Question 1 
What is the number of possible words that can be made using the word “EASYQUIZ” such that the vowels always come together?
120  
720  
2880  
4320 
Discuss it
Question 1 Explanation:
The word “EASYQUIZ” has 8 letters in which “EAUI” are vowels.
Since vowels always come together, we can assume “EAUI” as a single unit letter.
4+1 letter can be arranged in 5! ways. Also, vowels “EAUI” can be arranged in 4! ways.
Hence the total number of possible words = 5! * 4! = 2880.
Question 2 
What is the number of possible words that can be made using the word “QUIZ” such that the vowels never come together?
8  
12  
16  
24 
Discuss it
Question 2 Explanation:
The word “QUIZ” has 4 letters in which “UI” are vowels.
Total number of possible words = 4!
Treating “UI” as a single letter we can make words in 3! ways.
“UI” can be arranged in 2! ways.
Therefore, the number of words can be made using vowels together = 3! * 2! = 12 ways.
Hence, the number of words can be made such that vowels never come together = 24 – 12 = 12 ways.
Question 3 
How many words can be made from the word “APPLE” using all the alphabets with repetition and without repetition respectively?
1024, 60  
60, 1024  
1024, 1024  
240, 1024 
Discuss it
Question 3 Explanation:
The word “APPLE” has 5 letters in which “P” comes twice.
If repetition is allowed, the number of words we can form = 4*4*4*4*4 = 1024.
(This is because, when repetition is allowed, we can put any of the four unique alphabets at each of the five positions.)
If repetition is not allowed, the number of words we can form = 5!/2! = 60. (This is because "P" comes twice.)
Question 4 
How many ways a 6 member team can be formed having 3 men and 3 ladies from a group of 6 men and 7 ladies?
700  
720  
120  
500 
Discuss it
Question 4 Explanation:
We have to pick 3 men from 6 available men and 3 ladies from 7 available ladies.
Required number of ways = ^{6}C_{3} * ^{7}C_{3} = 20 * 35 = 700.
Question 5 
In how many ways can an interview panel of 3 members be formed from 3 engineers, 2 psychologists and 3 managers if at least 1 engineer must be included?
30  
15  
46  
45 
Discuss it
Question 5 Explanation:
The interview panel of 3 members can be formed in 3 ways by selecting 1 engineer and 2 other professionals, 2 engineers and 1 other professionals and all 3 engineers.
 1 engineer out of 3 engineers and 2 other professionals out of 5 professionals can be selected as = ^{3}C_{1} * ^{5}C_{2} = 3 * 10 = 30 ways.
 2 engineers out of 3 engineers and 1 other professional out of 5 professionals can be selected as = ^{3}C_{2} * ^{5}C_{1} = 3 * 5 = 15 ways.
 3 engineers out of 3 engineers and 0 other professional out of 5 professionals can be selected as = ^{3}C_{3} * ^{5}C_{0} = 1 way.
Question 6 
How many 4digit numbers can be formed from the digits 1, 2, 3, 4, 5, 6 and 7 which are divisible by 5 when none of the digits are repeated?
120  
35  
24  
720 
Discuss it
Question 6 Explanation:
A number is divisible by 5 if and only if its last digit is either 5 or 0. But, 0 is not available here. So, we have to fix 5 as a last digit of 4digit number and fill 3 places with remaining 6 digits.
Number of ways to choose 3 digits = ^{6}C_{3} = 20.
Number of ways to arrange the chosen digits = 3!
Hence, total number of required ways = ^{6}C_{3} * 3! = ^{6}P_{3} = 120.
Question 7 
In how many ways can 20 boys and 18 girls make a queue such that no two girls are together?
190  
3540  
1230  
1330 
Discuss it
Question 7 Explanation:
There are a total of 21 possible places available between boys such that no 2 girls can be placed together (alternate sequence of boys and girls, starting and ending positions for girls).
Therefore, the 18 girls can stand at these 21 places only.
Hence, the number of ways = ^{21}C_{19} = 21*20*19/3*2 = 1330.
Question 8 
There are 5 floating stones on a river. A man wants to cross the river. He can move either 1 or 2 steps at a time. Find the number of ways in which he can cross the river?
11  
12  
13  
14 
Discuss it
Question 8 Explanation:
The man needs to take 6 steps to cross the river. He can do this in the following ways:
 Crossing the river by 6 unit steps = 1 way.
 Crossing the river by 4 unit steps and 1 double step = ^{5}C_{1} = 5C4 = 5 ways.
 Crossing the river by 2 unit steps and 2 double steps = ^{4}C_{2} = 6 ways.
 Crossing the river by 3 double steps = 1 way.
Question 9 
Out of 7 boys and 4 girls, how many queues of 3 boys and 2 girls can be formed?
120  
25200  
24800  
1440 
Discuss it
Question 9 Explanation:
Number of ways to choose 3 boys out of 7 = ^{7}C_{3}.
Number of ways to choose 2 girls out of 4 = ^{4}C_{2}.
Therefore, number of ways to choose the required groups = ^{7}C_{3} * ^{4}C_{2} = 35 * 6 = 210.
Number of ways to arrange the 3 boys and 2 girls in a queue = 5! = 120.
Therefore, the required number of queues = 210 * 120 = 25200.
Question 10 
A box contains 2 red coins, 3 green coins and 4 blue coins. In how many ways can 3 coins be chosen such that at least one coin is green?
16  
32  
64  
128 
Discuss it
Question 10 Explanation:
There are three cases:
 3 green coins
 2 green coins + 1 nongreen coin
 1 green coin + 2 nongreen coins
Question 11 
Out of 6 engineers and 4 doctors, how many groups of 4 professionals can be formed such that at least 1 engineer is always there?
129  
109  
229  
209 
Discuss it
Question 11 Explanation:
There are four cases:
 4 engineers = 6C4 = 15
 3 engineers and 1 doctor = ^{6}C_{3}*^{4}C_{1} = 20*4 = 80
 2 engineers and 2 doctors = ^{6}C_{2}*^{4}C_{2} = 15*6 = 90
 1 engineer and 3 doctors = ^{6}C_{1}*^{4}C_{3} = 24
Question 12 
Out of 8 boys and 10 girls, how many groups of 5 boys and 6 girls can be formed?
11760  
25200  
720
 
120960 
Discuss it
Question 12 Explanation:
Required number of ways = ^{8}C_{5} * ^{10}C_{6} = 56 * 210 = 11760.
Question 13 
In how many ways can the alphabets of the word ‘DERAIL’ be arranged so that the vowels come at the odd positions only?
12  
18  
24  
36 
Discuss it
Question 13 Explanation:
There are 3 consonants (D,R,L) and 3 vowels (E,A,I).
There are 3 odd positions and 3 even positions: O E O E O E.
Therefore, required number of arrangements = ^{3}P_{3} * ^{3}P_{3} = 6 * 6 = 36.
Question 14 
In how many different ways can the alphabets of the word ‘SCORING’ be arranged so that the vowels always come together?
120  
720  
240  
1440 
Discuss it
Question 14 Explanation:
We have 5 consonants and 2 vowels.
Since, the vowels must always come together, we can treat them as a single alphabet.
Then, we have to arrange 6 alphabets.
Number of ways to arrange 6 alphabets = 6! = 720.
The two vowels can be arranged in 2! Ways. So, the required number of ways = 6! * 2! = 1440.
Question 15 
The value of ^{75}C_{2} is:
1215
 
2315
 
2775
 
1675

Discuss it
Question 15 Explanation:
^{75}C_{2} = 75! / (73! * 2!) = 75*74/2 = 2775.
There are 15 questions to complete.