Tag: permutations and combinations

Questions Related to permutations and combinations

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

If $\displaystyle ^{n+5}P _{n+1} = \frac{11\left ( n-1 \right )}{2}.^{n+3}P _n$ then the value of n is

  1. 7

  2. 8

  3. 6

  4. 5


Correct Option: A,C
Explanation:

Applying the given condition, we get
$\dfrac{(n+5)!}{4!}=\dfrac{11(n-1)}{2}\dfrac{(n+3)!}{3!}$
$\dfrac{(n+4)(n+5)}{4}=\dfrac{11(n-1)}{2}$
$(n+4)(n+5)=22(n-1)$
$n^{2}+9n+20=22n-22$
$n^{2}-13n+42=0$
$(n-7)(n-6)=0$
$n=7$ $n=6$

Find the value of $n$ when:

  1. $^{n-1}P _{3}:^{n+1}P _{3}=5:12$

  2. $^{n}P _{6}=10.^{n}P _{5}$

  3. $^{56}P _{n+6}:^{54}P _{n+3}=30800$

  4. $^{6+n}P _{2}:^{6+n}P _{2}=56:12$


Correct Option: A

If P(n, n) denotes the number of permutations of n different things taken all at a time then P(n, n) is also identical to

  1. n.P(n 1, n 1)

  2. P(n, n 1)

  3. n! 

  4. (n r) . P(n, r)


Correct Option: C
Explanation:
$P(n,n)=\dfrac{n!}{0!}=n!$
No one correct, answer is $n.$

$\displaystyle ^{n}P {n}=$___.

  1. $n!$

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

  3. $1$

  4. $(n+1)6$


Correct Option: A
Explanation:

$^nP _r = \dfrac { n! }{ (n-r)! } $

So, $ ^nP _n = \dfrac { n! }{ (n-n)! } =\dfrac { n! }{ 0! }  = n ! $ (Since $ 0 ! = 1 )$