Online exam in Permutation and Combination For Quantitative Aptitude for Competitive exams like IBPS BANK PO/Clerical,SBI,RRB,SSC,LIC,UPSC-CSAT,SCRA.MAT,CMAT,MBA,SNAP,CAT,NTSE,CLAT
Permutation and Combination-Test 1
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Subject :- Quantitative Aptitude
Chapter :- Permutation and Combination-Test 1
Questions :- 25
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How many 3 digit number can be formed with the digits 5, 6, 2, 3, 7 and 9 which are
divisible by 5 and none of its digit is repeated?
Correct
first two places can be filled in 5 and 4 ways respectively so, total number of 3 digit number =5*4*1 = 20
Incorrect
first two places can be filled in 5 and 4 ways respectively so, total number of 3 digit number =5*4*1 = 20
Question 2 of 25
2. Question
1 points
In how many different ways can the letter of the word ELEPHANT be arranged so that
vowels always occur together?
Correct
Vowels = E, E and A. They can be arranged in 3!/2! Ways
so total ways = 6!*(3!/2!) = 2160
Incorrect
Vowels = E, E and A. They can be arranged in 3!/2! Ways
so total ways = 6!*(3!/2!) = 2160
Question 3 of 25
3. Question
1 points
There are 4 bananas, 7 apples and 6 mangoes in a fruit basket. In how many ways can a
person make a selection of fruits from the basket.
Correct
Zero or more bananas can be selected in 4 + 1 = 5 ways (0 orange, 1 orange, 2 orange, 3 orange and 4 orange)
similarly apples can be selected in 7 +1 = 8 ways
and mangoes in 6 +1 = 7 ways
so total number of ways = 5*8*7 = 280
but we included a case of 0 orange, 0 apple and 0 mangoes, so we have to subtract this, so 280– 1 = 279 ways
Incorrect
Zero or more bananas can be selected in 4 + 1 = 5 ways (0 orange, 1 orange, 2 orange, 3 orange and 4 orange)
similarly apples can be selected in 7 +1 = 8 ways
and mangoes in 6 +1 = 7 ways
so total number of ways = 5*8*7 = 280
but we included a case of 0 orange, 0 apple and 0 mangoes, so we have to subtract this, so 280– 1 = 279 ways
Question 4 of 25
4. Question
1 points
There are 15 points in a plane out of which 6 are collinear. Find the number of lines that
can be formed from 15 points.
Correct
From 15 points number of lines formed = 15c2
6 points are collinear, number of lines formed by these = 6c2
So total lines = 15c2 – 6c2 + 1 = 91
Incorrect
From 15 points number of lines formed = 15c2
6 points are collinear, number of lines formed by these = 6c2
So total lines = 15c2 – 6c2 + 1 = 91
Question 5 of 25
5. Question
1 points
In how many ways 4 Indians, 5 Africans and 7 Japanese be seated in a row so that all
person of same nationality sits together
Correct
4 Indians can be seated together in 4! Ways, similarly for Africans and Japanese in 5! and 7!
respectively. So total ways = 4! 5! 7! 3!
Incorrect
4 Indians can be seated together in 4! Ways, similarly for Africans and Japanese in 5! and 7!
respectively. So total ways = 4! 5! 7! 3!
Question 6 of 25
6. Question
1 points
In how many ways 5 Americans and 5 Indians be seated along a circular table, so that
they are seated in alternative positions
Correct
First Indians can be seated along the circular table in 4! Ways and now Americans can be
seated in 5! Ways. So 4! 5! Ways
Incorrect
First Indians can be seated along the circular table in 4! Ways and now Americans can be
seated in 5! Ways. So 4! 5! Ways
Question 7 of 25
7. Question
1 points
4 matches are to be played in a chess tournament. In how many ways can result be
decided?
Correct
Every chess match can have three result i.e. win, loss and draw
so now of ways = 3*3*3*3 = 81 ways
Incorrect
Every chess match can have three result i.e. win, loss and draw
so now of ways = 3*3*3*3 = 81 ways
Question 8 of 25
8. Question
1 points
In a group of 6 boys and 5 girls, 5 students have to be selected. In how many ways it can
be done so that at least 2 boys are included
Correct
6c2*5c3 + 6c3*5c2 + 6c4*5c1 + 6c5
Incorrect
6c2*5c3 + 6c3*5c2 + 6c4*5c1 + 6c5
Question 9 of 25
9. Question
1 points
A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both hearts. Find the Probability of the lost card being a heart?
