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Question (1):
From a class of 32 students, 4 are to be chosen for a competition. In how many ways can this be done?
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Answer:
We are to select 4 students from 32.
This selection can done in 
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Question (2):
A sportsteam of 11 students is to be constituted choosing at least 5 from class XI and 5 at least from XII. If there are 20 students in each of these classes, in how many ways can the team be constituted?
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Answer:
Number of students in each class is 20.
 Total number of selections = 2 x 600935040 = 1201870080
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Question (3):
From a class of 12 boys and 10 girls, 10 students are to be chosen for a competition, including atleast 4 boys and 4 girls. 2 girls who won prizes last year should be included. In how many ways the selection can be made?
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Answer:
There are 12 boys and 10 girls in the class. We are to select 10 students for a competition atleast including 4 boys and 4 girls. The 2 girls who won prizes last year are to be included as a condition. Since 2 girls are already selected, now we are left with 8 girls of which at least 2 girls are to be selected. Thus, we can effect the selection in the following manner:
Adding all selections, 104874 are the ways in which the students can be selected.
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Question (4):
A committee of 6 is 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?
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Answer:
Let the two ladies who refuse to work on the same committee be denoted by L1 and L2. Let the remaining five ladies be denoted by L3,
L4, L5, L6 and L7.
Let us first find out the number of ways in which committees of choices I and II can be formed without restrictions given.
 Total number of committees
Let us find the number of committees of these choices in which L1 and L2 serve on the same committee. For type I committee, 3 men can be chosen in 10C3 ways. After having done that, we have to choose 3 ladies. But it has been decided to choose L1 and L2 already. Therefore, from remaining 5 ladies, we can select one more lady in 5C1 ways.
 The number of ways in which a committee of 6 can be choosen including atleast 3 men and 2 women such that L1 and L2 serve on the same committee
 The number of ways in which a committee of 6 can be formed including alteast 3 men and 2 ladies, when two particular ladies refuse to work on the same committee = 8610 - 810 = 7800.
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Question (5):
Given five different green dyes, four different blue dyes and three different red dyes, how many combinations of dyes can be chosen taking atleast one green and one blue dye?
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Answer:
The least number of dyes that a combination can have is 2.
(one blue and one green).
Maximum number of dyes that a combination can have is 12
(5G, 4B, 6R).
Atleast one green dye can be selected out of 5 green dyes. The number of ways is
After selecting one or more green dyes, we can select atleast one blue dye out of 4 different blue dyes. The number of ways is
After selecting atleast one green dye and atleast one blue dye, at least one red dye or no red dye can be selected in
= 1 + 3 + 3 + 1 = 8 ways
The total number of combinations (31)(15)(8) = 3720.
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Question (6):
Five balls of different colours are to be placed in three boxes of different sizes. Each box can hold all five balls. In how many different ways can we place the balls so that no box remains empty?
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Answer:
Given 5 balls can be placed in 3 boxes as follows:
i. 1 ball can be placed in each of two boxes and 3 balls are placed in the third box.
ii. Two balls are placed in each of the boxes and 1 ball is placed in the third box.
The number of ways in both cases = 20 + 30 = 50
There are exactly 3 such arrangements.
Hence, required number of ways = (50)(3) = 150
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Question (7):
Ten different letters of an alphabet are given. Words with five letters are formed from these given letters. Find the number of words which have atleast one letter repeated.
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Answer:
Total number of 5 letter words from given 10 letters = 105
(if every letter may be repeated any number of times)
Total number of words formed from given 10 letters by taking 5 at a time = 10P5 = 30240
\Required number of words = 105- 30240 = 69760
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Question (8):
Find the number of diagonals in a polygon of n sides.
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Answer:
In a polygon, no three points are collinear. To draw a line, we require two distinct points. The polygon of n sides has n distinct points.
Number of diagonals that can be drawn
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Question (9):
Find the number of triangles that can be drawn through 12 distinct points, no three of which 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 is
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Question (10):
From 12 books, in how many ways can 5 be chosen
i. to surely include a particular book
ii. never to include a particular book
iii. to have no restrictions at all?
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Answer:
i. To include a particular book, one particular book has already been chosen. We have to select 4 books from the remaining (12-1) = 11 
ii. A particular book not to be included i.e. we have to select 5 books out of (12-1) = 11 books.
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Question (11):
Out of 10 consonants and 4 vowels, how many words can be formed each containing 6 constants and 3 vowels?
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Answer:
6 consonants can be chosen out of 10 in 10C6 ways and 3 vowels can be drawn out of 4 in 4C3 ways.
Total number of selection of 6 consonants and 3 vowels is
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Question (12):
There are n concurrent lines and another line parallel to one of them. How many different triangles can be formed by (n+1) lines?
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Answer:
Number of triangles = number of selection of 2 lines from the (n-1) lines which are cut by the last line.
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Question (13):
A man has 7 relatives, 4 of them are ladies and 3 gentlemen. His wife has 7 relatives, 4 of them gentlemen and 3 ladies. In how many ways can they invite a dinner party of 3 ladies and 3 gentlemen, so that there are 3 of the man's relatives and 3 of the wife's relatives?
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Answer:
We can express all possibilities in the following table:
\The total number of ways = 16 + 324 + 144 + 1 = 485
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Question (14):
How many four digit numbers can be formed using the digits 3,4,6,7 and 8 without repetition?
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Answer:
Unit's place can be filled in 5 ways (First Digit).
Ten's place can be filled in 4 ways (Second Digit).
Hundred's place can be filled in 3 ways (Third Digit).
Thousand's place can be filled in 2 ways (Fourth digit).
\By fundamental principle of counting, the number of four digit numbers are 2 x 3 x 4 x 5 = 120.
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Question (15):
In a railway compartment, there are 6 seats available on a bench. How many ways can 4 passengers occupy the six seats?
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Answer:
The first passenger can occupy any of the 6 vacant seats. The second passenger can occupy any of the 5 remaining seats. The third passenger can occupy any of the 4 remaining seats and the fourth passenger can occupy any of the 3 remaining seats.
By the fundamental principle of counting, the total number of ways is 6 x 5 x 4 x 3 = 360.
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Question (16):
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Answer:
We know that the (r + 1)th term of (x + a)n is given by
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Question (17):
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Expanding by binomial theorem, we get
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Question (18):
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Answer:
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Question (19):
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To find the two middle terms:
Since n = 9, the two middle terms are
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Question (20):
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Answer:
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Question (21):
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The greatest value of 'r' can be 8. Then, the greatest term in the expansion is the 9th term.
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Question (22):
Find the greatest term in the expansion of
(2a + b)14 when a = 4, b= 5.
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Answer:
 
Hence the greatest term is the 6th term.
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Question (23):
If A and B be the sum of the odd terms and the sum of the even terms separately in the expansion of (x + a)n, then
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Again,
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Question (24):
In the expansion (x + y)n the coefficients of 4th and 13th terms are equal. Find n.
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Question (25):
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Here y = 4x, x = 1, n = 8
6th term is the greatest.
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Question (26):
Evaluate (0.998)8 correct to 6 decimal places.
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Answer:
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Question (27):
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Answer:
Put x = -1
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Question (28):
Prove that
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Question (29):
Use the Binomial Theorem to (expand) evaluate.
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Answer:
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Question (30):
Write and simplify the term involving  in the expansion 
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Question (31):
Simplify the term independent of x in the expansion of 
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