Tag: principles of counting

Questions Related to principles of counting

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?

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  1. $14$

  2. $4$

  3. $40$

  4. $45$

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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

statistics information processing fundamental principle of addition fundamental principles of counting principles of counting

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

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

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

  4. none of these

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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? 

statistics information processing fundamental principle of addition fundamental principles of counting principles of counting

  1. 15

  2. 6

  3. 28

  4. 56

  5. 64

Show answer
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

statistics information processing fundamental principle of addition fundamental principles of counting principles of counting

  1. $7$

  2. $8$

  3. $9$

  4. $10$

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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

statistics information processing fundamental principle of addition fundamental principles of counting principles of counting

  1. 12

  2. 11

  3. 10

  4. none of these

Show answer
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

statistics information processing fundamental principle of addition fundamental principles of counting principles of counting

  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

Show answer
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.

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  1. $35$

  2. $12$

  3. $40$

  4. $30$

Show answer
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.

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  1. $1825$

  2. $3801$

  3. $4845$

  4. $300$

Show answer
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

statistics information processing fundamental principle of addition fundamental principles of counting principles of counting

  1. $12$

  2. $6$

  3. $9$

  4. $3$

Show answer
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

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  1. $66$

  2. $30$

  3. $24$

  4. $15$

Show answer
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 $15.$