OP Malhotra Class 11 Maths Solutions Chapter 12 Permutations and Combinations Ex 12(e)

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S Chand Class 11 ICSE Maths Solutions Chapter 12 Permutations and Combinations Ex 12(e)

Question 1.
In how many ways can a committee of 8 be chosen from 10 individuals ?
Solution:
The required no. of ways in which a committee of 8 be chosen from 10 individuals = ¹⁰C⁸
= \(\frac{10 !}{2 ! 8 !}\) = \(\frac{10 \times 9}{2}\) = 45

Question 2.
In how many ways can a committee of five persons be formed out of 8 members when a particular member is taken every time ?
Solution:
Since we have to form a committee of five in which one particular member is taken every time so we have to form committee of 4 members out of 7.
Thus required no. of ways = ⁷C₄ = \(\frac{7 !}{4 ! 3 !}\) = \(\frac{7 \times 6 \times 5 \times 4 !}{4 ! \times 3 !}\) = 35

Question 3.
In how many ways can a committee of 4 be selected out of 12 persons so that a particular person may (i) always be taken, (ii) never be taken ?
Solution:
(i) Since we have to form a committee of 4 out of 12 in which a particular person may always be taken so we have to form a committee of 3 out of 11
∴ required no. of ways = ¹¹C₃ = \(\frac{11 !}{3 ! 8 !}\) = \(\frac{11 \times 10 \times 9}{6}\) = 165

(ii) Now we have to form a committee of 4 out of 12 in which a particular person never be taken so we have to form a committee of 4 out of 11.
Thus required no. of ways = ¹¹C₃ = \(\frac{11 !}{7 ! 4 !}\) = \(\frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2}\) = 330

Question 4.
In how many ways can a team of 11 players be selected from 14 players when two of them can play as goalkeepers only ?
Solution:
Now we have to form a team of 11 players out of 14 in which two of them can play as goalkeeper only. So we have to select one goalkeeper out of 2 and 10 players out of 12.
∴ required no. of ways = ²C₁ × ¹²C₁₀ = 2 × \(\frac{12 !}{10 ! 2 !} \) = 2 × \(\frac{12 \times 11}{2}\) = 132

Question 5.
A person has got 12 friends of whom 8 are relatives. In how many ways can he invite 7 guests such that 5 of them may be relatives ?
Solution:
Given a person has 12 friends of whom 8 are relatives and he invite 7 guests s.t 5 of them are relatives.
Total no. of ways of inviting 5 relatives out of 8 be ⁸C₅ and remaining 2 guests out of 4 be ⁴C₂.
Thus total required no. of ways = ⁸C₅ × ⁴C₂
= \(\frac{8 !}{5 ! 3 !}\) × \(\frac{4 !}{2 ! 2 !}\) = \(\frac{8 \times 7 \times 6 \times 5 !}{5 ! \times 6}\) × \(\frac{3 \times 3 \times 2 \times 1}{2 \times 2}\) = 56 × 6 = 636

Question 6.
How many diagonals are there in a polygon of (i) 8 sides, (ii) 10 sides ?
Solution:
(i) Here n = 8
∴ required no. of diagonals = ⁿC₂ – n = ⁸C₂ – 8 = \(\frac{8 !}{6 ! 2 !}\) – 8 \(\frac{8 \times 7}{2}\) – 8 = 20
(ii) Here n = 10
∴ required no. of diagonals = ⁿC₂ – 10 = \(\frac{10 !}{8 ! 2 !}\) – 10 = \(\frac{10 \times 9 \times 8 !}{8 ! \times 2 !}\) – 10 = 45 – 10 = 35

Question 7.
In how many ways can a committee, consisting of a chairman, secretary, treasurer and four ordinary members be chosen from eight persons ?
(Committees with different chairmen, secretaries, treasurers count as different committees.)
Solution:
Total no. of ways of selecting chairman out of 8 = ⁸C₁
Total no. of ways of selecting secretary out of 7 = ⁷C₁
no. of ways of selecting treasurer out of 6 = ⁶C₁
no. of ways of selecting 4 persons out of 5 = ⁵C₄
Thus, required no. of ways = ⁸C₁ × ⁷C₁ × ⁶C₁ × ⁵C₄ = 8 × 7 × 6 × 5 = 1680

