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Question (1):
Prove that the number of ways in which (m+n) dissimilar things can be divided into two groups containing m and n 
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Answer:
If we select m things out of (m+n) things, then n things are left out . Then, this gives (m+n) that can be divided into two groups containing m and n things.
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Question (2):
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Answer:
i.
 
The value of C(n, 25) = C(25, 25) = 1
ii.
iii.
 
iv.
v.
\n = 8
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Question (3):
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Answer:
= 5 + 10 + 10 + 5 + 1 = 31 = RHS
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Question (4):
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Answer:
Aliter:
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Question (5):
A boy has 3 library tickets and 8 books of his interest in the library. Of these 8, he does not want to borrow Chemistry Part II, unless Part I is also borrowed. In how many ways can he choose the three books to be borrowed?
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Answer:
There are two ways of selecting 3 books:
i) If Chemistry Part II is selected, then we have to select Chemistry Part I and in this case, the third book is selected from the remaining books.
Hence,the first selection can be made in C(6,1) ways = 6 ways.
ii) If Chemistry Part II is not selected, then we can select 3 books from the remaining 7 books. Hence second choice can be made in 
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Question (6):
At an election, three wards of a town are canvassed by 4, 5 and 8 men respectively. If 20 men volunteer, in how many ways can they be allotted to the different wards?
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Answer:
Let the three wards be labelled as A, B and C respectively. Let us select 4 men for ward A. 4 men can be selected out of 20 in 20C4 ways. Having selected 4 men from ward A in one of 20C4 ways, 5 men can be selected out of remaining 16 in 16C5 ways for ward B. After the selection, we are left with 11 men. Now, 8 men have to be selected out of 11 in 11C8 ways for ward C.
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Question (7):
A candidate is required to answer 6 out of 10 questions, which are divided into two groups, each containing 5 questions, and he is not permitted to attempt more than 4 from each group. In how many ways can he make his choice?
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Answer:
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Question (8):
For the post of 5 teachers, there are 23 applicants. 2 posts are reserved for S.C. candidates and there are 7 S.C. candidates among applicants. In how many ways can the selection be made?
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Answer:
There are 23 applicants. Among this, 7 are S.C. candidates. We have to select 2 S.C. candidates among 7 S.C. applicants.
So, there are 23 - 7 = 16 non-S.C. candidates.
We can select 2 S.C. canditates from 7 and the remaining 3 teachers from 16 applicants.
Hence, the number of ways in which the selections can be made is
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Question (9):
How many words of different 4 letters can be formed out of 7 capital letters, 3 vowels and 8 consonants, if each word starts with a capital letter and contains at least one vowel?
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Answer:
Any capital letter can be chosen as the first letter of each word. So there are 7 ways of choosing the first letter of each word. Next, we are to choose 3 letters containing atleast 1 vowel as follows:
i. By choosing one vowel and two consonants
ii. By choosing 2 vowels and 1 consonant
iii. By choosing 3 vowels i.e.,3C3 ways = 1 way.
\Total number of ways of choosing these 3 letters = 30+15+1 = 46
Further, the number of ways of arranging these 3 letters among themselves = 3! = 6 ways.
\The number of required words = (7)(6)(46) = 1932
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Question (10):
The sides AB, BC, CA of a triangle ABC have 3, 4 and 5 interior points respectively on them. Find the number of triangles that can be drawn using these points as vertices.
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Answer:
The points 3+4+5=12 points lie in a plane.
To construct a triangle, we require three non-collinear points.
The number of selecting three points from these 12 points
=12C3 = 220.
Number of selecting three points from three collinear points = 3C3 = 1
(on AB)
Number of selecting three points from four collinear points = 4C3 = 4
(on BC)
Number of selecting three points from five collinear points = 5C3 = 10
(on AC)
Hence, the required number of  = 220 - 1 - 4 - 10 = 205
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Question (11):
In how many ways can a picnic party of 16 people be conveyed in a van, if the first holds not more than 8 and the second one holds not more than 10 people?
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Answer:
16 people can be conveyed by
i. Taking 8 peopel in first van and 8 people in second van.
This can be done in 16C8 x 1 = 16C8 ways.
ii. Accommodating 7 people in first and then 9 people in second van.
This can be done in 16C7 x 1 = 16C7 ways.
iii. Taking 6 people in first and 10 people in second van.
This can be done in 16C6 x 1 = 16C6 ways.
\The total number of ways = 16C8 + 16C7 +16C10
= 12870 + 11440 + 8008
= 32318
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Question (12):
Find the number of lines that can be drawn through 12 points of which 5 are collinear.
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Answer:
To draw a line, we need two distinct points.
The number of lines that can be drawn through 12 points is 
There are 5 points which are collinear i.e., we can draw only one line through them.