Correct
Total cards = 52
Drawn cards(Heart) = 2
Present total cards = total cards-drawn cards =52-2=50
Remaining Card 13-2 = 11
Probability = 11/50
Incorrect
Total cards = 52
Drawn cards(Heart) = 2
Present total cards = total cards-drawn cards =52-2=50
Remaining Card 13-2 = 11
Probability = 11/50
Question 10 of 25
10. Question
1 points
There are three boxes each containing 3 Pink and 5 Yellow balls and also there are 2 boxes each containing 4 Pink and 2 Yellow balls. A Yellow ball is selected at random. Find the probability that Yellow ball is from a box of the first group?
A fruit basket contains 10 Guavas and 20 Bananas out of which 3 Guavas and 5 Bananas are defective. If two fruits selected at random, what is the probability that either both are Bananas or both are non-defective?
A committee of five persons is to be chosen from a group of 10 people. The probability that a certain married couple will either serve together or not at all is?
Correct
Five persons is to be chosen from a group of 10 people = 10C5 = 252
Couple Serve together = 8C3 * 2C2 = 56
Couple does not serve = 8C5 = 56
Probability = 102/252 = 51/126
Incorrect
Five persons is to be chosen from a group of 10 people = 10C5 = 252
Couple Serve together = 8C3 * 2C2 = 56
Couple does not serve = 8C5 = 56
Probability = 102/252 = 51/126
Question 13 of 25
13. Question
1 points
Out of 14 applicants for a job, there are 6 women and 8 men. It is desired to select 2 persons for the job. The probabilty that atleast one of selected persons will be a Woman is?
Correct
Man only = 8C2 = 14
Probability of selecting no woman = 14/91
Probability of selecting atleast one woman = 1 – 14/91 = 77/91
Incorrect
Man only = 8C2 = 14
Probability of selecting no woman = 14/91
Probability of selecting atleast one woman = 1 – 14/91 = 77/91
Question 14 of 25
14. Question
1 points
Three Bananas and three oranges are kept in a box. If two fruits are chosen at random, Find the probability that one is Banana and another one is orange?
Correct
Total probability = 6C2 = 15
Probability that one is Banana and another one is orange = 3C1 * 3C1 = 9
probability = 9/15 = 3/5
Incorrect
Total probability = 6C2 = 15
Probability that one is Banana and another one is orange = 3C1 * 3C1 = 9
probability = 9/15 = 3/5
Question 15 of 25
15. Question
1 points
A basket contains 6 White 4 Black 2 Pink and 3 Green balls. If three balls picked up random, What is the probability that all three are White?
Correct
Total Balls = 15
Probability = 6c3 / 15c3 = 4/91
Incorrect
Total Balls = 15
Probability = 6c3 / 15c3 = 4/91
Question 16 of 25
16. Question
1 points
A basket contains 6 White 4 Black 2 Pink and 3 Green balls. If three balls are picked at random, what is the probability that two are Black and one is Green?
Correct
Total Balls = 15
Probability = 4c2 * 3c1/ 15c3 = 18/455
Incorrect
Total Balls = 15
Probability = 4c2 * 3c1/ 15c3 = 18/455
Question 17 of 25
17. Question
1 points
A basket contains 6 White 4 Black 2 Pink and 3 Green balls. If four balls are picked at random, what is the probability that atleast one is Black?
Correct
Total Balls = 15
Probability = 11c4/15c4 = 22/91
One is black = 1 – 22/91 = 69/91
Incorrect
Total Balls = 15
Probability = 11c4/15c4 = 22/91
One is black = 1 – 22/91 = 69/91
Question 18 of 25
18. Question
1 points
A basket contains 6 White 4 Black 2 Pink and 3 Green balls.If two balls are picked at random, what is the probability that either both are Pink or both are Green?
Correct
Probability both are Pink = 1/15C2
Probability both are Green = 3/15C2
Required Probability = 4/15C2 = 4/105
Incorrect
Probability both are Pink = 1/15C2
Probability both are Green = 3/15C2
Required Probability = 4/15C2 = 4/105
Question 19 of 25
19. Question
1 points
How many words of 4 letters with or without meaning be made from the letters of the word ‘NUMBER’, when repetition of letters is not allowed?
Correct
NUMBER is 6 letters. We have 4 places where letters are to be placed. For first letter there are 6 choices, since repetition is not allowed, for second, third and fourth letter also we have 5, 4, and 3 choices resp., so total of 6*5*4*3 ways = 360 ways.