Question 8.
(a) In how many ways can a student choose 5 courses out of 9 courses if 2 courses are compulsory for every student ?
(b) In how many ways ca we select a cricket eleven from 17 players in which 5 players can bowl ? Each cricket team must include 2 bowlers.
Solution:
(a) since 2 courses are compulsory for every student
So, we have to select 3 courses out of 7 for each student and this can be done in ⁷C₃
i.e. \(\frac{7 !}{4 ! 3 !}\) = \(\frac{7 \times 6 \times 5}{6}\) = 35 ways

(b) Total no. of ways of selecting 2 bankers out of 5 = ⁵C₂
and remaining 9 players have to select from 12 and this can be done in ¹²C₉
∴ required no. of ways = ¹²C₉ × ⁵C₂ = \(\frac{12 !}{9 ! 3 !}\) × \(\frac{5 !}{3 ! 2 !}\)
= \(\frac{12 \times 11 \times 10}{6}\) × \(\frac{5 \times 4}{2}\) = 220 × 10 = 2200

Question 9.
How many committees of 5 members each can be formed with 8 officials and 4 non-official members in the following cases :
(i) each consists of 3 officials and 2 non-official members ;
(ii) each contains at least two non-official members ;
(iii) a particular official member is never included ;
(iv) a particular non-official member is always included ?
Solution:
Given no. of official members = 8 and no. of non-official members = 4
(i) total no. of ways of selecting 3 officials out of 8 = ⁸C₃
Total no. of ways of selecting 2 non-official members out of 4 = ⁴C₂
∴ required no. of such committees = ⁸C₃ × ⁴C₂ = \(\frac{8 !}{5 ! 3 !}\) × \(\frac{4 !}{2 ! 2 !}\)
= \(\frac{8 \times 7 \times 6}{6}\) × \(\frac{4 \times 3 \times 2}{2 \times 2}\) = 56 × 6 = 336

(ii) We want to find no. of committees containing atieast two non-official members and this can be done in following ways.
(a) 2 non-official and 3 official members
(b) 3 non-official and 2 official members
(c) 4 non-official and 1 official member
∴ required no. of such committees = ⁴C₂ × ⁸C₃ + ⁴C₃ ×⁸C₂ + ⁴C₄ × ⁸C₁
= \(\frac{4 !}{2 ! 2 !}\) × \(\frac{8 !}{5 ! 3 !}\) + 4 × \(\frac{8 \times 7}{2}\) + 1 × 8
= 336 + 112 + 8 = 456

(iii) We want to find committees in which a particular official is never included.
This can be done in following ways.
(a) 1 non-official and 4 official
(b) 2 non-official and 3 official
(c) 3 non-official and 2 official
(d) 4 non-official and 1 official
(e) 0 non-official and 5 official
∴ required no. of ways = ⁴C₁ × ⁷C₄ + ⁴C₂ × ⁷C₃ + ⁴C₃ × ⁷C₂ + ⁴C₄ × ⁷C₁ + ⁴C₀ × ⁷C₅
= 4 × \(\frac{7 !}{4 ! 3 !}\) + \(\frac{4 !}{2 ! 2 !}\) × \(\frac{7 !}{4 ! 3 !}\) + 4 × \(\frac{7 !}{2 ! 5 !}\) + 1 × 7 + 1 × 21
= 4 × \(\frac{7 \times 6 \times 5}{6}\) + \(\frac{24}{4}\) × \(\frac{7 \times 6 \times 5}{6}\) + 4 × \(\frac{7 \times 6}{2}\) + 7 + 21
= 140 + 210 + 84 + 7 + 21 = 462

(iv) We can’t to find committees in which a particular non-official member is always included. This can be done in following ways.
i.e. we have to selecting remaining 4 members out of 11 and this can be done in ¹¹C₄
= \(\frac{11 !}{7 ! 4 !}\) = \(\frac{11 \times 10 \times 9 \times 8}{24}\) = 330 ways

