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

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. ${3}^{5}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A one-to-one function from a set of 3 elements to a set of 5 elements is equivalent to choosing 3 elements from 5 and arranging them, which is 5P3.

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

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
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a garland (necklace where clockwise and counterclockwise arrangements are distinct or identical depending on convention, but standard formula applies), the number of ways to arrange 8 objects with 4 specific objects always together is ((8 - 4 + 1 - 1)! * 4!) / 2 = (4! * 4!) / 2 = (24 * 24) / 2 = 288.

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

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

  1. $10$
  2. $4$
  3. $5$
  4. $9$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 } $
$n-3=7$
$n=10$
$\therefore n=10$
Multiple choice exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

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
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

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