0

Fundamental principles of counting - class-XI

Attempted 0/24 Correct 0 Score 0

The greatest possible number of points of intersection of 8 straight lines and $4$ circles is $104$.

  1. True

  2. False


Correct Option: A
Explanation:

According to question,

There are 8 lines, for two lines meet in a point,

=>  $^8C _2\times 1= \dfrac{8.7}{1.2}=28$

Line and circle meet in two points,

=>$(^8C _1\times ^4C _1) \times 2 =64$

Two circles meet in two points,

=>  $(^4C _2)\times2 =\dfrac{4.3}{1.2}.2= 104$

Solve:$\dfrac{2}{2}+\dfrac{3}{3}+\dfrac{4}{4}+$...... + upto $1000$ terms= ?

  1. $1000$

  2. twice of $500$

  3. four times of $250$

  4. All of these


Correct Option: D
Explanation:

All the terms simplify to $1$ which is being added $1000$ times.
Therefore, the final answer is $1000$ and it can be seen that all the options are correct.

There are 6 equally spaced points A, B, C, D, E and F marked on a circle with radius R. How many convex pentagons of distinctly different areas can be drawn using these points advertises?

  1. $^6P _5$

  2. $1$

  3. $55$

  4. $42$


Correct Option: B

From 0 to 9 , four digited numbers can be formed such that
the digits  are in ascending order is

  1. ${}^{10}{P _4}$

  2. ${}^{10}{C _4}$

  3. ${}^{10}{P _4} - {}^9{P _3}$

  4. ${}^{10}{C _4} - {}^9{C _3}$


Correct Option: B
Explanation:

Selection of$ 4$ digits out of$ 10$ (including 0)=$ ^{10}C _4$


A point $(a, b)$ is called a good point if both $a$ and $b$ are integers. Number of good points on the curve $xy$ $=$ $225$ are

  1. 20

  2. 18

  3. 16

  4. 14


Correct Option: B
Explanation:

The order pair $(x, y)$ satisfying $xy=225$ are $(1, 225), (3, 75) (5, 45), (9, 25), (15, 15)$. Order can be changed in the first four pairs and both $x$ and $y$ can be negative also, so the no. of pairs $=2(2\times 4+1)=18$

A graph may be defined as a set of points connected by lines called edges. Every edge connects a pair of points. Thus, a triangle is a graph with 3 edges and 3 points. The degree of a point is the number of edges connected to it. For example, a triangle is agraph with three points of degree 2 each. Consider a graph with 12 points. It is possible to reach any point from any other point through a sequence of edges. The number of edges "e" in the graph must satisfy the condition

  1. $11 \leq e \leq 66$

  2. $10 \leq e \leq 66$

  3. $11 \leq e \leq 65$

  4. $0 \leq e \leq 11$


Correct Option: A
Explanation:

(A) Since every edge connects a pair of points, the given 12 points have to be joined using lines. We may have minimum number of edges if all the 12 points are collinear.
No. of edges in this particular case 
$=12-1=11$
Maximum number of edges are possible when all the 12 points are non-collinear. In this particular case number of different straight lines that can be formed using 12 points which is equal to $^12C _{2}$
$=\frac{12\times 11}{2}=66$
Therefore, following inequality holds for "e"
$11 \leq e  \leq 66$

Total number of ways of selecting two numbers from the set ${1,2,3,...90}$ so that their sum is divisible by $3$ is

  1. $885$

  2. $1335$

  3. $1770$

  4. $3670$


Correct Option: B

If the letter of word $MOTHER$ are written in all possible orders and these words are written out as in a dictionary. Find the rank of the word $MOTHER$

  1. $307$

  2. $308$

  3. $309$

  4. $120$


Correct Option: A

There are $6$ boxes numbered $1, 2 ....... 6$. Each box is to be filled up either with a red or a green ball in such a way that at least $1$ box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is:

  1. $5$

  2. $21$

  3. $33$

  4. $60$


Correct Option: B
Explanation:

(B) The number of ways in which 1 green ball can be put $=6$ . The number of ways in which two green balls can be put such that the boxes are consecutive 
$=5$ $(i.e., (1, 2),(2, 3),(3, 4),(4, 5),(5, 6))$

Similarly, the number of ways in which three green balls can be put 
$=4( i.e. (1, 2, 3),(2, 3, 4),(3, 4, 5),(4, 5, 6))$
$\cdots \cdots \cdots \cdots \cdots $ and so on.
$\therefore $ Total number of ways of doing this
$=6+5+4+3+2+1=21$

The sides of a quadrilateral are all positive integers and three of them are $5, 10, 20.$ How many possible value are there for the fourth side?

