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( 1 ) How many words can be formed from the letters of the word "AFTER", so that the vowels never comes together.
 1) 48
 2) 52
 3) 72
 4) 120

Show Answer Report Discussion in forumAnswer : 3) 72
Solution : We need to find the ways that vowels NEVER come together.
Vowels are A, E
Let the word be FTR(AE) having 4 words.
Total ways = 4! = 24
Vowels can have total ways 2! = 2
Number of ways having vowel together = 48
Total number of words using all letter = 5! = 120 Number of words having vowels never together = 12048 = 72
discussion
Answer : 3) 72



( 2 ) Out of eight crew members three particular members can sit only on the left side. Another two particular members can sit only on the right side. Find the number of ways in which the crew can be arranged so that four men can sit on each side.
 1) 864
 2) 863
 3) 865
 4) 1728

Show Answer Report Discussion in forumAnswer : 4) 1728
Solution : Required number of ways = ^{3}C_{2} * 4! * 4! = 1728
discussion
Answer : 4) 1728



( 3 ) There are 12 yes or no questions. How many ways can these be answered?
 1) 4096
 2) 2048
 3) 1024
 4) 144

Show Answer Report Discussion in forumAnswer : 1) 4096
Solution : Each of the questions can be answered in 2 ways (yes or no)
Therefore, no. of ways of answering 12 questions =2^{12} = 4096 ways.
discussion
Answer : 1) 4096



( 4 ) Four dice are rolled simultaneously. What is the number of possible outcomes in which at least one of the die shows 6?
 1) 6! / 4!
 2) 625
 3) 671
 4) 1296

Show Answer Report Discussion in forumAnswer : 3) 671
Solution : When 4 dice are rolled simultaneously, there are 64 = 1296 outcomes. The converse of what is asked in the question is that none of the dice show '6'. That is all four dice show any of the other 5 numbers. That is possible in 54 = 625 outcomes.
Therefore, in 1296  625 = 671 outcomes at least one of the dice will show 6.
discussion
Answer : 3) 671



( 5 ) How many alphabets need to be there in a language if one were to make 1 million distinct 3 digit initials using the alphabets of the language?
 1) 26
 2) 50
 3) 100
 4) 1000

Show Answer Report Discussion in forumAnswer : 3) 100
Solution : 1 million distinct 3 digit initials are needed.
Let the number of required alphabets in the language be 'n'.
Therefore, using 'n' alphabets we can form n * n * n = n^{3} distinct 3 digit initials.
Note distinct initials is different from initials where the digits are different.
For instance, AAA and BBB are acceptable combinations in the case of distinct initials while they are not permitted when the digits of the initials need to be different.
This n^{3} different initials = 1 million
i.e. n^{3} = 10^{6} (1 million = 106)
=> n^{3} = (10^{2})^{3}
=> n = 10^{2} = 100
Hence, the language needs to have a minimum of 100 alphabets to achieve the objective.
discussion
Answer : 3) 100



( 6 ) There are three places P, Q and R such that 3 roads connects P and Q and 4 roads connects Q and R. In how many ways can one travel from P to R?
 1) 8
 2) 10
 3) 12
 4) 14

Show Answer Report Discussion in forumAnswer : 3) 12
Solution : The number of ways in which one can travel from P to R = 3 * 4 = 12
discussion
Answer : 3) 12



( 7 ) Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed ?
 1) 210
 2) 1050
 3) 25200
 4) 21400

Show Answer Report Discussion in forumAnswer : 3) 25200
Solution : Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)
= (^{7}C_{3} * ^{4}C_{2}) = (7*6*5/3*2*1*4*3/2*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 * 4 * 3 * 2 * 1) = 120.
∴ Required number of words = (210 * 120) = 25200.
discussion
Answer : 3) 25200



( 8 ) Find out the number of ways in which 6 rings of different types can be worn in 3 fingers?
 1) 120
 2) 720
 3) 125
 4) 729

Show Answer Report Discussion in forumAnswer : 4) 729
Solution : The first ring can be worn in any of the 3 fingers
=> There are 3 ways of wearing the first ring
Similarly each of the remaining 5 rings also can be worn in 3 ways
Hence total number of ways
= 3 * 3 * 3 * 3 * 3 * 6 = 3^{6} =729
discussion
Answer : 4) 729



( 9 ) 25 buses are running between two places P and Q. In how many ways can a person go from P to Q and return by a different bus?
 1) None of these
 2) 600
 3) 576
 4) 625

Show Answer Report Discussion in forumAnswer : 2) 600
Solution : He can go in any bus out of the 25 buses.
Hence He can go in 25 ways.
Since he can not come back in the same bus that he used for travelling,
He can return in 24 ways.
Total number of ways = 25 * 24 = 600
discussion
Answer : 2) 600



( 10 ) An event manager has ten patterns of chairs and eight patterns of tables. In how many ways can he make a pair of table and chair?
 1) 100
 2) 80
 3) 110
 4) 64

Show Answer Report Discussion in forumAnswer : 2) 80
Solution : He has has 10 patterns of chairs and 8 patterns of tables
Hence, A chair can be arranged in 10 ways and A table can be arranged in 8 ways
Hence one chair and one table can be arranged in 10 * 8 ways = 80 ways
discussion
Answer : 2) 80
