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Question (1):
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Answer:
As n represents all positive integers, we have
Multiplying the above terms of both sides respectively, we get
Multiplying both sides of inequality by n!, we get
or n < 2n
Multiplying the above terms of both sides respectively, we get
Multiplying both sides by n!, we get
From (1) and (2), we get
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Question (2):
vi) (n! + 1) is not divisible by any natural number between 2 and n.
vii) Simplify
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Answer:
= (32.16.8.4.2)(33.31.30.18.17.15.14.13.12.11.10....3.1)
= (25.24.23.22.21)(33.31.30.....3.1)
which is divisible by 215.
vi) (n! + 1) is not divisible by any natural number between 2 and n.
Clearly, n! is divisible by every natural number from 2 to n.
 n! + 1 will leave remainder 1, when divided by any natural number
between 2 and n.
Hence, (n! + 1) is not divisible by any natural number from 2 to n.
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Question (3):
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Answer:
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Question (4):
Prove that n!(n + 2) = n! + (n + 1)!.
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Answer:
L.H.S = n!(n + 2)
= n![(n+1)+1]
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Question (5):
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Answer:
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Question (6):
Find n, if P(n-1,3): P(n+1,3):: 5 : 12.
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Answer:
n = 8 (as n cannot be a fraction)
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Question (7):
How many three letter words (with or without meaning) can be formed out of the letters of the word LOGARITHMS if repetition of letters is not allowed?
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Answer:
Number of letters in the word LOGARITHMS is 10.
The three letter words formed by using these letters will be
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Question (8):
There are three different rings to be worn in 5 fingers with at most one in each finger. In how many ways can this be done?
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Answer:
For the first ring, there are 4 options as one finger cannot have more
than one ring. The remaining two fingers will have 3 and 2 options respectively. Hence, the number of ways = 4.3.2 = 24 ways.
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Question (9):
How many six digits telephone numbers can be made if each number starts with 45 and no digit appears more than once?
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Answer:
The first two places are reserved for 4 and 5.
The remaining 8 digits can be arranged in P(8,4) = 8.7.6.5
= 1680 ways
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Question (10):
Twelve students are participating in a competition. In how many ways can the first 3 prizes be won?
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Answer:
Number of students participating in the competition = 12
The number of ways in which the first three prizes can be won
= P (12,3) = 12 x 11 x 10 = 1320
Hence, 12 students can win three prizes in 1320 ways.
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Question (11):
There are 8 objects in column A and 8 objects in column B. A boy is asked to match each item in column A with each item in column B. How many possible (correct or wrong) answers are there for these questions?
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Answer:
Each answer to this question is an arrangement of 8 items of column
B so as to correspond to the given arrangement of items in column A.
Therefore, there are as many arrangements as there are answers.
Therefore, number of answers required = P(8,8) = 8!
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Question (12):
Find n; r if
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Answer:
n = 10 is only valid.
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Question (13):
If there are 8 periods in each working day in a school, in how many ways can 6 subjects be arranged such that each subject is allowed atleast one period?
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Answer:
Out of 8 periods, 6 periods be arranged in 8P6 ways and the remaining 2 periods be arranged in 6P2 ways.
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Question (14):
Four alphabets T, A, L and K, one each were purchased from a plastic warehouse. How many ordered pairs of alphabets to be used as initials, can be formed from them?
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Answer:
Four letters A, K, L, T are purchased. For an initial, two alphabets are required.
Thus, the number of initials that can be formed is P (4,2) = 4x3 = 12.
 The number of initials that can be formed are 12.
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Question (15):
How many three digit numbers are there with distinct digits with each digit even, i) with 0 and ii) without 0?
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Answer:
Here we are required to form three-digit numbers with distinct each digit even.
The even digits are 0, 2, 4, 6, and 8.
The digits used are 0, 2, 4, 6, and 8.
The hundred's place can be filled in 4 ways (excluding zero).
The ten's place can be filled in again in 4 ways (including zero).
The unit's place can be filled in 3 ways.
