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Combination Questions
1. In how many ways can 21 books on English and 19 books on Hindi be placed in a row on a shelf so that no two books on Hindi may not be together?
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Correct Ans:1540
Explanation:
Given:
No two Hindi books are together.
Number of English books = 21
Number of Hindi books = 19
Hindi books can be placed in the gaps between the English books.
Since the total number of English books = 21, It can be placed in 22 gaps including the ends.
Hence the number of combinations is given by =
^{22}C_{19}
We know that ^{n}C_{r} = n! / [(n - r)! * r!]
^{22}C_{19}= (22*21*20*19! )/(19! * 3! )
= (22*21*20)/6
= 1540 ways.
No two Hindi books are together.
Number of English books = 21
Number of Hindi books = 19
Hindi books can be placed in the gaps between the English books.
Since the total number of English books = 21, It can be placed in 22 gaps including the ends.
Hence the number of combinations is given by =
^{22}C_{19}
We know that ^{n}C_{r} = n! / [(n - r)! * r!]
^{22}C_{19}= (22*21*20*19! )/(19! * 3! )
= (22*21*20)/6
= 1540 ways.
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2. A question has two parts, Part A and Part B, each containing 8 questions. If the students have to choose 6 from part A and 5 questions from Part B, in how many ways can he choose the questions?
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Correct Ans:1568
Explanation:
Given : Total 8 questions in part A of which 6 questions are chosen
= â¸Câ‚†
Total 8 questions in part B of which 5 questions are chosen = â¸Câ‚…
So the total number of ways = â¸Câ‚† * â¸Câ‚…
= (8! / ((8-6)! * 6! )) * (8! / ((8 - 5)! * 5! ))
= (8! / (2! * 6!)) * (8! / (3! * 5! ))
= ((8*7*6!) / (2*6! )) * ((8*7*6*5! ) / (3*2*5! ))
= 28*56
= 1568 ways.
= â¸Câ‚†
Total 8 questions in part B of which 5 questions are chosen = â¸Câ‚…
So the total number of ways = â¸Câ‚† * â¸Câ‚…
= (8! / ((8-6)! * 6! )) * (8! / ((8 - 5)! * 5! ))
= (8! / (2! * 6!)) * (8! / (3! * 5! ))
= ((8*7*6!) / (2*6! )) * ((8*7*6*5! ) / (3*2*5! ))
= 28*56
= 1568 ways.
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3. At an election there are 5 candidates and 3 members are to be elected. A voter is entitled to vote for any number of candidates not greater than the number to be elected. In how many ways a voter can vote?
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Correct Ans:25
Explanation:
A voter can give either 1 vote, 2 votes or 3 votes.
Number of ways to give only 1 vote = 5C1 = 5
Number of ways to give only 2 vote = 5C2 = 10
Number of ways to give all 3 vote = 5C3 = 10
so, a voter can cast his vote by total : 5+10+10 = 25 ways.
Number of ways to give only 1 vote = 5C1 = 5
Number of ways to give only 2 vote = 5C2 = 10
Number of ways to give all 3 vote = 5C3 = 10
so, a voter can cast his vote by total : 5+10+10 = 25 ways.
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4. In how many different ways can the letters of the word ‘VIRTUAL’ be arranged such that all the vowels come together?
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Correct Ans:720
Explanation:
The vowels in the word 'VIRTUAL' are I, U, A.
Number of ways to arrange the letters of the word'VIRTUAL'such that vowels always come together
= 5! x 3!
= (5 x 4 x 3 x 2 x 1) x (3 x 2 x 1)
= 120 x 6
= 720.
Number of ways to arrange the letters of the word'VIRTUAL'such that vowels always come together
= 5! x 3!
= (5 x 4 x 3 x 2 x 1) x (3 x 2 x 1)
= 120 x 6
= 720.
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5. In how many ways the five boys can be seated among six girls in such a way that no two boys sit together?
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Correct Ans:2520
Explanation:
Possible arrangement of boys and girls
B G B G B G B G B G B G B
where, B -- Boy
G --- Girl
Required number of ways, 7P_{5}= 7!/2!
= (7 x 6 x 5 x 4 x 3 x 2)/2
= 2520
B G B G B G B G B G B G B
where, B -- Boy
G --- Girl
Required number of ways, 7P_{5}= 7!/2!
= (7 x 6 x 5 x 4 x 3 x 2)/2
= 2520
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6. In a language, there are six different words. A sentence can be formed by at least 2 words. If order of words is changed in a sentence, we get a different sentence. How many different sentences can be formed in this language?
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Correct Ans:1950
Explanation:
Here, different order gives different sentence.
Different sentences that can be formed = 6Pâ‚‚ + 6Pâ‚ƒ + 6Pâ‚„ + 6Pâ‚… + 6Pâ‚†
= [6!/(6-2)!] + [6!/(6-3)!] + [6!/(6-4)!] + [6!/(6-5)!] +[6!/(6-6)!]
= [6!/4!] + [6!/3!] + [6!/2!] + [6!/1!] +6!
= 30 + 120 + 360 + 720 + 720
= 1950
Different sentences that can be formed = 6Pâ‚‚ + 6Pâ‚ƒ + 6Pâ‚„ + 6Pâ‚… + 6Pâ‚†
= [6!/(6-2)!] + [6!/(6-3)!] + [6!/(6-4)!] + [6!/(6-5)!] +[6!/(6-6)!]