Question 10.
In a college team there are 15 players of whom 3 are teachers. In how many ways can a team of 11 players be selected so as to include (i) only one teacher, (ii) at least one teacher ?
Solution:
Given there are 15 players of whom 3 are teachers.
(i) Total no. of ways of selecting 1 teacher out of 3 = ³C₁
Total no. of ways of selecting 10 players out of 12 = ¹²C₁₀
∴ required no. of ways = ³C₁ × ¹²C₁₀ = 3 × \(\frac{12 !}{2 ! 10 !}\) = 3 × \(\frac{12 \times 11}{2}\) = 198

(ii) We want to find no. of ways can a team of 11 players be selected so that it include one teacher. This can be done in following ways
(a) One teacher and 10 players
(b) Two teachers and 9 players
(c) Three teachers and 8 players
∴ required no. of ways = ³C₁ × ¹²C₁₀ + ³C₂ × ¹²C₉ + ³C₃ × ¹²C₈
= 3 × \(\frac{12 !}{2 ! 10 !}\) + 3 × \(\frac{12 !}{3 ! 9 !}\) + 1 × \(\frac{12 !}{8 ! 4 !}\)
= 3 × \(\) + 3 × \(\frac{12 \times 11 \times 10}{6}\) + \(\frac{12 \times 11 \times 10 \times 9}{24}\)
= 198 + 660 + 495 = 1353

Question 11.
How many different groups can be selected for playing tennis out of 4 ladies and 3 gentlemen, there being one lady and one gentleman on each side ?
Solution:
Total no. of ways of selecting 1 lady out of 4 = ⁴C₁
no. of ways of selecting 1 gentleman out of 3 = ³C₁
∴ Total no. of ways of selecting 1 lady and 1 gentleman for one side = ⁴C₁ × ³C₁ = 4 × 3 = 12
while for other side
Total no. of ways of selecting 1 lady and 1 gentleman = ³C₁ × ²C₁ = 3 × 2 = 6
Thus total no. of groups = 12 × 6 = 72
Let us assume that four ladies are L₁, L₂, L₃ and L₄ and Gentlemen are G₁, G₂, G₃.
In our above argument, I am assuming that L₁ and G₁ playing as a team L₂ against and G₂ playing as team 2 is a different case when L₂ and G₂ playing s team 1 against L₁ and G₁ playing as team 2.
Thus we have created a distinction between team 1 and team 2. Since only the teams matter not their team number.
∴ required no. of ways = \(\frac { 72 }{ 2 }\) = 36.

Question 12.
If ⁿC₁₀ = ⁿC₁₄, find the value of ⁿC₂₀ and ²⁵C_n.
Solution:
Given ⁿC₁₀ = ⁿC₁₄ ⇒ n = 10 + 14 = 24
[if ⁿC_r = ⁿC_s ⇒ r = s or r + s = n]
∴ ⁿC₂₀ = ²⁴C₂₀ = \(\frac{24 !}{4 ! 20 !}\) = \(\frac{24 \times 23 \times 22 \times 21}{4 \times 3 \times 2 \times 1}\) = 10626
∴ ²⁴C_n = ²⁵C₂₄ = ²⁵C_{25 – 24} = ²⁵C₁ = 25

Question 13.
In how many ways can I invite one or more of six friends to a dinner ?
Solution:
For every friend, person have two choices either to invite or not.
∴ required no. of ways in which he invite one or more of six friends to a dinner
= 2⁶ – 1 = 64 – 1 = 63
[In all possibilities, there is one and only one possibility when person does not invite any of his friends]

Question 14.
In how many ways can 10 marbles be divided between two boys so that one of them may get 2 and the other 8 ?
Solution:
Now 8 marbles out of 10 can be selected in ¹⁰C₈ ways
1 boy out of 2 boys can be selected in ²C₁ ways
Thus required no. of ways = ¹⁰C₈ × ²C₁ = \(\frac{10 !}{2 ! 8 !}\) × 2 = \(\frac{10 \times 9}{2}\) × 2 = 90