Hence, by the fundamental principle of counting, the number of three
digit numbers = 4 x 4 x 3 = 48.
ii) Without 0
The digits used are 2, 4, 6, and 8.
The hundred's place can be filled in 4 ways.
The ten's place can be filled in 3 ways and the unit's place can be filled
in 2 ways.
Hence, by the fundamental principle of counting, the number of three digit numbers = 4 x 3 x 2 = 24.
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Question (16):
m men and n women are to be seated in a row so that no two women sit together. If m>n, then show that the number 
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Answer:
m men can be arranged in m seats in m! ways. After the m men have been seated, we may get one seat in the beginning and one seat in the end and (m-1) seats in between each pair of men for the women.
Thus, the number of ways in which n women can be seated in
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Question (17):
Ten articles are to be placed in ten boxes, one in each box. Five of them are too big for four boxes. Find the number of possible arrangements.
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Answer:
5 of the ten articles which are too big for 4 boxes are to be kept in
the remaining 6 boxes. This can be done in 6P5 ways. Now there are 4
boxes and 5 articles left. This can be arranged in 5P5 ways.
Since, these two events are independent the total number of 
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Question (18):
How many permutations can be made out of the letters of the word TRIANGLE? How many of these begin with i and end with e?
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Answer:
i) There are 8 distinct letters in the word TRIANGLE. The number of permutations is P(8,8) = 8! = 40320
'i' is the first letter and 'e' is the last letter of each word. The remaining places can be filled in 6! ways = 720 ways.
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Question (19):
Find the number of arrangements that can be made out of the letters of the words:
i) INDEPENDENCE ii) SUPERSTITIOUS iii) INSTITUTIONS.
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Answer:
i) In the word INDEPENDENCE, I=1, D's=2, E's=4, N's=3, C=1, P=1.
\Total number of letters = 12
ii) The word SUPERSTITIOUS consists of S's=3, U's=2, P=1, E=1, R=1, T's=2, I's=2, O=1.
\Total number of letters = 13
iii) The word INSTITUTIONS consists of I's=3, N's=2, S's=2, T's=3, O=1, U=1.
\Total number of letters = 12
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Question (20):
In how many ways can 17 billiardballs be arranged if 7 of them are black, 6 red and 4 white.
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Answer:
Out of 17 balls, we have 7B, 6R and 4W.
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Question (21):
A letter lock consists of three rings, each marked with fifteen different letters. Find in how many ways it is possible to make an unsuccessful attempt to open the lock.
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Answer:
1st ring can be attempted in 15 ways.
1st and 2nd ring can be attempted in 15 x 15 = 152 ways
1st and 2nd and 3d ring can be attempted in 15 x 15 x 15 = 153 ways
Among these 153 attempts, one attempt will be a successful attempt.
Hence the number of unsuccessful attempts is 153-1= 3375 -1 = 3374.
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Question (22):
Find the sum of all four digit numbers formed by using 2, 3, 6, 9 in which no digit is repeated.
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Answer:
If 2 occupies unit's place, the remaining 3 digits can be arranged in 3!= 6ways. Similarly, if 2 occupies ten's place, hundred's place, thousand's place, in each of these cases we get 3! numbers. Thus, the positional value contributed by 2 to the sum when it occupies different
values is (3!)(2) + (3!)(20) + (3!)(200) + (3!)(2000) = 3!(2)(1111)
Similarly, the values contributed by 3, 6, 9 to the sum are
3! (3)(1111), 3! (6) (1111), 3! (9)(1111) respectively.
 The required sum is 3!(1111)(2+3+6+9) = 1,33,320
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Question (23):
Find n, r if
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Answer:
i. C(n,4) = 210
 ( 210 = 10 x 7 x 3)
ii. C(n,18) = C(n,12)
Using C(n,r) = C(n,n-r)
The value of C(32,n) = C(32,30) = C(32,2)
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Question (24):
If the ratio C(2n,3):C(n,3) = 11:1, find n.
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Answer:
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Question (25):
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Answer:
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