= [6!/4!] + [6!/3!] + [6!/2!] + [6!/1!] +6!
= 30 + 120 + 360 + 720 + 720
= 1950
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7. A team of 7 children is to be selected out of 7 girls and 5 boys such that it contains at least 5 girls. In how many different ways can the selection be made?
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Correct Ans:246
Explanation:
The team of 7 children can be selected with atleast 5 girls are
When 5 girls and 2 boys are selected = 7C_{5} x 5C_{2}
= [(7 x 6 x 5 x 4 x 3)/(1 x 2 x 3 x 4 x 5)] x [(5 x 4)/(1/2)]
= 21 x 10= 210
When 6 girls and 1 boys are selected = 7C_{6} x 5C_{1}
= [(7 x 6 x 5 x 4 x 3 x 2)/(1 x 2 x 3 x 4 x 5 x 6)] x [5/1]
= 7 x 5 = 35
When 7 girls and no boy are selected = 7C_{7}= 1
Required number of selection = (7C_{5} x 5C_{2}) + (7C_{6} x 5C_{1}) + 7C_{7}
= 210 + 35 + 1
= 246
Therefore, total number of ways = 246.
When 5 girls and 2 boys are selected = 7C_{5} x 5C_{2}
= [(7 x 6 x 5 x 4 x 3)/(1 x 2 x 3 x 4 x 5)] x [(5 x 4)/(1/2)]
= 21 x 10= 210
When 6 girls and 1 boys are selected = 7C_{6} x 5C_{1}
= [(7 x 6 x 5 x 4 x 3 x 2)/(1 x 2 x 3 x 4 x 5 x 6)] x [5/1]
= 7 x 5 = 35
When 7 girls and no boy are selected = 7C_{7}= 1
Required number of selection = (7C_{5} x 5C_{2}) + (7C_{6} x 5C_{1}) + 7C_{7}
= 210 + 35 + 1
= 246
Therefore, total number of ways = 246.
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8. Five students are to be arranged on five chairs for a photograph. Three of these are girls and the rest are boys. Find out the number of ways in which all three girls do not occupy consecutive seats.
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Correct Ans:84
Explanation:
5 students can be arranged among themselves = 5Pâ‚… ways
= 5*4*3*2*1
=120 ways.
Assume that the 3 girls are one entity.
And the total number of ways they arranged among themselves = 3!
= 3*2*1 = 6 ways
Also, set of three girls and other students are arranged among themselves = 3!
= 3*2*1 = 6 ways
So,the total number of ways in which three girls are together = 6 * 6 = 36
Thus, number of ways in which all 3 girls will not occupy consecutive seats
= 120 – 36 = 84
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9. How many four digits number can be formed by using the digits 0, 2, 4, 6, 7 if repetition of digits is allowed?
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Correct Ans:500
Explanation:
Total digits = 5
First place can be filled up by using only one of 4 digits (except 0, since 0 at the first place is meaningless).
Second place can be filled up by using all the five digits (as repetition is allowed).
Similarly, third and fourth place can be filled up by using all the five digits.
Thus,
Places: 0 0 0 0
Digit: 4 5 5 5
Total numbers = 4 Ã— 5 Ã— 5 Ã— 5 = 500
First place can be filled up by using only one of 4 digits (except 0, since 0 at the first place is meaningless).
Second place can be filled up by using all the five digits (as repetition is allowed).
Similarly, third and fourth place can be filled up by using all the five digits.
Thus,
Places: 0 0 0 0
Digit: 4 5 5 5
Total numbers = 4 Ã— 5 Ã— 5 Ã— 5 = 500
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10. There are 5 blue flags, 4 red flags and 3 green flags, in Debu"™s wardrobe. He has to select 4 flags from this set. In how many ways can he select these four flags such that there is at least one blue flag and exactly one green flag in them (Do not consider that the flags are in pairs)?
SHOW ANSWER
Correct Ans:240
Explanation:
No of selection can be in following ways = (3Câ‚*5Câ‚*4Câ‚‚) + (3Câ‚*5Câ‚‚*4Câ‚) + (3Câ‚* 5Câ‚ƒ)
= [3*5*((4*3)/(1*2))] + [3*((5*4)/(1*2))*4] + [3*((5*4*3)/(1*2*3))]
= (3*5*6) + (3*10*4) + (3*10)
= 90 + 120 + 30
= 240
= [3*5*((4*3)/(1*2))] + [3*((5*4)/(1*2))*4] + [3*((5*4*3)/(1*2*3))]
= (3*5*6) + (3*10*4) + (3*10)
= 90 + 120 + 30
= 240
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11. How many 3 - 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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Correct Ans:720
Explanation:
The word 'LOGARITHMS'' has 10 different alphabets
Hence, the number of 3-letter words (with or without meaning) formed by using these letters = 10P_{3}
= 10*9*8
=720
Hence, the number of 3-letter words (with or without meaning) formed by using these letters = 10P_{3}
= 10*9*8
=720
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12. 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
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
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13. 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
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
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14. 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
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15. 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
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
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16. 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.
Out of the remaining 5 people, they can be arranged in 5! ways.
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17. 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
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
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18. 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.
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.
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19. 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 non-black) or (2 black and 1 non-black) 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
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
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20. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
SHOW ANSWER
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
= (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
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