OP Malhotra Class 11 Maths Solutions Chapter 12 Permutations and Combinations Chapter Test

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S Chand Class 11 ICSE Maths Solutions Chapter 12 Permutations and Combinations Chapter Test

Question 1.
Find the value of (7, 3).
Solution:
P(7, 3) = ⁷P₃ = \(\frac{7 !}{(7-3) !}\) = \(\frac{7 !}{4 !}\) = \(\frac{7 \times 6 \times 5 \times 4 !}{4 !}\) = 210

Question 2.
How many numbers of S digits can be formed with the digits 0, 2, 5, 6, 7 without taking any of these digits more than once.
Solution:
Given digits are 0, 2, 5, 6, 7
Ten thousand place can be filled by 4 digits (excluding 0) in 4 ways.

Thousand place can be filled by 4 ways (including 0) since repetitions of digits is not allowed. Hundred place can be filled in 3 ways
Ten place can be filled by 2 ways and unit place is filled by 1 way.
∴ required no. of 5 digit numbers = 4 × 4 × 3 × 2 × 1 = 96

Question 3.
In how many ways can 8 persons sit in a round table.
Solution:
Required no. of ways in which 8 persons sits round a table = (8 – 1)! = 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

Question 4.
A man has to post 5 letters and there are 4 letter boxes in the locality, in how many ways can he post the letters ?
Solution:
For each letter, man have 4 choices i.e. he can post the letter to any of the four letter boxes since man has to post 5 letters.
Thus required no. of ways in which he post all the letters = 4 × 4 × 4 × 4 × 4 = 4⁵

Question 5.
If ⁴P₂ = n.⁴C₂, find the value of n.
Solution:
Given ⁴P₂ = n.⁴C₂ ⇒ \(\frac{4 !}{2 !}\) = n.\(\frac{4 !}{2 ! 2 !}\) ⇒ n = 2! = 2

Question 6.
A man has 6 friends. In how many ways may he invite one or more of them to dinner ?
Solution:
(i) Since a man can deal with his friend in two ways either invite him or not i.e. in 2 different ways.
∴ required number of ways he invite one or more of his friends to a dinner = 2⁶ – 1 = 64 – 1 = 63
[where one possibility that he does not invite any of his friend is to be subtracted]
Aliter: Since a man has 6 friends so he can invite 1 or 2 or 3 or 4 or 5 or 6 friends to a party.
∴ required no. of ways = ⁶C₁ + ⁶C₂ + ⁶C₃ + ⁶C₄ + ⁶C₅ + ⁶C₆ = 2⁶ – 1 = 64 – 1 = 63
[∵ ⁿC₀ + ⁿC₁ + ⁿC₂ + …… + ⁿC_n = 2ⁿ]

Question 7.
In how many ways 5 members forming a committee out of 10 be selected so that
(i) 2 particular members must be included.
(ii) 2 particular members must not be included.
Solution:
(i) We want to find the no. of ways in which 5 member committee contain 2 particular members i. e. we have to select remaining 3 members from remaining (10 – 2) members and this can be done in ⁸C₃ ways
i.e. \(\frac{8 !}{5 ! 3 !}\) ways = \(\frac{8 \times 7 \times 6}{6}\) ways i.e. 56 ways

(ii) We want to find the number of ways in which 5 member committee does not include 2 particular members.
i.e. we have to select 5 members out of (10 – 2) members and this can be done in ⁸C₃ ways i.e. 56 ways

Question 8.
Find the number of different words that can be formed from 12 consonants and 5 vowels by taking 4 consonants and 3 vowels in each word.
Solution:
Give no. of consonants = 12 and no. of vowels = 5
Thus total no. of combinations formed by taking 4 consonants and 3 vowels = ¹²C₄ × ⁵C₃
each of these combination give rise to 7! words.
Thus required no. of words = ¹²C₄ × ⁵C₃ × 7! = \(\frac{12 !}{8 ! 4 !}\) × \(\frac{5 !}{3 ! 2 !}\) × 7!
= \(\frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1}\) × \(\frac{5 \times 4}{2}\) × 7 × 6 × 5 × 4 × 3 × 2 = 24948000

Question 9.
In an examination there are three multiple choice questions and each question has 4 choices. Find the number of ways in which a student can fail to get all answers correct.
Solution:
Since it is given that, each multiple choice question has 4 choices. Thus each multiple choice question can be answered in 4 ways and we have 3 multiple choice questions. Further, all questions can be answered correctly in only one way.
∴ required no. of ways in which a student can fail to get all answers correct
= (4 × 4 × 4) – 1 = 63.

