Tag: exponent of a prime in n!

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

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

  1. $\dfrac{71}{4}$
  2. $14$
  3. $6$
  4. $8$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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 

  1. $7$
  2. $5$
  3. $10$
  4. $8$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given $T _n=\ ^nC _3$

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

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

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

  1. $x=n+1\ ;\ y=r$
  2. $x=n\ ;\ y=r+1$
  3. $x=n\ ;\ y=r$
  4. $x=n+1\ ;\ y=r+1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the hockey stick identity (or repeated application of Pascal's identity C(n,r) + C(n,r+1) = C(n+1,r+1)), the sum of combinations simplifies to C(n, r+1). Therefore, x = n and y = r + 1.

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

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 

  1. $2\times 10!$
  2. $\dfrac{1}{2}\times10!$
  3. $^{10}P _{2}$
  4. $^{10}C _{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By symmetry, for any linear arrangement of 10 distinct elements, the condition that A1 appears before (or above) A2 is equally likely as A2 appearing before A1. Thus, exactly half of the total 10! arrangements satisfy the given condition.

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

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

  1. $n = 2$
  2. $n = 3$
  3. $n = 4$
  4. $n = 5$
Reveal answer Fill a bubble to check yourself
C Correct answer
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,$n=4$.