There are 5 doors to a lecture hall.The number of ways that a student can enter the hall and leave it by a different door is:
✓
20
B
16
C
19
D
21
Answer
Correct option: A.
20
A student has 5 different doors to enter,
Every door a student enters through, there are 4 more doors to leave through.
So,
Total no. of ways $=5\times4=20$
Amy and Adam are making boxes of truffles to give out as wedding favors. They have an unlimited supply of 5 different types of truffles. If each box holds 2 truffles of different types, how many different boxes can they make?
A
12
✓
10
C
15
D
20
Answer
Correct option: B.
10
10 boxes In every combination, 2 types of truffles will be in the box, and 3 types of truffles will not.
Therefore, this problem is a question about the number of anagrams that can be made from the "word" YYNNN:
$\frac{5!}{2!3!}=\frac{5\times4\times3\times2\times1}{3\times2\times1\times2\times1}=5\times2=10$
A group consists of 4 couples in which each of the 4 persons have one wife each.In how many ways could they be arranged in a straight line such that the men and women occupy alternate positions:
The total number of 9 digit numbers of different digits is:
A
99!
B
9!
C
8 × 9!
✓
9 × 9!
Answer
Correct option: D.
9 × 9!
Given digit in the number = 9
1st place can be filled = 9 ways = 9 (from 1 - 9 any number can be placed at first position)
2nd place can be filled = 9 ways (from 0 - 9 any number can be placed except the number which is placed at the first position)
3rd place can be filled = 8 ways.
4th place can be filled = 7 ways.
5th place can be filled = 6 ways.
6th place can be filled = 5 ways.
7th place can be filled = 4 ways.
8th place can be filled = 3 ways.
9th place can be filled = 2 ways.
So total number of ways = 9 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2
= 9 × 9!
If the letters of the word KRISNA are arranged in all possible ways and these words are written out as in a dictionary, then the rank of the word KRISNA is:
✓
324
B
341
C
359
D
None of these
Answer
Correct option: A.
324
When arranged alphabetically, the letters of the word KRISNA are A, I, K, N, R and S.
Number of words that will be formed with A as the first letter = Number of arrangements of the remaining 5 letters = 5!
Number of words that will be formed with I as the first letter = Number of arrangements of the remaining 5 letters = 5!
$\therefore$ The number of words beginning with KA = Number of arrangements of the remaining 4 letters = 4!
The number of words beginning with KA = Number of arrangements of the remaining 4 letters = 4!
The number of words starting with KN = Number of arrangements of the remaining 4 letters = 4!
Alphabetically, the next letter will be KR.
Number of words starting with KR followed by A, i.e. KRA = Number of arrangements of the remaining 3 letters = 3!
Number of words starting with KRI followed by A, i.e. KRIA = Number of arrangements of the remaining 2 letters = 2!
Number of words starting with KRI followed by N, i.e. KRIN = Number of arrangements of the remaining 2 letters = 2!
The first word beginning with KRIS is the word KRISAN and the next word is KRISNA.
$\therefore$ Rank of the word KRISNA = 5! + 5! + 4! + 4! + 4! + 3! + 2! + 2! + 2
= 324
If in a group of n distinct objects, the number of arrangements of 4 objects is 12 times the number of arrangements of 2 objects, then the number of objects is:
A
10
B
8
✓
6
D
None of these
Answer
Correct option: C.
6
According to the question:
$^\text{n}\text{P}_4=12\times\ ^\text{n}\text{P}_2$
$\Rightarrow \frac{\text{n!}}{(\text{n}-4)!}=12\times\frac{\text{n!}}{\text{(n-2)!}}$
$\Rightarrow \frac{(\text{n}-2)!}{(\text{n}-4)!}=12$
$\Rightarrow (\text{n}-2)(\text{n}-3)=4\times 3$
$\Rightarrow \text{n}-2=4$
$\Rightarrow \text{n}=6$
A garrison of nn men had enough food to last for 30 days. After 10 days, 50 more men joined them. If the food now lasted for 1616 days, what is the value of n?
✓
200
B
240
C
280
D
320
Answer
Correct option: A.
200
After 10 days, the food for n men is there for 20 days.
This food can be eaten by $(\text{n}+50)$ men in 16 days.
$\therefore20\text{n}=16(\text{n}+5)$
$\therefore\text{n}=200$
The number of permutations of n different things taking r at a time when 3 particular things are to be included is:
A
$^{\text{n}-3}\text{P}_{\text{r}-3}$
B
$^{\text{n}-3}\text{P}_{\text{r}}$
C
$^{\text{n}}\text{P}_{\text{r}-3}$
✓
$\text{r! }^{ \text{n}-3}\text{C}_{\text{r}-3}$
Answer
Correct option: D.
