Tag: combinatorics and mathematical induction

Quizzes (3) Questions (65)

Questions Related to combinatorics and mathematical induction

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

There are m apples and n oranges to be placed in a line such that the two extreme fruits being both oranges. Let P denotes the number of arrangements if the fruits of the same species are different and Q the corresponding figure when the fruits of the same species are alike, then the ratio P/Q has the value equal to :

  1. $^{ n }{ P } _{ { 2 }^{ . } }\quad ^{ m }{ P } _{ { m }^{ . } }(n-2)!$
  2. $^{ m }{ P } _{ { 2 }^{ . } }\quad ^{ n }{ P } _{ { n }^{ . } }(n-2)!$
  3. $^{ n }{ P } _{ { 2 }^{ . } }\quad ^{ n }{ P } _{ { n }^{ . } }(m-2)!$

  4. none
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a duplicate of 459936 and 459940.

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

Number of ways in which these $16$ players can be divided into equal groups, such that when the best player is selected from each group, ${P} _{6}$ is one among them, is $(k)\dfrac{12!}{{4!}^{3}}$. The value of $k$ is:

  1. $36$
  2. $24$
  3. $18$
  4. $20$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This involves grouping 16 players into 4 groups of 4. The number of ways to divide them is 16! / ((4!)^4 * 4!). The condition regarding the best player involves combinations.

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

The number of one one functions that can be defined from \$A={a,b,c}\$ into \$B=1,2,3,4,5}\$ is

  1. $^{5}{P} _{3}$
  2. $^{5}{C} _{3}$
  3. ${5}^{3}$
  4. ${3}^{5}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A one-to-one function from a set of 3 elements to a set of 5 elements is equivalent to choosing 3 elements from 5 and arranging them, which is 5P3.

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

The number of ways in which $8$ different flowers can be strung to form a garland so that $4$ particulars flowers are never separated, is?

  1. $4!\cdot 4!$
  2. $\dfrac{8!}{4!}$
  3. $288$

  4. None
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a garland (necklace where clockwise and counterclockwise arrangements are distinct or identical depending on convention, but standard formula applies), the number of ways to arrange 8 objects with 4 specific objects always together is ((8 - 4 + 1 - 1)! * 4!) / 2 = (4! * 4!) / 2 = (24 * 24) / 2 = 288.

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

If $\displaystyle ^{n}P _{3}:^{n}P _{6}=1:210$, find $n$.

  1. $10$
  2. $4$
  3. $5$
  4. $9$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$^{ n }{ P } _{ 3 }:^{ n }{ P } _{ 6 }=1:210$
$\cfrac { \cfrac { n! }{ \left( n-3 \right) ! }  }{ \cfrac { n! }{ \left( n-6 \right) ! }  } =\cfrac { 1 }{ 210 } $
$\cfrac { (n-6)! }{ \left( n-3 \right) ! }=\cfrac { 1 }{ 210 } $
$\cfrac { 1 }{ \left( n-3 \right) \left( n-4 \right) \left( n-5 \right)  } =\cfrac { 1 }{ 7\times 6\times 5 } $
$n-3=7$
$n=10$
$\therefore n=10$