Incorrect
NUMBER is 6 letters. We have 4 places where letters are to be placed. For first letter there are 6 choices, since repetition is not allowed, for second, third and fourth letter also we have 5, 4, and 3 choices resp., so total of 6*5*4*3 ways = 360 ways.
Question 20 of 25
20. Question
1 points
In how many ways the letters of the word ‘ALLIGATION’ be arranged taking all the letters?
Correct
ALLIGATION contains 10 letters, so total 10! ways. There are 2 As, 2 Ls, 2 Is
So 10!/(2!*2!*2!)
Incorrect
ALLIGATION contains 10 letters, so total 10! ways. There are 2 As, 2 Ls, 2 Is
So 10!/(2!*2!*2!)
Question 21 of 25
21. Question
1 points
In how many ways all the letters of the word ‘MINIMUM’ be arranged such that all vowels are together?
Correct
Take vowels in a box together as one – IIU, M, N, M, M
So there are 5 that to be placed for this 5!, now 3 Ms, so 5!/3!, so arrangement of vowels inside box gives
3!/2!
So total = 5!/3! * 3!/2!
Incorrect
Take vowels in a box together as one – IIU, M, N, M, M
So there are 5 that to be placed for this 5!, now 3 Ms, so 5!/3!, so arrangement of vowels inside box gives
3!/2!
So total = 5!/3! * 3!/2!
Question 22 of 25
22. Question
1 points
In how many ways a group of 4 men and 3 women be made out of a total of 8 men and 5
women?
Correct
Total ways = 8C4*5C3
Incorrect
Total ways = 8C4*5C3
Question 23 of 25
23. Question
1 points
How many 3 digit numbers are divisible by 4?
Correct
A number is divisible by 4 when its last two digits are divisible by 4
For this the numbers should have their last two digits as 00, 04, 08, 12, 16, … 96
By the formula, an = a + (n-1)d
96 = 0 + (n-1)*4
n = 25
so there are 25 choices for last 2 digits and 9 choices (1-9) for the 1st digit
so total 9*25=225
Incorrect
A number is divisible by 4 when its last two digits are divisible by 4
For this the numbers should have their last two digits as 00, 04, 08, 12, 16, … 96
By the formula, an = a + (n-1)d
96 = 0 + (n-1)*4
n = 25
so there are 25 choices for last 2 digits and 9 choices (1-9) for the 1st digit
so total 9*25=225
Question 24 of 25
24. Question
1 points
How many 3 digits numbers have exactly one digit 2 in the number?
Correct
0 cannot be placed at first digit to make it a 3 digit number.
3 cases:
Case 1: 2 is placed at first place
1 choice for the first place, 9 choices each for the 2nd and 3rd digit (0-9 except 2)
So numbers = 1*9*9 = 81
Case 2: 2 is placed at second place
8 choices for the first place (1-9 except 2), 1 choice for the 2nd digit and 9 choices for the 3rd digit (0-9
except 2)
So numbers = 8*1*9 = 72
Case 3: 2 is placed at third place
8 choices for the first place (1-9 except 2), 9 choices for the 2nd digit (0-9 except 2) and 1 choice for the
3rd digit
So numbers = 8*9*1 = 72
So total numbers = 81+72+72 = 225
Incorrect
0 cannot be placed at first digit to make it a 3 digit number.
3 cases:
Case 1: 2 is placed at first place
1 choice for the first place, 9 choices each for the 2nd and 3rd digit (0-9 except 2)
So numbers = 1*9*9 = 81
Case 2: 2 is placed at second place
8 choices for the first place (1-9 except 2), 1 choice for the 2nd digit and 9 choices for the 3rd digit (0-9
except 2)
So numbers = 8*1*9 = 72
Case 3: 2 is placed at third place
8 choices for the first place (1-9 except 2), 9 choices for the 2nd digit (0-9 except 2) and 1 choice for the
3rd digit
So numbers = 8*9*1 = 72
So total numbers = 81+72+72 = 225
Question 25 of 25
25. Question
1 points
There are 8 men and 7 women. In how many ways a group of 5 people can be made such that the particular woman is always to be included?
Correct
Total 15 people, and a particular woman is to be taken to form a group of 5, so choice is to be done from
14 people of 4 people
Ways are 14C4.
Incorrect
Total 15 people, and a particular woman is to be taken to form a group of 5, so choice is to be done from
14 people of 4 people
Ways are 14C4.