Tag: combinatorics and mathematical induction

Quizzes (3) Questions (65)

Questions Related to combinatorics and mathematical induction

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is an arithmetic progression where a = 22 and d = 1. The number of seats in row n is a + (n - 1)d. For row 44: 22 + (44 - 1) * 1 = 22 + 43 = 65.

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

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. $380$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $= 8 \times 9 \times 5 + 5 = 365$
Multiple choice maths combinatorics and mathematical induction fundamental principle of addition fundamental principles of counting principles of counting

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
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

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

  1. $11!$
  2. $\dfrac{11!}{2!+2!+2!}$
  3. $\dfrac{11!}{(2!)^{3}}$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The word MATHEMATICS has 11 letters, with M appearing twice, A twice, and T twice. The number of distinct permutations is 11! divided by the factorial of the counts of repeated letters, which is 11! / (2! * 2! * 2!).

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

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

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The given relation has garbled notation with P)(n+1,n+1)-3 which appears to be a typo. For the standard identity sum of k×P(k,k) from k=1 to n, the right side should be P(n+1,n+1)-1, not -3. Testing with n=3: LHS = 1×1 + 2×2 + 3×3 = 14, but RHS with -3 would give P(4,4)-3 = 24-3 = 21 ≠ 14. The statement is FALSE.

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

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

  1. $39$
  2. $41$
  3. $28$
  4. $43$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$^{ 56 }{ P } _{ r+6 }:^{ 54 }{ P } _{ r+3 }=30800$
$\cfrac { \cfrac { 56! }{ \left( 50-r \right) ! }  }{ \cfrac { 54! }{ \left( 51-r \right) ! }  } =30800$
$\cfrac { 56!\times \left( 51-r \right) ! }{ 54!\left( 50-r \right) ! } =30800$
$56\times 55\times \left( 51-r \right) =30800$
$\left( 51-r \right) =\cfrac { 30800 }{ 56\times 55 }$
$\left( 51-r \right) =10$
$r=41$
Multiple choice exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

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

  1. $\dfrac{10!}{2!2!3!}$
  2. $\dfrac{7!}{2!2!}. ^8P _3$
  3. $^8P _3$
  4. $\dfrac{7!}{2!2!}. \dfrac{^8P _3}{3!}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $\cfrac{10!}{\left( 3! \right) \left( 2! \right) \left( 2! \right)}$.