Tag: exponent of a prime in n!

Questions Related to exponent of a prime in n!

If $^{15}{P _{r - 1}}:{\,^{15}}{P _{r - 2}} = 3:4$ then $r=$

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

  1. $\dfrac{71}{4}$

  2. $14$

  3. $6$

  4. $8$

Show answer
Correct Option: A
Explanation:

Consider the given value of permutation,$^{n}{{P} _{r-1}}{{:}^{15}}{{P} _{r-2}}=3:4$

  $ \,\,\,\,\,\,\dfrac{^{15}{{P} _{r-1}}}{^{15}{{P} _{r-2}}}=\dfrac{3}{4} $

 $ \,\,\,\dfrac{\dfrac{15!}{(15-r+1)!}}{\dfrac{15!}{\left( 15-r+2 \right)!}}=\dfrac{3}{4} $

 $  $

 $ \,\,\,\,\,\,\dfrac{\left( 15-r+2 \right)!}{(15-r+1)!}=\dfrac{3}{4} $

 $ \,\,\,\,\,\dfrac{\left( 17-r \right)!}{(16-r)!}=\dfrac{3}{4} $

 $ \,\,\,\dfrac{\left( -17+r \right)!}{(-16+r)!}=\dfrac{3}{4} $

 $ \dfrac{\left( r-17 \right)!}{(r-16)!}=\dfrac{3}{4} $

 $ \dfrac{\left( r-17 \right)\left( r-16 \right)!}{(r-16)!}=\dfrac{3}{4} $

 $ r-17=\dfrac{3}{4} $

 $ r=\dfrac{71}{4} $

This is the required answer.

Let ${T _n}$ be the number of all possible triangles formed by joining vertices of an $n$-sided regular polygon. If ${T _{n + 1}} - {T _n} = 10$. then the value of $n$ is 

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

  1. $7$

  2. $5$

  3. $10$

  4. $8$

Show answer
Correct Option: B
Explanation:

Given $T _n=\ ^nC _3$

$T _{n+1}=\ ^{n+1}C _3$
therefore,
$T _{n+1}-T _n=\ ^{n+1}C _3-\ ^nC _3=10$
$\Rightarrow\ ^nC _2+\ ^nC _3-\ ^nC _3=10$        $[\because\ ^nC _r+\ ^nC _{r-1}=\ ^{n+1}C _r]$
$\Rightarrow\ ^nC _2=10$
$\Rightarrow\ ^nC _2=\ ^5C _2$
$\Rightarrow n=5$

The number of all possible different arrangements of the word $"BANANA"$ is

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

  1. $6!$

  2. $6!\times2!\times3!$

  3. $\dfrac{6!}{2!3!}$

  4. none of these

Show answer
Correct Option: C

If $\displaystyle \overset{n-r}{\underset{k=1}{\sum }}\ ^{n-k}C _r=^{x}C _y$ then-

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

  1. $x=n+1\ ;\ y=r$

  2. $x=n\ ;\ y=r+1$

  3. $x=n\ ;\ y=r$

  4. $x=n+1\ ;\ y=r+1$

Show answer
Correct Option: C

If ${}^{15}{P _{r - 1}}\,:\,{}^{15}{P _{r - 2}} = 3:4$, then $r =$

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

  1. $10$

  2. $14$

  3. $20$

  4. $15$

Show answer
Correct Option: C
Explanation:
${}^{15}{P _{r - 1}}\,:\,{}^{15}{P _{r - 2}} = 3:4$

$=>\dfrac{\dfrac{15!}{(15-(r-1))!}}{\dfrac{15!}{(15-(r-2))!}}=\dfrac{3}{4}$


$=>\dfrac{17-r}{16-r}=\dfrac{3}{4}$

$=>68-4r=48-3r$

$=>r=20$

The number of arrangements of $A _{1},A _{2},..A _{10}$ in a line so that $A _{1}$ is always above then $A _{2}$, is 

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

  1. $2\times 10!$

  2. $\dfrac{1}{2}\times10!$

  3. $^{10}P _{2}$

  4. $^{10}C _{2}$

Show answer
Correct Option: A

The number of arrangements of ${ A } _{ 1 },{ A } _{ 2 },\dots ,{ A } _{ 10 }$ in a line so that  ${ A } _{ 1 }$ is always above than ${ A } _{ 2 }$. Is

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

  1. $2x10!$

  2. $\dfrac { 1 }{ 2 } \times 10!$

  3. $^{ 10 }{ P _{ 2 } }$

  4. $^{ 10 }{ C _{ 2 } }$

Show answer
Correct Option: A

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

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

  1. $n = 2$

  2. $n = 3$

  3. $n = 4$

  4. $n = 5$

Show answer
Correct Option: C
Explanation:

  Given $^{2n+1}P _{n-1}: ^{2n-1}P _n = 3 : 5$ which can be further simplified as
$ \displaystyle \frac {\frac { (2n+1)! }{ (n+2)! }  }{ \frac { (2n-1)! }{ (n-1)! }  }= \frac { 3 }{ 5 } $
$\Rightarrow \displaystyle \frac {(2n+1)(2n)}{(n+2)(n+1)(n)} = \displaystyle\frac{3}{5} $
$\Rightarrow $ $3{n}^{2}-11n-4=0$ .
On solving this quadratic equation we get roots as  $ 4 $ and $ -\displaystyle \frac{1}{3} $.
Since $n$ is an integer,$n=4$.

The number of ways od arranging 9 persons around a circle of there are two other persons between two particular persons is 

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

  1. $2\times (7!)$

  2. $3\times 7!$

  3. $9\times ^{ 8 }{ P } _{ 2 }$

  4. $4\times 7!$

Show answer
Correct Option: A

The number of 7 digit numbers which can be formed using the digits 1,2,3,2,3,3,4 is _.

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

  1. 420

  2. 840

  3. 2520

  4. 5040

Show answer
Correct Option: A