1. How many Permutations of the letters of the word APPLE are there?
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Correct Ans:60
Explanation:
In the given word “APPLE”, letter "P" is written twice.
So word APPLE contains 1A, 2P, 1L and 1E.
Required Permutations = 5! / 2!
= (5 x 4 x 3 x 2 x 1) / (2 x 1)
= 60
2. Using all the letters of the word "NOKIA", how many words can be formed, which begin with N and end with A?
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Correct Ans:6
Explanation:
There are five letters in the given word “NOKIA”
Consider 5 blanks ....
The first blank and last blank must be filled with N and A all the remaining three blanks can be filled with the remaining 3 letters in 3! ways.
The number of words = 3! = 3 X 2 X 1 = 6
3. In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
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Correct Ans:209
Explanation:
Chances for selecting 4 children 
From 6 Boys (with atleast 1 boy) 
From 4 Girls 

Chance 1 
1 
3 
> 6C1 X 4C3
= 6 X 4
= 24 
Chance 2 
2 
2 
> 6C2 X 4C2
= 15 X 6
= 90 
Chance 3 
3 
1 
> 6C3 X 4C1
= 20 X 4
= 80 
Chance 4 
4 
0 
> 6C4
= 15 
Different ways to select 4 children (with atleast 1 boy) = 24 + 90 + 80 + 15 =
209
4. In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?
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Correct Ans:720
Explanation:
The vowels in the word 'LEADING' are E, A, I
Number of ways to arrange the letters of the word 'LEADING' such that vowels always come together = LDNG(EAI)
= 5! X 3!
= (5 x 4 x 3 x 2 x 1) X (3 x 2 x 1)
= 120 X 6
= 720
5. In how many ways 6 people can be arranged in row, if one particular person always wants to stand in the rightmost corner?
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Correct Ans:5!
Explanation:
Out of 6 person one wants to be in the right most corner, so ignore him
Out of the remaining 5 people, they can be arranged in 5! ways.
6. In how many ways a four digit even number can be formed by using the digits 4,5,9,8 exactly once.
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Correct Ans:12
Explanation:
Given
Four digit even number can be formed by using the digits 4,5,9,8
Since the number has to be a even digit number,the units digit has to be 4 or 8. First three places can be filled by remaining three digits.
Hence it is totally 6 x 2 = 12 ways
Answer is 12
7. In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?
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Correct Ans:63
Explanation:
Given
Total Men = 7
Total Women = 3
From total, agroup of 5 men and 2 women be made out
Required number of ways
=>(7C5 x 3C2)
=>( 7 * 6 * 5 * 4 * 3 * 2 * 1 ) / (5 * 4 * 3 * 2 * 1) x ( 3 * 2) / ( 2 * 1 )
=>( 7 * 3 ) x ( 3 )
=> 21 * 3
=> 63.
8. A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?
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Correct Ans:64
Explanation:
We may have(1 black and 2 nonblack) or (2 black and 1 nonblack) or (3 black).
Required number of ways = (3C1 x 6C2) + (3C2 x 6C1) + (3C3)
= ( 3 x 6 x 5 / 2 x 1 ) + ( 3 x 2 / 2 x 1 x 6) + 1
= (45 + 18 + 1)
= 64
9. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
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Correct Ans:25200
Explanation:
Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)
= (7C3 x 4C2)
= ( 7 x 6 x 5 / 3 x 2 x 1) x (4 x 3 / 2 x 1 )
= 210
Number of groups, each having 3 consonants and 2 vowels = 210.
Each group contains 5 letters.
Number of ways of arranging 5 letters among themselves = 5! = 5 x 4 x 3 x 2 x 1 = 120.
Therefore, Required number of ways = (210 x 120) = 25200
10. In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?
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Correct Ans:50400
Explanation:
In the word 'CORPORATION', we treat the vowels OOAIO as one letter.
Thus, we have CRPRTN (OOAIO).
This has (6 + 1) = 7 letters of which R occurs 2 times and the rest are different.
Number of ways arranging these letters = 7!/2! = 2520.
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged in 5! /3! = 20 ways.
Required number of ways = (2520 x 20) = 50400
11. In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?
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Correct Ans:720
Explanation:
The word 'LEADING' has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LNDG (EAI).
Now, 5 letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720
12. There are 5 boys and 3 girls in a family. They are photographed in groups of 2 boys and one girl. The number of different photographs will be
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Correct Ans:180
Explanation:
The number of ways of forming the groups 5C2 x 3C1 = 30
Number of each groups can be arranged among themselves is 3! Ways = 6.
Therefore, the number of photographs = 30 x 6 = 180.
13. Seven points lie on a circle. How many chords can be drawn by joining these points?
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Correct Ans:21
Explanation:
7C2 = 7 x 6 /2 = 21
14. A box contains 10 balls out of which 3 are red and the rest are blue. In how many ways can a random sample of 6 balls be drawn from the bag so that at the most 2 red balls are included in the sample and no sample has all the 6 balls of the same color ?
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Correct Ans:168
Explanation:
Given that, Out of 10 balls, 3 balls are red and remaining 7 balls are blue.
The possible ways are as follows:
(i) 1 red ball out of the three and 5 blue balls out of the seven (OR)
(ii) 2 red balls out of the three and 4 blue balls out of the seven
Total number of ways in which a random sample of six balls can be drawn = (^{3}C_{1} x ^{7}C_{5}) + (^{3}C_{2} x ^{7}C_{4})
= [3 * (7*6) / 2] + [3 * (7*6*5) /(3*2)]
= [126 / 2] + [ 630 / 6]
= [63] + [105]
= 168.
15. In an examination paper there are two groups, each containing 4 questions. A candidate is required to attempt 5 questions but not more than 3 questions from any group. In how many ways can 5 questions be selected?
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Correct Ans:48
Explanation:
A candidate is required to attempt 5 questions but not more than 3 questions from any group.
= 4C3 x 4C2+4C2 x 4C3
= (4 x 6) + (4 x 6)
= 48.
16. What is the remainder left after dividing 1! + 2! + 3! + ....... + 100! by 7?
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Correct Ans:5
Explanation:
From 7! onwards all terms are divisible by 7 as 7 is one of the factor.
So there is no remainder left for those terms i.e. remainder left after dividing 7! + 8! + 9! + ... + 100! is 0.
The only part to be consider is
= 1! + 2! + 3! + 4! + 5! + 6!
= 1 + 2 + 6 + 24 + 120 + 720
= 873
The remainder left after dividing 873 by 7 is 5
Hence, the remainder is 5
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