $r = \_\_\_\_$.

Solution. ${}^nP_r = r!\,\cdot{}^nC_r$, so $r! = \dfrac{840}{35} = 24 = 4!$ and $r = 4$. (Then ${}^nP_4 = 840 = 7 \cdot 6 \cdot 5 \cdot 4$ gives $n = 7$, and ${}^7C_4 = 35$ confirms it.)

Question 42

${}^{15}C_8 + {}^{15}C_9 – {}^{15}C_6 – {}^{15}C_7 = \_\_\_\_$.

Solution. Use ${}^nC_r = {}^nC_{n-r}$ with $n = 15$: ${}^{15}C_9 = {}^{15}C_6$ and ${}^{15}C_8 = {}^{15}C_7$. The first and fourth terms cancel, and so do the second and third.

Question 43

The number of permutations of $n$ different objects, taken $r$ at a line, when repetitions are allowed, is $\_\_\_\_$.

Solution. Each of the $r$ places in the line can be filled by any of the $n$ objects, whatever was used before, so the choices multiply: $n \times n \times \cdots \times n$ ($r$ times).

Question 44

The number of different words that can be formed from the letters of the word INTERMEDIATE such that two vowels never come together is $\_\_\_\_$.

Solution. INTERMEDIATE has 12 letters. The vowels are I, E, I, A, E, E, that is 6 vowels (I twice, E three times, A once), and the consonants are N, T, R, M, D, T, that is 6 consonants with T repeated twice.

Arrange the consonants first: $\dfrac{6!}{2!} = 360$ ways. They create $7$ gaps, and putting at most one vowel in each gap keeps all vowels apart. Six of the seven gaps are used, and the vowels are placed in order with repeats: ${}^7P_6 \times \dfrac{1}{3!\,2!} = \dfrac{5040}{12} = 420$ ways (the division removes the 3 identical E’s and 2 identical I’s). Multiply: $360 \times 420 = 151200$.

Question 45

Three balls are drawn from a bag containing 5 red, 4 white and 3 black balls. The number of ways in which this can be done if at least 2 are red is $\_\_\_\_$.

Solution. At least 2 red means exactly 2 red (with one non-red from the other $4 + 3 = 7$) or all 3 red.

$${}^5C_2 \times {}^7C_1 + {}^5C_3 = 70 + 10 = 80.$$

Question 46

The number of six-digit numbers, all digits of which are odd is $\_\_\_\_$.

Solution. Each of the six places takes any one of the five odd digits 1, 3, 5, 7, 9, with repetition allowed, so $5^6 = 15625$.

Question 47

In a football championship, 153 matches were played. Every two teams played one match with each other. The number of teams, participating in the championship is $\_\_\_\_$.

Solution. One match per pair of teams: ${}^nC_2 = 153$, so $n(n-1) = 306 = 18 \times 17$ and $n = 18$.

Question 48

The total number of ways in which six ‘+’ and four ‘$-$’ signs can be arranged in a line such that no two signs ‘$-$’ occur together is $\_\_\_\_$.

Solution. The six plus signs are identical, so there is only one way to line them up. They leave $7$ gaps (before, after and between). To keep the minus signs apart, put each in a different gap; the four minus signs are identical, so we only choose which 4 gaps: ${}^7C_4 = 35$.

Question 49

A committee of 6 is to be chosen from 10 men and 7 women so as to contain atleast 3 men and 2 women. In how many different ways can this be done if two particular women refuse to serve on the same committee.

Solution. First count all committees of 6 with at least 3 men and at least 2 women. With 6 members the splits are 4 men and 2 women, or 3 men and 3 women:

$${}^{10}C_4 \times {}^7C_2 + {}^{10}C_3 \times {}^7C_3 = 210 \times 21 + 120 \times 35 = 4410 + 4200 = 8610.$$

Now remove the committees that contain both of the two particular women. With 4 men and those 2 women: ${}^{10}C_4 = 210$. With 3 men, those 2 women and one more woman from the other 5: ${}^{10}C_3 \times {}^5C_1 = 600$. Together $810$ committees break the rule. The answer is $8610 – 810$.

Question 50

A box contains 2 white balls, 3 black balls and 4 red balls. The number of ways three balls be drawn from the box if at least one black ball is to be included in the draw is $\_\_\_\_$.

Solution. This is the same problem as Question 14 of this exercise. All draws of three from 9 balls: ${}^9C_3 = 84$. Draws with no black ball, taken from the other 6: ${}^6C_3 = 20$. So $84 – 20$.

Common mistakes

• Question 26 (${}^nC_{12} = {}^nC_8$): Students write $n = 12 – 8 = 4$ or set the lower indices equal. The identity ${}^nC_r = {}^nC_{n-r}$ says the two lower indices must add to $n$.
• Question 30 (words from vowels and consonants): Picking the letters gives 60, but a word is an arrangement, not a set, so you must multiply by $5!$. Option (A) is put there for students who stop at the selection.
• Question 36 (telephone numbers): Students bar a leading 0 and compute $9 \times 10^4$ or $9 \times {}^9P_4$, as they would for a five-digit number. A telephone number may start with 0, so the totals are $10^5$ and ${}^{10}P_5$.
• Question 38 (nine-digit numbers, all different): Choosing the first digit as 10 choices and getting $10!$ lets 0 start the number. Fix the first digit from 9 non-zero options before counting the rest.
• Question 40 (dyes): Using $2^5 \times 2^4 \times 2^3 – 1$ or forgetting that the red dyes are optional. “At least one” applies to green and blue only, so those two get $2^k – 1$ and red gets $2^3$.
• Question 44 (INTERMEDIATE): Treating the repeated letters as different multiplies the answer by $2!\,3!\,2!$. Divide for the repeated T in the consonants and for the repeated E and I in the vowels.

Practise next

• The other pages of this unit: Short and Long Answer Type, Q1–25 and True or False and Matching, Q51–64.
• NCERT textbook: Exercise 6.2 for arrangements with restrictions and repeated letters, and Exercise 6.3 for combinations.
• Exercise 6.4 for committee-type selections, and the Miscellaneous Exercise on Chapter 6 for mixed problems like Questions 31 and 49.