$\text{r! }^{ \text{n}-3}\text{C}_{\text{r}-3}$
Here, we have to permute n things of which 3 things are to be included. So, only the remaining (n − 3) things are left for permutation, taking (r − 3) things at a time. This is because 3 things have already been included. But, these r things can be arranged in r! ways.
$\therefore$ Total number of permutations $=\text{r! }^{ \text{n}-3}\text{C}_{\text{r}-3}$
15 buses operate between Hyderabad and Tirupathi.The number of ways can a man go to Tirupathi from Hyderabad by a bus and return by a different bus is:
A
15
B
150
✓
210
D
225
Answer
Correct option: C.
210
210
Solution:
While going ,number of ways to choose 1 bus out of 15 is ${ }^{15} \mathrm{C}_1$.
While return trip, in order to come with different bus, number of buses left $=14$.
Number of ways to choose 1 bus for return trip $={ }^{14} \mathrm{C}_1$
So, the required number of ways for going and return $={ }^{15} \mathrm{C}_1 \times{ }^{14} \mathrm{C}_1=210$
The number of words that can be formed out of the letters of the word "ARTICLE" so that vowels occupy even places is:
A
574
B
36
C
754
✓
144
Answer
Correct option: D.
144
The word ARTICLE consists of 3 vowels that have to be arranged in the three even places. This can be done in 3! ways. And, the remaining 4 consonants can be arranged among themselves in 4! ways.
$\therefore$ Total number of ways = 3! × 4! = 144
A coin is tossed n times, the number of all the possible outcomes is:
A
$2 n$
✓
$2^n$
C
$C(n, 2)$
D
$P(n, 2)$
Answer
Correct option: B.
$2^n$
$2^n$
Solution:
We know that, when a coin is tossed, we will get either head or tail.
Therefore, the number of all possible outcomes when a coin is tossed n times is $2^n$.
Seven different lecturers are to deliver lectures in seven periods of a class on a particular day. A B, and C are three of the lecturers.The umber of ways in which a routine for the day can be made such that A delivers his lecture before B and B before C, is:
A batsman can score 0, 1, 2, 3, 4 or 6 runs from a ball. The number of different sequences in which he can score exactly 30 runs in an over of six balls is:
Choose the correct answer. The number of triangles that are formed by choosing the vertices from a set of 12 points, seven of which lie on the same line is.
A
105
B
15
C
175
✓
185
Answer
Correct option: D.
185
185
Solution:
Total number of triangle formed from 12 points taking 3 at a time $={ }^{12} \mathrm{C}_3$
But given that out of 12 points 7 are collinear. So, these seven points will form no triangle.
$\therefore$ The required number of triangles $={ }^{12} \mathrm{C}_3-{ }^7 \mathrm{C}_3$
$=\frac{12!}{3!\ 9!}-\frac{7!}{3!\ 4!}=\frac{12\times11\times10\times9}{3\times2\times1\times9!}-\frac{7\times6\times5\times4!}{3\times2\times1\times4!}$
$=\frac{12\times11\times10}{3\times2}-\frac{7\times6\times5}{3\times2}=220-35=185$
The number of ways to arrange the letters of the word CHEESE are:
✓
120
B
240
C
720
D
6
Answer
Correct option: A.
120
Total number of arrangements of the letters of the word CHEESE = Number of arrangements of 6 things taken all at a time, of which 3 are of one kind $=\frac{6!}{3!}=120$
The number of ways in which four particular persons A, B, C, D, and six more persons can stand in a queue so that A always stands before B B, before C and C before D, is:
Find the number of permutations if n = 12 and r = 2.
A
24
B
60
C
106
✓
132
Answer
Correct option: D.
132
132
Solution:
The solution is here:
$\text{n}=12$
$\text{r}=2$
Using the formula given above:
Permutation:
$\ ^\text{n}\text{p}_\text{r}=\frac{\text{(n})!}{\text{(n-r)}!}=\frac{(12)!}{(12-2)}=\frac{12!}{10!}=\frac{(12\times11\times10!)}{10!}=132$
How many factors are $2^5 \times 3^6 \times 5^2$ are perfect squares:
✓
24
B
12
C
16
D
22
Answer
Correct option: A.
24
24
Solution:
Any factors of $2^5 \times 3^6 \times 5^2$ which is a perfect square will be of the form $2^a \times 3^b \times 5^c$ where a can be 0 or 2 or 4 , So there are 3 ways.
b can be 0 or 2 or 4 or 6 , So there are 4 ways.
a can be 0 or 2 , So there are 2 ways.