  1. $29$

  2. $31$

  3. $32$

  4. $34$


Correct Option: D

In Hyderabad there are 5 routes to Begumpet from Kukatpally and 9 routes to Dilsukhnagar from Begumpet In how many ways can a person travel from Kukatpally to Dilsukhnagar via Begumpet?

  1. $14$

  2. $4$

  3. $40$

  4. $45$


Correct Option: D
Explanation:
This is an implication of AND principal so multiplication shall be done.
So, number of ways $=5\times9$
                                  $=45$
Hence, the answer is $45.$

Rajdhani Express going from Bombay to Delhi stops at five intermediate stations, 10 passengers enter the train during the journey with 10 different ticket of two classes. The number of different sets of tickets they may have is

  1. $^{15}C _{10}$

  2. $^{20}C _{10}$

  3. $^{30}C _{10}$

  4. none of these


Correct Option: C
Explanation:

For a particular class, the total number of different tickets from first intermediate station is $5.$ 
Similarly, number of different tickets from second intermediate station is $4.$ 
So the total number of different tickets is $5+4+3+2+1=15$.
And same number of tickets for another class is equal to total number of different tickets, 

which is equal to $30$ and number of selection is $^{30}C _{10}$.

There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played? 

  1. 15

  2. 6

  3. 28

  4. 56

  5. 64


Correct Option: C
Explanation:

Matches played between 8 teams =$ 7+6+5+4+3+2+1 = 21$

In these 8 matches, 4 winners will emerge.
So, total number of matches between these 4 winners = $ 3+2+1= 6$
Two winners will emerge from these 4 winners who will play the final match.
so, total matches played in the tournament = $21 + 6+1 = 28$ (option C)

The number of $n$ digit numbers which consists of the digits $1$ & $2$ only if each digits is to be used atleast once, is equal to $510$  then $n$ is equal to

  1. $7$

  2. $8$

  3. $9$

  4. $10$


Correct Option: B

In a test there were n questions. In the test $\displaystyle 2^{n-i}$ students gave wrong answers to i questions where $\displaystyle i=1,2,3...,n$. If the total number of wrong answers given is 2047 then n is

  1. 12

  2. 11

  3. 10

  4. none of these


Correct Option: B
Explanation:

Total number of wrong answers $= 2^{n-1}+2^{n-2}+...+2^{2}+2+1$
$=2^{n-1}+2^{n-2}+...+2^{4}+2^{3}+2^{2}+2+1$
$\Rightarrow 2^{n}-1=2043$  [Using formula for sum of G.P]
$ \Rightarrow 2^{n}=2048$
$ \Rightarrow 2^{n}=2^{11}$
$\Rightarrow n = 11$
Hence, option 'B' is correct.

The number of ways in which three numbers in A.P. can be seleced from the set of first n natural number if n is odd is

  1. $ \displaystyle \frac{n\left ( n-2 \right )}{4} $

  2. $ \displaystyle \frac{n\left ( n-1 \right )^2}{4} $

  3. $ \displaystyle \frac{\left ( n-1 \right )^2}{4} $

  4. None of these


Correct Option: C
Explanation:

In order to solve this question, we must observe the number of ways in which we can select the first term of the required A.P. for different values of the common difference($r$) starting from $r=1$ given that there are only $3$ terms before $n$.
For $r=1$, the number of ways in which we can select the first term of the A.P. $=n-2$
For $r=2$, the number of ways to select the first term $=n-4$
For $r=3$, the number of ways to select the first term $=n-6$
Now we see a pattern emerging, we also realize from this that $r<=\dfrac{n-1}{2}$ for an A.P. with 3 terms to exist in the given interval.
$\therefore$ The final answer $=n-2+n-4+n-6+...+5+3+1$
Now we use the formula to find the sum of an A.P. which is $S _n=\dfrac{n}{2}[a _1+a _n]$ 
$\therefore$ Answer $=\dfrac{n-1}{4}[1+n-2]=\dfrac{(n-1)^2}{4}$

A college offers $7$ courses in the morning and $5$ courses in the evening. Find the number of ways a student can select exactly one course either in the morning or in the evening.

