TCS Permutations and Combinations Quiz-1

Question 1

Time: 00:00:00
Find no of ways in which 4 particular persons a,b,c,d and 6 more persons can stand in a queue so that A always stand before B. B always stand before C, And C always stand before D?

6!

6!

7!

7!

^{5}\textrm{C}_{2}

^{5}\textrm{C}_{2}

^{10}\textrm{C}_{4}\ast 4!

^{10}\textrm{C}_{4}\ast 4!

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it is not mention that B needs to be just after A.. like A can be in 1 in line B in 5 position still A is before.. They have just assumed the case where they are consequently arranged in line

7!

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Question 2

Time: 00:00:00
There are 10 points on a straight line AB and 8 on another straight line AC none of them being point A. how many triangles can be formed with these points as vertices?

680

680

720

720

816

816

640

640

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10 points on AB and 8 on AC 10C2 * 8C1 + 10C1 * 8C2

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Question 3

Time: 00:00:00
Find the number of ways a batsman can score a double century only in terms of 4's & 6's?

32

32

16

16

64

64

256

256

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It is a trick question most of the answers will try to convince you to find all the possible combinations. The correct answer somehow can be determined by using the LCM of the 2 numbers. Here LCM(4,6)=12 Now divide 200 by 12 to see how many possible arrangements we have. Now we will have the answer as 16.6666….. So since we get 16 as a whole number so the answer will be 16. If we were told that we could also not use one type,i.e If the batsman only scores using 4s or only using 6s then the answer would have been the next whole number, i.e 17.

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Question 4

Time: 00:00:00

How many positive integer numbers not more than 4300 can be formed with the digits 0, 1, 2, 3, 4 if repetitions are allowed?

625

625

560

560

565

565

575

575

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total number minus all the numbers greater than 4300. 625-50

1 digit no. = 4 2 digits no. = 20 3 digits no. = 375 for calculating 4 digits we can three cases - 1st where 4 and 0 won\'t come at thousand place = 3*5*5*5 = 375 2nd where 4 comes at thousand but 3 and 4 won\'t come at hundred place = 1*3*5*5=75 sum up all and you will get 564 well is it the correct answer. do the answers provided in the options are wrong? well, the questions says \'not more than\' means we can get at max 4300 which is our last case (3rd) which we didn\'t cover yet in 4 digits no. = 1*1*1*1*1

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Question 5

Time: 00:00:00
In a 3x3 square grid comprising 9 tiles, each tile can be painted in red or blue colour. When the tile is rotated by 180 degrees, there is no difference which can be spotted. How many such possibilities are there?

16

16

64

64

32

32

256

256

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Question 6

Time: 00:00:00
How many 6 digit even numbers can be formed from digits 1 2 3 4 5 6 7 so that the digit should not repeat and the second last digit is even?

720

720

320

320

2160

2160

6480

6480

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Question 7

Time: 00:00:00
How many vehicle registration plate numbers can be formed with digits 1,2,3,4,5 (no digit being repeated)if it's given that registration number can have 1 to 5 digits?

325

325

205

205

100

100

105

105

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Question 8

Time: 00:00:00
There are 20 persons sitting in a circle. In that, there are 18 men and 2 sisters. How many arrangements are possible, in which the two sisters are always separated by a man?

2

2

17!*2

17!*2

17!

17!

18!*2

18!*2

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Question 9

Time: 00:00:00
A number plate can be formed with two alphabets followed by two digits with no repetition. then how many possible combinations can we get?

67600

67600

64320

64320

58500

58500

65000

65000

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Question 10

Time: 00:00:00
In how many ways a team of 11 must be selected a team 5 men and 11 women such that the team must comprise of not more than 3 men.?

1234

1234

2256

2256

2456

2456

1565

1565

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["0","40","60","80","100"]
["Need more practice! \r\n","Keep trying! \r\n","Not bad! \r\n","Good work! \r\n","Perfect! \r\n"]

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