Question 15.
In how many ways can a selection be made out of 5 oranges, 8 mangoes and 7 plantains ?
Solution:
Suppose out of(p + q + r +….) things, p are alike of one kind, q are alike of second kind, r alike of third kind and rest different.
Then total no. of selections = (p + 1) (q + 1) (r + 1) 2^t – 1
∴ Required no. of selections = (5 + 1) (8 + 1) (7 + 1) – 1 = 6 × 9 × 8 – 1 = 431

Question 16.
In how many ways can 20 articles be packed in the three parcels so that the first contains 8 articles, the second 7 and the third 5 ?
Solution:
Total no. of ways selecting 8 articles out of 20 for parcel-I = ²⁰C₈
No. of ways of selecting 7 articles out of 12 for parcel-II = ¹²C₇
No. of ways of selecting of 5 articles out of 5 for parcel-III = ⁵C₅
Thus required no. of ways = ²⁰C₈ × ¹²C₇ × ⁵C₅ = \(\frac{20 !}{8 ! 12 !}\) × \(\frac{12 !}{7 ! 5 !}\) × 1 = \(\frac{20 !}{8 ! 7 ! 5 !}\)

Question 17.
Find the number of four letter arrangements of the letters of the word ‘SHOOT’. How many of them begin with O?
Solution:
There are 5 letters in word SHOOT in which there are 2O ‘s, 1S, 1H and 1T.
The following combinations are possible
(i) 1 alike, 2 different = ¹C₁ × ³C₂ = 1 × 3 = 3
(ii) 4 different ⁴C₄ = 1
∴ Total no. of combinations = 3 + 1 = 4
Total no. of permutations = 3 × \(\frac{4 !}{2 ! 1 ! 1 !}\) + 1 × 4! = 3 × 12 + 24 = 60
To fix beginning with O, the no. of 4-letter arrangement in which 4 places to be filled by 3 letters in ⁴C₃ ways i.e. 24 ways.

Question 18.
(i) Find the number of permutations of the letters of the word ‘MATHEMATICS’, taken 4 at a time.
(ii) How many four-letter words can be formed using the letters of the word ‘INEFFECTIVE’ ?
(iii) Find the number of ways in which (a) a selection, (6) an arrangement of four letters can be made from the letters of the word ‘PROPORTION’ ?
Solution:
(i) There are 11 letters in word MATHEMATICS in which there are 2M’s, 2A’s, 2T’s, 1H, IE, 1I, 1C and 1S.
The following combinations are possible :
(a) 2 alike = ³C₂ = 3
(b) 1 alike, 2 different = ³C₁ × ⁷C₂ = 3 × \(\frac{7 !}{5 ! 1 ! 2 !}\) = 63
(c) 4 different = ⁸C₄ = \(\frac{8 !}{4 ! 4 !}\) = \(\frac{8 \times 7 \times 6 \times 5}{24}\) = 70
∴ Total no. of combinations = 3 + 63 + 70 = 136
Thus total no. of permutations = 3 × \(\frac{4 !}{2 ! 2 !}\) + 63 × \(\frac{4 !}{2 !}\) + 70 × 4!
= 3 × \(\frac{24}{4}\) + 63 × 12 + 70 × 24 = 18 + 756 + 1680 = 2454

(ii) There are 11 letters in word INEFFECTIVE with 2 I’s, 3 E’s, 2F’s, 1N, 1C, 1T and 1V.
The following combinations are possible.
(a) ABCD : ⁷C₄ = \(\frac{7 !}{4 ! 3 !}\) = \(\frac{7 \times 6 \times 5}{6}\) = 35
(b) AACD : ³C₁ × ⁶C₂ = 3 × \(\frac{6 !}{4 ! 2 !}\) = 3 × \(\frac{6 \times 5}{2}\) = 45
(c) AABB : ³C₂ = 3
(d) AAAB : ¹C₁ × ⁶C₁ = 1 × 6 = 6
∴ Total no. of permutations =