  1. $35$

  2. $12$

  3. $40$

  4. $30$


Correct Option: B
Explanation:

$7$ Courses in morning 

$5$ courses in evening 
Total number of courses $=12$
Selecting any one of the course 
Number of ways $^{ 12 }{ C } _{ 1 }$
$=\cfrac { 12! }{ 1!\times 1! } $ 
$=12$ Ways 
Therefore total ways $=12$

If $^nC _3=^nC _{13}$, then $^{20}C _n$ is.

  1. $1825$

  2. $3801$

  3. $4845$

  4. $300$


Correct Option: C
Explanation:

We know that $ { { n } _{ C } } _{ r }=  { { n } _{ C } } _{n- r }$

Given, $ { { n } _{ C } } _{3}= { { n } _{ C } } _{13} $
$ => r = 3 $ and $ n -r = 13 $
$ => n - 3 = 13 $
$ => n = 16 $

Also, $ { { n } _{ C } } _{ r }=\dfrac { n! }{ r!(n-r)! }  $
So, $ { { 20 } _{ C } } _{ 16 }=\dfrac { 20! }{ 16!(20-16)! } = \dfrac { 20! }{ 16! \times 4! }  = \dfrac { 20 \times 19 \times 18 \times 17 \times 16! }{ 16! \times 4 \times 3 \times 2 } = 4845   $

Let $A$ be the set of all $3 \times  3$ symmetric matrices all of whose entries are either $0$ or $1$. Five of these entries are $1$ and four of them are $0$.
The number of matrices in $A$ is

  1. $12$

  2. $6$

  3. $9$

  4. $3$


Correct Option: A
Explanation:

If two zeros are the entries in the diagonal, then
$^{3}\mathrm{C} _{2}\times^{3}\mathrm{C} _{1}$
If all the entries in the principle diagonal is 1, then
$^{3}\mathrm{C} _{1}$
$\Rightarrow$  Total matrix $= 12.$

The number of rectangles that can be obtained by joining four of the twelve vertices of a $12$ sided regular polygon is

  1. $66$

  2. $30$

  3. $24$

  4. $15$


Correct Option: D
Explanation:
The first vertex can be choosed in $12$ ways and diagonally opposite to it is $1$ vertex. Now for $3rd$ vertex we have $10$ choices and for $4th$ $1.$
However, each rectangle is counted $8$ times.
$\therefore$ No. of ways $=\dfrac{12\times1\times10\times1}{8}$  $=15$ ways.
Hence, the answer is $15.$

The 30 members of a club decided to playa badminton singles tournament. Every time a member loses a game he is out of tournament. There is no ties. What is the minimum number of matches that must be played to determine the winner?

  1. 15

  2. 29

  3. 61

  4. 435


Correct Option: B
Explanation:

Clearly, every member except one (i.e. the winner) must lose one game to decide the winner. Thus, minimum number of matches to be played = 30 - 1 = 29.

Each  section of soccer stadium has 44 rows with 22 sets in first row, 23 in the second row, 24 in the third row, and so on. How many seats are there in row 44.

  1. 65

  2. 1914

  3. 43

  4. None of these


Correct Option: A

The number of all three digit even number such that if $3$ is one of the digits, then next digit is $5$, is 

  1. $359$

  2. $360$

  3. $365$

  4. $380$


Correct Option: C
Explanation:
We need even numbers so the last digit must be $0, 2, 4, 6, 8$. So $5$ possibilities
Now the ten's place can be any number between $0$ to $9$ except $3$ because in that case, the $3$ is to be followed by $5$ in the units place then it won't be an even no., so there are $9$ possibilities.
Now in hundred's place can be anything between $1$ to $9$ except $3$ because if $3$ is present then the next digit must be $5$ so there are $8$ possibilities.
Now if there is $3$ in hundred's place then $5$ will be in ten's place and the numbers must be $350, 352, 354, 356, 358$ which will be treated seperately. So $5$ possibilities
Therefore, 
Total no. of even numbers $= 8 \times 9 \times 5 + 5 = 365$

In an election, the number of candidates is one more than the number of members to be elected. A voter can cast any number of the vote but not more than the candidates to be elected. If a voter can cast his vote in $30$ ways, then the number of the candidates is 

  1. $4$

  2. $5$

  3. $6$

  4. None of these


Correct Option: A
- Hide questions