Tag: combinatorics and mathematical induction

Questions Related to combinatorics and mathematical induction

If $^{2n+1}P _{n-1}:^{2n-1}P _n=7:10$, then $^nP _3$ equals

  1. 60

  2. 24

  3. 120

  4. 6


Correct Option: A

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


Correct Option: A

If $^{2n + 1}P _{n -1} : ^{2n - 1}P _n = 3 : 5$, then n is equal to 

  1. 4

  2. 6

  3. 8

  4. 3


Correct Option: A

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. $20$


Correct Option: A

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. ${3}^{5}$


Correct Option: A

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


Correct Option: A

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

  1. $10$

  2. $4$

  3. $5$

  4. $9$


Correct Option: A
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 } $
$n-3=7$
$n=10$
$\therefore n=10$

Total number of $6-$digit numbers in which all the odd digits and only odd digits appears, is

  1. $\dfrac {5}{2}(6\ !)$

  2. $6!$

  3. $\dfrac {1}{2}(6\ !)$

  4. $\dfrac {3}{2}(6\ !)$


Correct Option: A

Total number of $6-$ digit numbers in which all the odd digits and only odd digits appear, is

  1. $\dfrac{5}{2}\left(6!\right)$

  2. $6!$

  3. $\dfrac{1}{2}\left(6!\right)$

  4. $none$


Correct Option: A

The number of many one functions from $A=}1,2,3}$ to $B={a,b,c,d}$ is 

  1. $64$

  2. $24$

  3. $40$

  4. $0$


Correct Option: A