(iii) There are 10 letters in word PROPORTION with 2 P’s, 2 R’s, 3 O’s, 1T, 1I, 1N.
The following combinations are possible
(a) 2 alike, = ³C₂ = 3
(b) 1 alike 2 different, = ³C₁ × ⁵C₂ = 3 × \(\frac{5 !}{3 ! 2 !}\) = 3 × 10 = 30
(c) 4 different, = ⁶C₄ = \(\frac{6 !}{2 ! 4 !}\) = \(\frac{6 \times 5}{2}\) = 15
(d) 3 Same, 1 different = ¹C₁ × ⁵C₁ = 5
Thus, total no. of combinations = 3 + 30 + 15 + 5 = 53
and Total no. of permutations =3 × \(\frac{4 !}{2 ! 2 !}\) + 30 × \(\frac{4 !}{2 !}\) + 15 × 4! + 5 × \(\frac{4 !}{3 !}\)
= 3 × \(\frac{24}{4}\) + 30 × 12 + 15 × 24 + 5 × 4
= 18 + 360 + 360 + 20 = 758

Question 19.
A table has 7 seats, 4 being on one side facing the window and 3 being on the opposite side. In how many ways can 7 people be seated at the table.
(i) if 2 people, X and Y, must sit on the same side ;
(ii) X and Y must sit on opposite sides ;
(iii) if 3 people, X, Y and Z, must sit on the side facing the window ?
Solution:
Given table has 7 seats, 4 being on one side facing the window and 3 being on opposite side.
(i) Case-I. If X and Y sit on facing window seat.
∴ possible no. of ways are ⁴C₂
also both persons can be arranged among themselves in \(\lfloor 2\) ways
remaining 5 persons can be seated on the table in \(\lfloor 5\) ways.
∴ no. of ways can 7 people be seated at the table if X and Y must sit on side with 4 chairs
= ⁴C₂ × \(\lfloor 2\) × \(\lfloor 5\) = 6 × 2 × 120 = 1440

Case-II: When two persons X and Y be seated on side with 3 chairs.
∴ possible no. of ways be ³C₂
Now X and Y can arrange among themselves in \(\lfloor 2\) ways.
∴ remaining 5 persons can be seated on table in \(\lfloor 5\) ways.
Thus no. of ways, when two persons X and Y are seated on side with 3 chairs
= ³C₂ × \(\lfloor 2\) × \(\lfloor 5\) = 3 × 2 × 120 = 720
∴ required no. of ways = 1440 + 720 = 2160

(ii) When X and Y must sit on opposite side.
No. of ways of picking 1 seat out of 3 chairs = ³C₁
No. of ways of picking 1 seat out of 4 chairs = ⁴C₁
Two persons X and Y can arrange themselves in \(\lfloor 2\) ways.
Remaining 5 persons can be seated on table in \(\lfloor 5\) ways
Then by fundamental principle of counting
Required no. of ways = ³C₁ × ⁴C₁ × 2 × \(\lfloor 5\) = 3 × 4 × 2 × 120 = 2880

(iii) When 3 people X, Y, Z must sit on the side facing the window.
Total no. of ways of seating of three person on the window side containing 4 chairs = ⁴C₃
These three people can arrange themselves in \(\lfloor 3\) ways.
Remaining 4 peoples be seated on table in \(\lfloor 4\) ways.
Then by fundamental principle of counting
required no. of ways = ⁴C₃ × \(\lfloor 3\) × \(\lfloor 4\) = 4 × 6 × 24 = 576

Question 20.
Seven cards, each bearing a letter, can be arranged to spell the word “DOUBLES”. How many three-letter code-words can be formed from these cards ?
How many of these words
(i) contain the letter S ;
(ii) do not contain the letter O ;
(iii) consist of a vowel between two consonants ?
Solution:
No. of letters in word DOUBLES = 7
required no. of arrangements (three letter code words) = ⁷P₃ = \(\frac{7 !}{4 !}\) = 7 × 6 × 5 = 210

(i) Now we want to find 3 letter words that contains S.
i.e. Total no. of such arrangements = Total no. of arrangements – No. of arrangements which does not contains S
= 210 – ⁶P₃ =210 – \(\frac{6 !}{3 !}\) = 210 – 120 = 90

