Online Quiz 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 3
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Subject :- Quantitative Aptitude
Chapter :- Permutation and Combination-Test 3
Questions :- 25
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In how many ways can 7 beads be strung into necklace ?
Correct
No of way in Necklace = (n-1)!/2 = 6!/2
= 720/2 = 360
Incorrect
No of way in Necklace = (n-1)!/2 = 6!/2
= 720/2 = 360
Question 3 of 25
3. Question
1 points
Find the no of 3 digit numbers such that atleast one of the digit is 6 (with repetitions) ?
Correct
Total no of 3 digit number = 9*10*10 = 900
No of 3 digit number- none of the digit is 6 = 8*9*9 = 648
No of 3 digit number – atleast one digit is 6 = 900-648 = 252
Incorrect
Total no of 3 digit number = 9*10*10 = 900
No of 3 digit number- none of the digit is 6 = 8*9*9 = 648
No of 3 digit number – atleast one digit is 6 = 900-648 = 252
Question 4 of 25
4. Question
1 points
In how many ways can 7 girls and 4 boys stand in a row so that no 2 boys are together ?
Correct
No of ways = 7!*8P4
7! = 5040
8P4 = 8*7*6*5 = 1680
No of ways = 5040*1680 = 8467200
Incorrect
No of ways = 7!*8P4
7! = 5040
8P4 = 8*7*6*5 = 1680
No of ways = 5040*1680 = 8467200
Question 5 of 25
5. Question
1 points
In how many ways the letters of the word PERMUTATION be arranged ?
Correct
Incorrect
Question 6 of 25
6. Question
1 points
How many numbers can be formed with the digits 1, 7, 2, 5 without repetition ?
Correct
1 digit number = 4
2 digit no = 4*3 = 12
3 digit no = 4*3*2 = 24
4 digit no = 4*3*2*1 = 24
Total = 4+12+24+24 = 64
Incorrect
1 digit number = 4
2 digit no = 4*3 = 12
3 digit no = 4*3*2 = 24
4 digit no = 4*3*2*1 = 24
Total = 4+12+24+24 = 64
Question 7 of 25
7. Question
1 points
There are 3 boxes and 6 balls. In how many ways these balls can be distributed if all the balls and all the boxes are different?
Correct
${3}^{6}=729$
Incorrect
${3}^{6}=729$
Question 8 of 25
8. Question
1 points
In how many ways can 4 books be selected out of 10 books on different subjects ?
Correct
16C4 = 10*9*8*7/4*3*2*1 = 5040/24 = 210
Incorrect
16C4 = 10*9*8*7/4*3*2*1 = 5040/24 = 210
Question 9 of 25
9. Question
1 points
A six-digit is to be formed from the given numbers 1, 2, 3, 4, 5 and 6. Find the probability that the number is divisible by 4.
Correct
For a number to be divisible by 4, the last two digit should be divisible by 4.
So possible cases – 12, 16, 24, 32, 36, 52, 56, 64 (last two digits)
So favourable outcomes = 24 +24 +24 +24 + 24+ 24+24+24 = 192
So p = 192/720 = 4/15
Incorrect
For a number to be divisible by 4, the last two digit should be divisible by 4.
So possible cases – 12, 16, 24, 32, 36, 52, 56, 64 (last two digits)
So favourable outcomes = 24 +24 +24 +24 + 24+ 24+24+24 = 192
So p = 192/720 = 4/15
Question 10 of 25
10. Question
1 points
A bag contains 6 red balls and 7 white balls. Another bag contains 5 red balls and 3 white balls. One ball is selected from each. Find the probability that one ball is red and one is white?
Correct
(6/13)*(3/8) + (7/13)*(5/8) = 53/104
Incorrect
(6/13)*(3/8) + (7/13)*(5/8) = 53/104
Question 11 of 25
11. Question
1 points
A lottery is organised by the college ABC through which they will provide scholarship of rupees one lakhs to only one student. There are 100 fourth year students, 150 third year students, 200 second year students and 250 first year students. What is the probability that a second year student is choosen.
Correct
Second year students = 200
so, P = 200/700 = 2/7
Incorrect
Second year students = 200
so, P = 200/700 = 2/7
Question 12 of 25
12. Question
1 points
A card is drawn from a pack of 52 cards. The card is drawn at random; find the probability that it is neither club nor queen?
Correct
1 – [13/52 + 4/52 – 1/52] = 9/13
Incorrect
1 – [13/52 + 4/52 – 1/52] = 9/13
Question 13 of 25
13. Question
1 points
A box contains 50 balls, numbered from 1 to 50. If three balls are drawn at random with
replacement. What is the probability that sum of the numbers are odd?
Correct
There are 25 odd and 25 even numbers from 1 to 50.
