View in English View in Hindi 
By using digits 1 to 9 how many 4 digit numbers can be formed which has 5 as a unit digit? Repetition of digits is are not allowed. 
A) 336
B) 1400
C) 504
D) 251
Correct Answer : 336 Explanation : 5 at the unit place or divisible by 5. So, there is 1 way of doing it. 
In how many ways can 5 prizes be given away to 7 boys when each boy is eligible for all the prizes? 
A) 5!
B) 7!
C) 5^{7}
D) 7^{5}
Correct Answer : 7^{5} Explanation : A prize can be given to any one of the 7 boys in (7/1)=7 ways 
How many different words, can be formed by rearranging the letters of the word ‘SUCCESS’? 
A) 5040
B) 840
C) 420
D) 1680
Correct Answer : 420 Explanation : The word "SUCCESS" has total 7 elements, there two repeated elements S(thrice) and C(twice). The formula for permutations with repeated elements is as follows, When k out of n elements are indistinguishable, e.g. k copies of the same book, the number of different permutations is n!/k!. 
In how many different ways can the letters of the word ‘ARCHIVE’ be arranged? 
A) 2630
B) 5040
C) 1680
D) 3456
Correct Answer : 5040 Explanation : There are 7 unique letters in ARCHIVE which means that there are 7!=7*6*5*4*3*2*1=5040 different ways to rearrange them. 

In how many different ways can the letters of the word ‘PUBLIC’ be arranged so that the vowels are at the two ends? 
A) 48
B) 72
C) 36
D) None of these
Correct Answer : 48 Explanation : PUBLIC has 2 vowels(UI) and 4 consonants (PBLC). All the letters are distinct. 
If ^{56}P_{r + 6} : ^{54}P_{r + 3} = 30800 : 1 then the value of r is ? 
40 Explanation : ^{56}P_{r + 6} : ^{54}P_{r + 3} = 30800 : 1 As 56! = 56*55*54! So, 
How many groups of 6 persons can be formed from 8 men and 7 women ? 
5000 Explanation : Total number of person = 8 men + 7 women = 15 

If 10 persons meet at a meeting and shake hands exactly once with each other, what is the total number of hand shakes? 
A) 35
B) 35
C) 50
D) 45
Correct Answer : 45 Explanation : Combinations of n objects taken r at a time = ^{n}C_{r}= n! / [r!(nr!)] 
In a chess competition involving some boys and girls of a school, every student had to play exactly one game with every other student. It was found that in 45 games both the players were girls and in 190 games both were boys. The number of games in which one player was a boy and the other was a girl is 
A) 200
B) 216
C) 235
D) 256
Correct Answer : 200 Explanation : Let there were g girls and b boys. Total number of games in which both players were men ^{b}C_{2} = 190 Total number of games= ^{(g + b)}C_{2} = ^{30}C_{2} = (30 *29) / (2* 1) = 435 Number of games in which one player is a boy and the other is a girl = 435  45  190 = 200 