Tag: permutations and combinations

Questions Related to permutations and combinations

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

How many $4$-letter words, with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?

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

  1. $5040$

  2. $1000$

  3. $2500$

  4. $2060$

Show answer
Correct Option: A
Explanation:

There are $10$ letters in the word 'LOGARITHMS'.
So, the number of $4$-letter word$=$Number of arrangements of $10$ letters, taken $4$ at a time
$=$ $^{10}P _4=5040$.

If $^{10}P _r\,= 5040$, then find the value of $r$.

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

  1. $2$

  2. $4$

  3. $6$

  4. $8$

Show answer
Correct Option: B
Explanation:

Given, $^{10}P _r\,= 5040$
$\Rightarrow \cfrac{10!}{(10-r)!} =5040 $

$\Rightarrow (10-r)! =\cfrac{9!}{504}=\cfrac{9\times 8\times 7\times 6!}{504} = 6!$
$\therefore r=4$

If $ {^n}P _r $ $=$ 5040, then $(n, r)$ $= $

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

  1. (9,4)

  2. (10,4)

  3. (11,3)

  4. (11,4)

Show answer
Correct Option: B
Explanation:

$:^{n}P _{r}$ $=\dfrac{n!}{(n-r)!}$.
Now $5040$ $=10(9)(8)(7)$
$=\dfrac{10!}{(6)!}$
$=\dfrac{10!}{(10-4)!}$
$=:^{10}P _{4}$
Hence, $n=10$ and $r=4$

If the last four letters of the word 'CONCENTRATION' are written in reverse order followed by next two in the reverse order and next three in the reverse order and then followed by the first four in the reverse order counting from the end which letter would be eighth in the new arrangement? 

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

  1. N

  2. T

  3. E

  4. R

Show answer
Correct Option: D
Explanation:

The new letter sequence is 
NOITARTNECNOC
The eighth letter from the end is R