(ii) Total no. of 3 letter word that does not contain letter O = ⁶P₃ = \(\frac{6 !}{3 !}\) = 6 × 5 × 4 =120

(iii) We want to find 3 letter code words that contain a vowel between two consonants.
Total no. of ways in which two consonants can be arranged out of 4 = ⁴P₂ = \(\frac{4 !}{2 !}\) = 12
Total no. of ways in which vowels can be arranged = 3
∴ required no. of ways = 12 × 3 = 36

Question 21.
How many triangles may be formed by joining any three of the nine points when
(i) no three of them are collinear;
(ii) five of them are collinear ?
Solution:
(i) Required no. of triangles formed from 9 points, when no three of them are collinear
= ⁹C₃ = \(\frac{9 !}{6 ! 3 !}\) = \(\frac{9 \times 8 \times 7}{6}\) = 84

(ii) Here total no. of triangles formed from 9 points = 84 but these also include the triangles that are formed from 5 collinear points which does not form any triangle.
∴ No. of triangles formed from 5 collinear points be ⁵C₃ and we have to subtract these triangles from 84.
Thus required no. of triangles = 84 – \(\frac{5 !}{3 ! 2 !}\)
= 84 – \(\frac{5 \times 4}{2}\) = 74

Question 22.
A committee of 5 is to be formed from a group of 12 students consisting of 8 boys and 4 girls. In how many ways can be committee be formed if it
(i) consists of exactly 3 boys and 2 girls ;
(ii) contains at least 3 girls ?
Solution:
Given no. of boys = 8
no. of girls = 4
(i) Total no. of ways of selecting 3 boys out of 8 = ⁸C₃
Total no. of ways of selecting 2 girls out of 4 = ⁴C₂
∴ required no. of committees that contain exactly 3 boys and 2 girls = ⁸C₃ × ⁴C₂
= \(\frac{8 !}{5 ! 3 !}\) × \(\frac{4 !}{2 ! 2 !}\) = \(\frac{8 \times 7 \times 6}{6}\) × 6 = 336

(ii) We want to find the no. of committees that contain atleast 3 girls.
Thus can be done in following ways
(a) 3 girls and 2 boys
(b) 4 girls and 1 boy
∴ required no. of such committees = ⁴C₃ × ⁸C₂ + ⁴C₄ × ⁸C₁
= 4 × \(\frac{8 \times 7}{2}\) + 1 × 8 = 112 + 8 = 120

Question 23.
There are 5 gentlemen and 4 ladies to dine at a round table. In how many ways can they seat themselves so that no two ladies are together ?
Solution:
The total no. of ways in which 5 gentlemen can sits round a table
= \(\lfloor 5 – 1\) = \(\lfloor 4\) = 24
The four ladies will occupy the places marked ‘X’ so that no two ladies sit together and this can be done in ⁵P₄ ways

Thus required no. of ways = 24 × 120 = 2880

Question 24.
There are 12 points in a plane, of which 5 are collinear. Find
(i) the number of triangles that can be formed with vertices at these points ;
(ii) the number of straight lines obtained by joining these points in pairs.
Solution:
(i) Since there are 12 points in a plane of which 5 are collinear. Since collinear points lies on one straight line and does not form any triangle.
∴ Required no. of triangles = ¹²C₃ – ⁵C₃ = \(\frac{12 !}{9 ! 3 !}\) – \(\frac{5 !}{3 ! 2 !}\)
= \(\frac{12 \times 11 \times 10}{6}\) – \(\frac{5 \times 4}{2}\) = 220 – 10 = 210

(ii) We know that, a straight line is formed by joining two points.
∴ No. of straight lines formed from given 12 points = ¹²C₂
But these include also the straight lines formed from 5 collinear points i.e. ⁵C₂
and we have to subtract it from total no. of straight lines. Further, 5 collinear points make only one straight line.
Thus, required no. of straight lines = ¹²C₂ – ⁵C₂ + 1 = \(\frac{12 !}{2 ! 10 !}\) – \(\frac{5 !}{3 ! 2 !}\) + 1
= \(\frac{12 \times 11}{2}\) – \(\frac{5 \times 4}{2}\) + 1 = 66 – 10 + 1 = 57