Number of lines that could have been drawn through 5 points
= 5C2 = 10
\ Total number of lines = 66 - 10 - 1 = 57
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Question (13):
Find the number of triangles that can be drawn through 12 distinct points of which 5 are collinear.
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Answer:
To draw a triangle, we require three non-collinear points.
Hence, the number of triangles that can be drawn through 12 points
= 12C3 = 220
But there are 5 points which are collinear.
Hence, we cannot draw any triangle through 5 points.
We can draw 5C3 triangles through them.
\Total number of triangles = 12C3 - 5C3 = 220 - 10 = 210
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Question (14):
From 4 officers and 8 jawans, how many ways can 6 be chosen
i. To include exactly one officer ?
ii. To include at least one officer ?
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Answer:
The number of ways of choosing 5 jawans from 8 
\Total number of ways of choosing an officer and 5 jawans 
ii.
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Question (15):
In a plane, there are n lines, no two of which are parallel and no three are concurrent. How many triangles can be formed with their points of intersection on their vertices?
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Answer:
The number of points to take as vertices  number of intersection of n 
But on each line there are (n-1) collinear points.
(which are the points of intersection of the line with (n-1) lines)
The number of selections of three points for each set of (n-1) collinear points, no triangle can be drawn i.e., n-1C3 triangles are not obtained.
\The required number of triangles
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Question (16):
Six X's have to be placed in the squares of the figure shown, such that each row contains at least one X. In how many different ways can this be done?
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Answer:
All X's are identical, we have to select 6 squares out of 8. Without 
But there are two possibility which are to be excluded
(i) 1st row - 2, 2nd row - 4
(ii) 3rd row -2 , 2nd row - 4
In this case either 1st row is empty or second row is empty. But according to the problem, no row should be empty.
\Required number of selections = 28 - 2 = 26
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Question (17):
Eighteen guests have 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 the other side. Determine the number of ways in which the sitting arrangements can be made.
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Answer:
9 seats are there on either side of the table. 4 guests desire to sit on one side and 3 others on the opposite side. Thus, we have to choose 5 guests out of 11 who can sit along the four guests already occupying seats on one particular side. The number of ways in which 5 guests out of 11 can be chosen is given by 11C5. For any one of the above combinations, the number of ways in which the seats can be arranged on the other side will be 9! ways. Thus, for two sides of the table, the number of ways will be (9!)2. Hence, for 11C5 ways of choosing 5 guests out of 11, the total number of arrangements will be (9!)2 x 11C5.
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Question (18):
How many three digit numbers can be formed without using the digits 0,2,3,5 and 7?
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Answer:
Since, we are not to use the digits 0,2,3,5,and 7, we are left to use the digits 1,4,6,8 and 9, clearly repetition of digits is allowed. Unit's place can be filled in 5 ways. Ten's place can be filled in 5 ways. Hundredth's place can be filled in 5 ways.
By fundamental principle of counting, total three digits numbers are
5 x 5 x 5 = 125.
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Question (19):
How many four digit numbers can be formed using the digits 0,2,3,5 and 7, repetition being allowed.
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Answer:
In this problem, we shall begin filling the places from thousand's place.
Since, 0 cannot occupy thousand's place, this place can be filled in 4 ways (First Digit).
Since, 0 can occupy hundred's place, this place can be filled in 5 ways (Second Digit).
Ten's place can be filled in 5 ways (third digit).
Unit's place can also be filled in 5 ways (Fourth digit).
By fundamental principle of counting, the number of four digit numbers are 4 x 5 x 5 x 5 = 500
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Question (20):
A team consists of 7 boys and 5 girls and the other team consists of 6 boys and 4 girls. How many single matches can be arranged between the two teams when a boy plays against a boy and a girl against a girl.
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Answer:
Since, a boy plays against a boy only, the number of single matches between boys of two teams = 7 x 6 = 42
Similarly, number of single matches between girls of two teams 
Total number of events (number of single matches) = 42 + 20 = 62
( The two events does not depend on each other)
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Question (21):
Evaluate:
i) 7!
iii) 8! + 6!
iv) 9! - 7!
viii) L.C.M of [5!6!7!]
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Answer:
viii) L.C.M of [5!6!7!]
6! is a multiple of 5!
7! is a multiple of 6!
 L.C.M. = 7!
= 5040
or
 L.C.M = 24.32.5.7. = 5040
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Question (22):
Prove that
i) (2n)! = [1.3.5.7...(2n - 1)]2n (n!)
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Answer:
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Question (23):
Compute
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Answer:
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Question (24):
If (n +1)! = 12(n-1)!, find n.
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Answer:
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Question (25):
When n = 8, r = 3, and when n = 8, r = 5, find the value of 
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Answer:
When n = 8, r = 3
When n = 8, r = 5
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