Question 10.
Find the number of ways in which 12 apples may be equally divided among 3 childrens.
Solution:
Since 12 apples may be equally divided among 3 childrens so that each child get 4 apples.
Thus required no. of ways in which this can be done in \(\frac{12 !}{4 ! 4 ! 4 !}\) = 34650

Question 11.
There are two urns. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each urn two balls are taken out at random and then transferred to the other. The number of ways in which this can be done is.
(a) 3
(b) 36
(c) 66
(d) 108
Solution:
No. of distinct red balls in Urn A = 3
No. of distinct blue balls in Urn B = 9
∴ No. of ways of selecting 2 red balls out of 3 = ³C₂
no. of ways of selecting 2 blue balls out of 9 = ⁹C₂
Thus required no. of ways = ³C₂ × ⁹C₂ = 3 × \(\frac{9 !}{7 ! 2 !}\) = 3 × \(\frac{9 \times 8}{2}\) = 108
∴ Ans. (d)

Question 12.
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 each group. The number of ways in which he can choose the 7 questions is
(a) 780
(b) 640
(c) 820
(d) None of these
Solution:
Total no. of questions = 12
no. of questions in each group = 6
and candidate is not permitted to attempt more than 5 questions from each group. We want to find the no. of ways in which he can choose 7 questions.
and this can be done in following ways
(a) 2 questions from group I and 5 from group-II
(b) 3 questions from group I and 4 from group-II
(c) 4 questions from group I and 3 from group-II
(d) 5 questions from group I and 2 questions from group-II
Thus required no. of ways =

Question 13.
The number of permutations by taking all letters and keeping the vowels of the word COMBINE in the odd places is.
(a) 96
(b) 144
(c) 512
(d) 576
Solution:
In word COMBINE, No. of consonants = 4 and no. of vowels = 3
First we fix the position of consonants, 4 consonants placed in 4 places in \(\lfloor 4\) ways.
Since vowels occupy odd places and we have 4 odd places
Thus the no. of ways in which vowels can be arranged = ⁴P₃

∴ required no. of permutations = 4 ! × ⁴P₃ = 24 × \(\frac{4 !}{(4-3) !}\) = 24 × 24 = 576
∴ Ans. (d)

Question 14.
if ⁵⁶P_r+6; ⁵⁴P_r+3 = 30800, then, r is
(a) 40
(b) 51
(c) 101
(d) 410
(e) 41
Solution:

Question 15.
For 2 ≤ r ≤ n, ⁿC_r + 2.ⁿP_r-1 + ⁿP_r-2 =
(a) ⁿ⁺¹P_r-1
(b) 2.ⁿ⁺²P_r+1
(c) 2.ⁿ⁺²P_r
(d) ⁿ⁺²P_r
Solution:
Given 2 ≤ r ≤ n
Now, ⁿC_r + 2. ⁿC_{r – 1} + ⁿC_{r – 2}
= (ⁿC_r + ⁿP_{r – 1}) + (ⁿC_{r – 1} + ⁿC_{r – 2})
= ⁿ⁺¹C_r + ⁿ⁺¹C_{r – 1} [∵ ⁿC_r + ⁿC_{r – 1} = ⁿ⁺¹C_r]
= ⁿ⁺¹C_r + ⁿ⁺²C_r
∴ Ans. (d)

Question 16.
How many different words can be formed by jumbling the letters in the word MISSISSIPPI in which no two S are adjacent ?
(a) 8 .⁶C₄ : ⁷C₄
(b) 6.7. ⁸C₄
(c) 6.8.⁷C₄
(d) 7.⁶C₄.⁸C₄
Solution:
In word MISSISSIPPI, we have 4 S, 4 I, 2 P’s and 1M. Other than S, remaining 7 letters can be arranged in a row in \(\frac{7 !}{2 ! 4 !}\) ways.

Since no two S’s are adjacent so letter S must be placed at position marked by ×.
Therefore 4 S ‘s can be placed in 8 places in ⁸C₄ ways.
∴ required no. of ways = \(\frac{7 !}{2 ! 4 !}\) × ⁸C₄
= \(\frac{7 \times 6 \times 5}{2}\) × \(\frac{8 !}{4 ! 4 !}\) = 7 × ⁶C₄ × ⁸C₄
∴ Ans. (d)

Question 17.
The number of diagonals of a polygon of 20 sides is
(a) 210
(b) 190
(c) 180
(d) 170
Solution:
Here no. of sides of polygon = n = 20
∴ required no. of diagonals of polygon = ⁿC₂ – n = ²⁰C₂ – 20 = \(\frac{20 !}{2 ! 18 !}\) – 20
= \(\frac{20 \times 19}{2}\) – 20 = 190 – 20 = 170
∴ Ans. (d)

Question 18.
The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by
(a) 6! × 5!
(b) 30
(c) 5! × 4!
(d) 7! × 5!
Solution:
The total no. of ways in which 6 men can sits in round table
=(6 – 1)! = 5!
five ladies will occupy the places marked by ‘×’ so that no two ladies sit together
and this can be done in ⁶P₅ ways.

