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Question (1):
Find n.
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Answer:
i)
ii)
\n = 81
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Question (2):
A lady wants to select one cotton saree and one polyster saree from a collection of 8 cotton sarees and 11 polyster sarees. In how many ways can the lady choose the two sarees?
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Answer:
Here, there are two events E1 and E2.
E1 = Selection of one cotton saree from 8 cotton sarees.
E2 = Selection of one polyster saree from 11 polyster sarees.
E1 = 8 ways E2 = 11 ways
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Question (3):
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Answer:
i)
ii)
iii)
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Question (4):
When n = 5 and r = 2, find the values of
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Answer:
i)
ii)
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Question (5):
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Answer:
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Question (6):
Convert the following into factorials:
i) 5.6.7.8.9.10 ii) 4.6.8.10.12
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Answer:
i)
ii)
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Question (7):
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Answer:
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Question (8):
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Answer:
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Question (9):
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Answer:
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Question (10):
Find n if P(n,4) = 20P(n,2)
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Answer:
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Question (11):
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Answer:
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Question (12):
In how many ways can 6 men and 5 women be seated in a line so that no two women sit together?
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Answer:
Let the women be denoted by W and the men be denoted by M.
Now the women can sit at the places marked by '.' .
We have seven places for 5 women. The women can sit in 7P5 ways. Also 6 men can be arranged in 6! ways.
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Question (13):
In how many ways can 10 examination papers be arranged so that best and the worst are never together?
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Answer:
10 papers can be arranged in 10! ways. When the best and the worst are together, the number of arrangements will be 9! x 2.
\When the best and the worst are not together, the number of arrangements will be 10! - (9!)2 = 9!(10-2) = (8) x (9!).
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Question (14):
In how many ways can three prizes be distributed among 4 boys when
i) No one gets more than one prize.
ii) A boy can get any number of prizes.
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Answer:
i) The first prize can be given in 4 ways as one cannot get more than one prize, the remaining two prizes can be given in 3 and 2 ways respectively.
\The total number of ways = 4 x 3 x 2 = 24.
ii) As there is no restriction, each prize can be given in 4 ways.
\The total number of ways = 43 = 64.
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Question (15):
How many three digit numbers are there, with distinct digits, with each digits odd?
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Answer:
Here, we have to form three digit numbers with distinct digits, each digit odd.
The digits used are 1, 3, 5, 7, 9. Number of odd digits is 5. From 5 odd digits, we have to take 3 at a time. This can be done in P(5,3) ways.
\Number of three digit numbers = P(5,3) = 5 x 4 x 3 = 60
Aliter
As we require the numbers with distinct digits, repetition of digits is not allowed.
Hundred's place can be filled in 5 ways.
Ten's place can be filled in 4 ways.
Unit's place can be filled in 3 ways.
By fundamental principle of counting, required number of 3-digit numbers = 5 x 4 x 3 = 60.
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Question (16):
Given 6 flags of different colours, how many different signals can be generated, if a signal requires the use of two flags one below the other?
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Answer:
Number of signals obtained by 6 flags taking 2 flags at a time is P(6,2).
P(6,2) = 6 x 5 = 30
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Question (17):
Find n; r if
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Answer:
i)
ii)
iii)
iv)
v)
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Question (18):
In how many ways can 6 men and 5 women sit in a row, so that women occupy the even places?
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Answer:
Total number of places = 6+5 = 11
These five even places can be filled by 5 women in 5! ways and the remaining six places by 6 men in 6! ways.
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Question (19):
In how many ways can six children stand in a queue?
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Answer:
6 children can stand in a queue in 6P6 ways.
6P6 = 6! = 720 ways.
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Question (20):
How many words with or without meanings can be formed using all the letters of the word EQUATION using each letter exactly once?
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Answer:
The number of distinct letters in the word EQUATION are 8.
We have to use all the 8 letters without repetition.
Hence, the number of words formed = 8P8 = P(8,8) = 8! = 40320.
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Question (21):
A family of 4 brothers and 3 sisters is to be arranged in a row for a photograph. In how many ways can they be seated if all the sisters are together?
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Answer:
Let B1,B2,B3,B4 denote the brothers and S1,S2,S3 denote the sisters. Since the sisters are to be seated together for a photograph, consider all the sisters as one unit or entity. Then B1,B2,B3,B4,S can be arranged to sit in 5! ways. The sisters can be arranged among themselves in 3! ways. Since the two events are independent, the total number of arrangements = 5!.3! = 720 ways.
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Question (22):
In how many ways can a consonant and a vowel be chosen out of the letters in the word COURAGE?
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Answer:
There are three consonants (C, R, G) and four vowels (A, E, O, U) in the word COURAGE.
With the consonants, we may choose any one of the 4 vowels. It can be done in 4 ways. There are three consonants.
\The total number of ways will be 4 x 3 = 12.
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Question (23):
How many arrangements can be made out of the letters of the word DRAUGHT, the vowels never being separated?
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Answer:
There are 7 letters in the word DRAUGHT, the two vowels are A and U. Since, the vowels are not to be separated, AU can be considered as one entity. Therefore, the number of letters will be 6 instead of 7. The permutations will be P(6,6) = 6! ways.
But the two vowels A and U can be arranged in two ways, i.e. AU and UA.
\The required number of arrangements = 2!.6! = 1440 ways.
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Question (24):
Find the number of arrangements that can be made out of the letters i) ASSASSINATION ii) GANESHPURI.
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Answer:
i) The word ASSASSINATION consists of
A's = 3, S's = 4, I's = 2,
N's = 2, T's = 1, O's = 1
\ The total number of letters is 13 letters.
ii) The word GANESHPURI consists of 10 distinct letters.
The number of permutations is 10!.
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Question (25):
How many different arrangements can be made out of the letters in the expression a3b2c4, when written at full length?
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Answer:
There are 3 + 2 + 4 = 9 letters.
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