Sum will be odd if = odd + odd + odd, odd + even + even, even + odd + even, even+ even + odd
P= (1/2)*(1/2)*(1/2) + (1/2)*(1/2)*(1/2) + (1/2)*(1/2)*(1/2) + (1/2)*(1/2)*(1/2)
=4/8 = ½
Incorrect
There are 25 odd and 25 even numbers from 1 to 50.
Sum will be odd if = odd + odd + odd, odd + even + even, even + odd + even, even+ even + odd
P= (1/2)*(1/2)*(1/2) + (1/2)*(1/2)*(1/2) + (1/2)*(1/2)*(1/2) + (1/2)*(1/2)*(1/2)
=4/8 = ½
Question 14 of 25
14. Question
1 points
From a pack of cards, if three cards are drawn at random one after the other with replacement, find the probability that one is ace, one is jack and one is queen?
Correct
(4c1 + 4c1 + 4c1)/(52c3)
Incorrect
(4c1 + 4c1 + 4c1)/(52c3)
Question 15 of 25
15. Question
1 points
A and B are two persons sitting in a circular arrangement with 8 other persons. Find the
probability that both A and B sit together.
Correct
Total outcomes = (10 -1)! = 9!
Favourable outcomes = (9 -1)!*2!
So p = 2/9
Incorrect
Total outcomes = (10 -1)! = 9!
Favourable outcomes = (9 -1)!*2!
So p = 2/9
Question 16 of 25
16. Question
1 points
Find the probability that in a random arrangement of the letter of words in the word
‘PROBABILITY’ the two I’s come together.
Correct
Total outcomes = 11!/(2!*2!)
favourable outcomes = (10!*2!)/(2!*2!)
p = 2/11
Incorrect
Total outcomes = 11!/(2!*2!)
favourable outcomes = (10!*2!)/(2!*2!)
p = 2/11
Question 17 of 25
17. Question
1 points
In a race of 12 cars, the probability that car A will win is 1/5 and of car B is 1/6 and that of car C is 1/3. Find the probability that only one of them won the race.
A bag contains 3 red balls and 8 blacks ball and another bag contains 5 red balls and 7 blacks balls, one ball is drawn at random from either of the bag, find the probability that the ball is red.
Correct
Probability = probability of selecting the bag and probability of selecting red ball
(1/2)*(3/11) + (1/2)*(5/12) = 91/264
Incorrect
Probability = probability of selecting the bag and probability of selecting red ball
(1/2)*(3/11) + (1/2)*(5/12) = 91/264
Question 19 of 25
19. Question
1 points
In how many ways can 5 boys and 4 girls can be seated in a row so that they are in alternate position.
Correct
First boys are seated in 5 position in 5! Ways, now remaining 4 places can be filled by 4 girls in 4! Ways,
so number of ways = 5! 4! = 2880
Incorrect
First boys are seated in 5 position in 5! Ways, now remaining 4 places can be filled by 4 girls in 4! Ways,
so number of ways = 5! 4! = 2880
Question 20 of 25
20. Question
1 points
In how many ways 5 African and five Indian can be seated along a circular table, so that they occupy alternate position.
Correct
First 5 African are seated along the circular table in (5-1)! Ways = 4!. Now Indian can be seated in 5! Ways, so 4! 5!
Incorrect
First 5 African are seated along the circular table in (5-1)! Ways = 4!. Now Indian can be seated in 5! Ways, so 4! 5!
Question 21 of 25
21. Question
1 points
There is meeting of 20 delegates is to be held in a hotel. In how many ways these delegates can be seated along a round table, if three particular delegates always seat together.
Correct
Total 20 persons, 3 always seat together, 17 + 1 =18 delegates can be seated in (18 -1)! Ways = 17! And
now that three can be arranged in 3! Ways. So, 17! 3!
Incorrect
Total 20 persons, 3 always seat together, 17 + 1 =18 delegates can be seated in (18 -1)! Ways = 17! And
now that three can be arranged in 3! Ways. So, 17! 3!
Question 22 of 25
22. Question
1 points
In how many 8 prizes can be given to 3 boys, if all boys are equally eligible of getting the prize.
Correct
Prizes cab be given in 8*8*8 ways = 512 ways
Incorrect
Prizes cab be given in 8*8*8 ways = 512 ways
Question 23 of 25
23. 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 24 of 25
24. Question
1 points
In party there is a total of 120 handshakes. If all the persons shakes hand with every other person. Then find the number of person present in the party.
Correct
Nc2 = 120 (N is the number of persons)
Incorrect
Nc2 = 120 (N is the number of persons)
Question 25 of 25
25. Question
1 points
There are 8 boys and 12 girls in a class. 5 students have to be chosen for an educational trip. Find the number of ways in which this can be done if 2 particular girls are always included