So, the required number of factors $=3 \times 4 \times 2=24$
The number of different ways in which 8 persons can stand in a row so that between two particular persons A and B there are always two persons, is:
✓
60 × 5!
B
15 × 4! × 5!
C
4! × 5!
D
None of these.
Answer
Correct option: A.
60 × 5!
60 × 5!
Solutions:
The four people, i.e A, B and the two persons between them are always together. Thus, they can be considered as a single person. So, along with the remaining 4 persons, there are now total 5 people who need to be arranged. This can be done in 5! ways. But, the two persons that have to be included between A and B could be selected out of the remaining 6 people in ${ }^6 P_2$ ways, which is equal to 30. For each selection, these two persons standing between A and B can be arranged among themselves in 2 ways.
$\therefore$ Total number of arrangements = 5! × 30 × 2 = 60 × 5!
The number of ways in which 6 men add 5 women can dine at a round table, if no two women are to sit together, is given by:
A
30
✓
5! × 5!
C
5! × 4!
D
7! × 5!
Answer
Correct option: B.
5! × 5!
Again, 6 girls can be arranged among themselves in 5! ways in a circle.
So, the number of arrangements where boys and girls sit attentively in a circle = 5! × 5!
Choose the correct answer.The number of words which can be formed out of the letters of the word ARTICLE, so that vowels occupy the even place is.
A
1440
✓
144
C
7!
D
$^4C_4 \times ^3C_3$
Answer
Correct option: B.
144
144
Solution:
Total number of letters in the 'ARTICLE' is 7 out which $A , E , I$ are vowels and $R , T , C , L$ are consonants
Given that vowels occupy even place
$\therefore$ Possible arrangement can be shown as below
C, V, C, V, C, V, C i.e. on $2^{\text {nd }}, 4^{\text {th }}$ and $6^{\text {th }}$ places
Therefore, number of arrangement $={ }^3 P _3=3!=6$ ways
Now consonants can be placed at $1,3,5$ and $7^{\text {th }}$ place
$\therefore$ Number of arrangement $={ }^4 P _4=4!=24$
So, the total number of arrangements $=6 \times 24=144$
The number of ways 4 boys and 3 girls can be seated in a row so that they are alternate is:
A
12
B
104
✓
144
D
256
Answer
Correct option: C.
144
Given that, there are 4 boys and 3 girls.
The only pattern 4 boys and 3 girls are arranged in an alternate way is BGBGBGB.
Therefore, the total number of ways is 4! × 3! = 144.
6 men and 4 women are to be seated in a row so that no two women sit together.The number of ways they can be seated is:
✓
604800
B
17280
C
120960
D
518400
Answer
Correct option: A.
604800
6 men can be sit as
$\times\text{M}\times\text{M}\times\text{M}\times\text{M}\times\text{M}\times\text{M}\times$
Now, there are 7 spaces and 4 women can be sit as $ \ ^7\text{P}_4 = \ ^7\text{P}_3 = \frac{7!}{3!} = \frac{(7\times6\times5\times4\times3!)}{3!}$
$=7\times6\times5\times4\times=840$
Now, total number of arrangement $= 6!\times840$
$= 720\times840$
$= 604800$
How many different signals can be transmitted by arranging 3 red, 2 yellow and 2 green flags on a pole: [Assume that all the 7 flags are used to transmit a signal].
✓
210
B
215
C
220
D
225
Answer
Correct option: A.
210
Here, $\text{n}=3+2+2=7$
$\text{P}_1=3,\text{P}_2=2$ and $\text{P}_3=2$
$\therefore$ Required number of different signals
$=\frac{\text{n}!}{\text{p}_1!\text{p}_1\text{p}_1}=\frac{7!}{3!2!1!}=\frac{7.6.5.4}{2.2}$
$=7.6.5=210$
On the eve of Diwali festival, a group of 12 friends greeted every other friend by sending greeting cards. Find the number of cards purchased by the group:
A
156
B
144
✓
132
D
72
Answer
Correct option: C.
132
There being 12 friends in the group, each friend must have purchased (12 - 1) i.e. 11 cards for sending greeting to rest of his 11 friends.
Thus total number of cards purchased by all the friends together is 12 x 11 i.e. 132.
The number of ways in which 8 students can be seated in a line is:
A
5040
B
50400
C
40230
✓
40320
Answer
Correct option: D.
40320
For the 1st position, there are 8 possible choices. For the 2nd position, there are 7 possible choices. For the 3rd position, there are 7 possible choices, etc.
And for the eighth position, there is only one possible choice.
Hence, this can be written as 8!