Question 25.
A committee of 5 is to be formed from a group of 10 people, consisting of 4 single men, 4 single women and a married couple, the committee is to consist of a chairman, who must be a single man, 2 other men and 2 women,
(i) Find the total number of committees possible.
(ii) HOW many of these would include the married couple ?
Solution:
Given no. of single men = 4
no. of single women = 4
no. of married couple = 1 (1 M + 1 W)

(i) The no. of ways of selecting chairman (which is a single man) out of 4 = ⁴C₁
The no. of ways of selecting 2 other men out (3 + 1) = 4 men = ⁴C₂
The no. of ways of selecting 2 women from (4 + 1) = 5 women = ⁵C₂
∴ required no. of such committees = ⁴C₁ × ⁴C₂ × ⁵C₂
= 4 × \(\frac{4 !}{2 ! 2 !}\) × \(\frac{5 !}{2 ! 3 !}\) = 4 × 6 × 10 = 240

(ii) We want to find no. of committees that include marked couple.
Total no. of ways of selecting chairman, which is a single man = ⁴C₁
Total no. of ways of selecting 1 man out of remaining 3 men = ³C₁
Total no. of ways of selecting 1 woman out of 4 = ⁴C₁
Thus, required no. of such committees = ⁴C₁ × ³C₁ × ⁴C₁ = 4 × 3 × 4 = 48

Question 26.
A committee of 5 persons is to be formed from a group of 6 gentlemen and 4 ladies. In how many ways can this be done if the committee is to include atleast one lady ?
Solution:
Given no. of gentlemen = 6 and no. of ladies = 4
we want to find no. of ways in which committee of 5 person is to be formed that contain atleast one lady.
Thus can be done in following ways
(a) one lady and 4 gentlemen
(b) two ladies and 4 gentlemen
(c) Three ladies and 2 gentlemen
(d) Four ladies and 1 gentlemen
∴ required no. of such committees =

Question 27.
Out of 3 books on Economics, 4 books on Political Science and 5 books on Geography, how many collections can be made, if each collection consists of
(i) exactly one book on each subject,
(ii) at least one book on each subject ?
Solution:
(i) The no. of ways of selecting one economics book from 3 economics books = ³C₁
The no. of ways of selecting one Political Science book from 4 books = ⁴C₁
The no. of ways of selecting one History book from 5 books = ⁵C₁
Thus the required no. of such collections = ³C₁ × ⁴C₁ × ⁵C₁ = 3 × 4 × 5 = 60

(ii) A selection of atleast one book on economics can be made out of 3 different books in ³C₁ + ³C₂ + ³C₃ = 2³ – 1 = 7 ways
Similarly a selection of atleast one book on Political Science can be made out of 4 books in ⁴C₁ + ⁴C₂ + ⁴C₃ + ⁴C₄ = 2⁴ – 1 = 16 – 1 = 15 ways
A selection of atleast one book on History can be made out of 5 books in
⁵C₁ + ⁵C₂ + ⁵C₃ + ⁵C₄ + ⁵C₅ = 2⁵ – 1 = 31 ways
Thus by fundamental principle of counting required number of ways of selecting atleast one books on each subject = 7 × 15 × 31 = 3255

Question 28.
Find the number of words which can be formed by taking two alike and two different letters from the word “COMBINATION”.
Solution:
In word COMBINATION we have 2 I’s, 2 O’s, 2 N’s, 1C, 1M, IB, 1A and IT.
The number of combinations taking 2 alike and two different letters = ³C₁ × ⁷C₂
= 3 × \(\frac{7 !}{5 ! 2 !}\) = 3 × \(\frac{7 \times 6}{2}\) = 63

Now each of these combination give rise to \(\frac{4 !}{2 ! 1 !}\) i.e. 12 words
∴ required no. of words = 63 × 12 = 756