Tag: combinatorics and mathematical induction

Questions Related to combinatorics and mathematical induction

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?

maths combinatorics and mathematical induction fundamental principle of addition fundamental principles of counting principles of counting

  1. 15

  2. 29

  3. 61

  4. 435

Show answer
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.

maths combinatorics and mathematical induction fundamental principle of addition fundamental principles of counting principles of counting

  1. 65

  2. 1914

  3. 43

  4. None of these

Show answer
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 

maths combinatorics and mathematical induction fundamental principle of addition fundamental principles of counting principles of counting

  1. $359$

  2. $360$

  3. $365$

  4. $380$

Show answer
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 

maths combinatorics and mathematical induction fundamental principle of addition fundamental principles of counting principles of counting

  1. $4$

  2. $5$

  3. $6$

  4. None of these

Show answer
Correct Option: A

The number of words that can be formed by using the letter of the word "MATHEMATICS", taken all at a time is

exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

  1. $11!$

  2. $\dfrac{11!}{2!+2!+2!}$

  3. $\dfrac{11!}{(2!)^{3}}$

  4. none of these

Show answer
Correct Option: C

Let $P _m$ stand for $^m P _m$, then,
$1 + P _1 + 2P _2 + 3 P _3 + ... + n.P _n$ is equal to 

exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

  1. $(n - 1)!$

  2. $n !$

  3. $(n + 1)! - 1$

  4. $(n + 1)!$

Show answer
Correct Option: A

The given relation is  $1.P(1,)+2.P(2,2)+3.P(3,3)+......+n.P(n,n)=P)(n+1,n+1)-3$.

exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

  1. True

  2. False

Show answer
Correct Option: B

If $^{ 56 }{ { P } _{ r+6 } }:^{ 54 }{ { P } _{ r+3 }}=30800$, then $r$ is

exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

  1. $39$

  2. $41$

  3. $28$

  4. $43$

Show answer
Correct Option: B
Explanation:
$^{ 56 }{ P } _{ r+6 }:^{ 54 }{ P } _{ r+3 }=30800$
$\cfrac { \cfrac { 56! }{ \left( 50-r \right) ! }  }{ \cfrac { 54! }{ \left( 51-r \right) ! }  } =30800$
$\cfrac { 56!\times \left( 51-r \right) ! }{ 54!\left( 50-r \right) ! } =30800$
$56\times 55\times \left( 51-r \right) =30800$
$\left( 51-r \right) =\cfrac { 30800 }{ 56\times 55 }$
$\left( 51-r \right) =10$
$r=41$

In how many ways unique can arrange the  letters  in the word "SUCCESSFUL" 

exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

  1. $\dfrac{10!}{2!2!3!}$

  2. $\dfrac{7!}{2!2!}. ^8P _3$

  3. $^8P _3$

  4. $\dfrac{7!}{2!2!}. \dfrac{^8P _3}{3!}$

Show answer
Correct Option: A
Explanation:
In the word 'SUCCESSFUL',
We have to arrange $10$ letters, out of which S occurs thrice, U and C occurs twice.
$\therefore$ Number of ways of arranging these letters $= \cfrac{10!}{\left( 3! \right) \left( 2! \right) \left( 2! \right)}$

Hence the correct answer is $\cfrac{10!}{\left( 3! \right) \left( 2! \right) \left( 2! \right)}$.

The given relation is  $1.P(1,1)+2.P(2,2)+3.P(3,3)++n.P(n,n)=P(n+1,n+1)-3$.

exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

  1. True

  2. False

Show answer
Correct Option: B