(i.e.) 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
= 40,320
Choose the correct answer. If ${}^n \mathrm{C}_{12}={ }^{\mathrm{n}} \mathrm{C}_8$ , then $n$ is equal to.
✓
20
B
12
C
6
D
30
Answer
Correct option: A.
20
20
Solution:
Give that $^\text{n}\text{C}_{12}=\ ^\text{n}\text{C}_8[\because\ ^\text{n}\text{C}_\text{r}=\ ^\text{n}\text{C}_\text{n-r}]$
$^\text{n}\text{C}_{12}=\ ^\text{n}\text{C}_\text{n-8}$
$\therefore\text{n}-8=12\Rightarrow\text{n}=12+8=20$
Each combination corresponds to many permutations:
✓
True
B
False
C
Either
D
Neither
Answer
Correct option: A.
True
In combination.
Each combination can be considered as a set of selection an order.
Each selection has a defined order.
They can be considered as a permutation.
Each cpmbination corresponds to many permutations.
Hence the above statement is true.
In a room there are 12 bulbs of the same wattage, each having a separate switch. The number of ways to light the room with different amounts of illumination is:
A
$12^2-1$
B
$2^{12}$
✓
$2^{12}-1$
D
None of these
Answer
Correct option: C.
$2^{12}-1$
$2^{12}-1$
Solutions:
Each of the bulb has its own switch, i.e each bulb will have two outcomes - it will either glow or not glow. Thus, each of the 12 bulbs will have 2 outcomes.
$\therefore$ Total number of ways to illuminate the room $=2^{12}$
Here, we have also considered the way in which all the bulbs are switched-off. However, this is not required as we need to find out only the number of ways of illuminating the room.
Hence, we subtract that one way from the total number of ways.
$=2^{12}-1$
A lady gives a dinner party for six guests. The number of ways in which they may be selected from among ten friends if two of the friends will not attend the party together is:
A
112
✓
140
C
164
D
None of these.
Answer
Correct option: B.
140
Suppose there are two friends, A and B, who do not attend the party together.
If both of them do not attend the party, then the number of ways of selecting 6 guests $={^\text{8}}\text{C}_{\text{6}}=28$
If one of them attends the party, then the number of ways of selecting 6 guests $=2.{^\text{8}}\text{C}_{\text{5}}=112$
Total number of ways = 112 + 28 = 140
A committee of 7 has to be formed from 9 boys and 4 girls.In how many ways can this be done when the committee consists of at least 3 girls.
✓
588
B
885
C
858
D
None of these
Answer
Correct option: A.
588
Given number of boys = 9
Number of girls = 4
Now, A committee of 7 has to be formed from 9 boys and 4 girls.
Now, the committee consists of atleast 3 girls:
$^4\text{C}_3\times\ ^9\text{C}_4+\ ^4\text{C}_4\times\ ^9\text{C}_3$
$= \bigg[\frac{4!}{(3! × 1!)}\times\frac{9!}{(4! × 5!)}\bigg] +\ ^9\text{C}_3$
$=[\frac{(4 × 3!)}{3!}\times\frac{(9\times8\times7\times6\times5!)}{(4!\times5!)]} +\frac{9!}{(3!\times6!)}$
$= \bigg[\frac{4\times(9\times8\times7\times6)}{4!}\bigg] + \frac{(9\times8\times7\times6!)}{(3!\times6!)}$
$= \bigg[\frac{4\times(9\times8\times7\times6)}{4\times3\times2\times1}\bigg] + \frac{(9\times8\times7)}{(3!)}$
$=(9\times8\times7) + \frac{(9\times8\times7)}{(3\times2\times1)}$
$=504+\big(\frac{504}{6}\big)$
$=504+84$
$=588$
Two persons entered a Railway compartment in which 7 seats were vacant.The number of ways in which they can be seated is:
A
30
✓
42
C
720
D
360
Answer
Correct option: B.
42
42
Solution:
$\rightarrow$ The $1^{\text {st }}$ person can take one of the 7 seats
$\rightarrow 2^{\text {nd }}$ person can take any one of the remaining 6 seats.
$\Rightarrow$ So, the total $=7 \times 6=42$
In a colony, there are 55 members. Every member posts a greeting card to all the members. How many greeting cards were posted by them?
A
990
✓
2970
C
1980
D
890
Answer
Correct option: B.
2970
First player can post greeting cards to the remaining 54 players in 54 ways. Second player can post greeting card to the 54 players.
Similarly, it happens with the rest of the players.The total numbers of greeting cards posted are.
54 + 54 + 54 …
54 (55times) = 54 × 55 = 2970.