∴ required no. of ways = 5! × ⁶P₅
= 5! × \(\frac{6!}{1!}\) = 5! × 6!
∴ Ans. (a)

Question 19.
Let T_n denote the number of triangles which can be formed using the vertices of a regular polygon of n sides. If T_n+1 – T_n = 21, then n =
(a) 5
(b) 7
(c) 6
(d) 4
Solution:
Since we know that, a triangle is formed by joining 3 points (non-collinear) in plane.
∴ T_n = no. of Δ’s formed using the vertices of regular polygon of n sides = ⁿP₃

Question 20.
If ⁴³P_r-6 = ⁴³P_3r+1, then the value of r is
(a) 12
(b) 8
(c) 6
(d) 0
Solution:
Given ⁴³P_r-6 = ⁴³P_3r+1
⇒ r – 6 = 3r + 1 or r – 6 + 3r + 1 = 43 [if ⁿC_r = ⁿC_s then either r = s or r + s = n]
⇒ 2r = -7 or 4r = 48
⇒ r = –\(\frac { 7 }{ 2 }\) or r = 12
But r ∈ N
∴ r = 12
∴ Ans. (a)

Question 21.
A committee of 7 members has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of
(a) exactly 3 girls
(b) at least 3 girls
(c) atmost three girls
Solution:
Given no. of boys = 9 and no. of girls = 4
(i) Now we want to find no. of ways in which 7 member committee consists of exactly 3 girls we have to select 4 members out of 9 boys and this can be done in ⁹C₄ ways and no. of ways of selecting 3 girls out of 4 be ⁴C₃.
∴ required no. of ways = ⁹C₄ × ⁴C₃ = \(\frac{9 !}{4 ! 5 !} \) × \(\frac{4 !}{3 !}\) = \(\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2}\) × 4 = 504

(ii) We want to find the no. of ways in which 7 member committee consists of atleast 3 girls and this can be done in following ways
(a) 3 girls and 4 boys
(b) 4 girls and 3 boys
∴ required no. of such committees = ⁴C₃ × ⁹C₄ + ⁴C₄ × ⁹C₃ = 4 × \(\frac{9 !}{5 ! 4 !}\) + 1 × \(\frac{9 !}{6 ! 3 !}\)
= 4 × \(\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}\) + 1 × \(\frac{9 \times 8 \times 7}{6}\) = 504 + 84 = 588

(iii) We want to find the no. of ways in which 7 member committee consists of atmost three girls. This can be done in following ways
(a) 7 boys and 0 girls
(b) 6 boys and 1 girls
(c) 5 boys and 2 girls
(d) 4 boys and 3 girls
∴ required no. of ways =

Question 22.
In how many ways can the letters of the word ‘PERMUTATIONS’ be arranged if the
(a) words start with P and end with S
(b) vowels are all together
Solution:
(a) Total no. of letters in word PERMUTATIONS = 12
Also there are 10 letters between letters P and S (which are fixed) with 2 T’s
∴ required no. of arrangements = \(\frac{10 !}{2 !}\) = \(\frac{10 \times 9 !}{2}\)
= 5 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1=18,14,400

(b) Here no. of vowels = 5 {E, U, A, I, 0} and no. of consonants = 7 with 2 T’s
Now five vowels forms a group and this group and remaining 7 consonants i.e. in total 8 letters can be arranged in \(\frac{8 !}{2 !}\) ways
Also, all the five vowels can be arranged in 5! ways.
∴ Total no. of arrangements = \(\frac{8 !}{2 !}\) × 5! = \(\frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2}{2}\) × 5 × 4 × 3 × 2 = 20160 × 120 = 24, 19, 200

Question 23.
In how many different ways can the letters of the word ‘SALOON’ be arranged
(i) if the two O’s must not come together ?
(it) if the consonants and vowels must occupy alternate places ?
Solution:
(i) Here total no. of ways in which the letters of the word SALOON be arranged = \(\frac{6 !}{2 !}\)
= \(\frac{6 \times 5 \times 4 \times 3 \times 2}{2}\) = 360
No. of arrangement when 2 O’s must come together = \(\lfloor 5\) = 120
[since 2 O’s forms a one group and this group and remaining four letters can be arranged in 5! ways]
∴ required no. of arrangements in which two O’s must not come together
= Total no. of arrangements – Total no. of arrangements in which 2 O’s come together = 360 – 120 = 240

(ii) No. of arrangements in which vowels occupy even places and consonants occupy odd places

Total no. of arrangements in which vowels occupy odd places and consonants occupy even places

Thus required no. of arrangements in which consonants and vowels must occupy alternate places = 18 + 18 = 36