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

How many words, with meaning or without meaning, can be formed by using the letters of the word $'MISSISSIPPI'$

  1. $11!$
  2. $\dfrac{11!}{2!.4!}$
  3. $\dfrac{11!}{2!.4!.4!}$
  4. $ 2!.4!.4! $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Number of letters in the word $=  11$. Among them $ 2P, 4S, 4I$ are there.
Total number of words that can be formed $= \dfrac{11!}{2!.4!.4!} $

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

If $ ^nP _{100} = ^nP _{99} $, then $n$ is equal to

  1. $100$
  2. $101$
  3. $99$
  4. $86$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given  $ ^nP _{100} = ^nP _{99} $ 


Formula: $^np _r=\dfrac{n!}{(n-r)!}$

$\Rightarrow \displaystyle \dfrac{n!}{(n-100)!}=\frac{n!}{(n-99)!}$


$\Rightarrow (n-99)!=(n-100)!$

$\Rightarrow (n-99)(n-100)!=(n-100)!$

$\Rightarrow {n}-99=1$

$\Rightarrow {n}=100$

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

How many words can be formed by taking $4$ letters at a time of the letters of word $MATHEMATICS$ 

  1. $2234$
  2. $2542$
  3. $2346$
  4. $2454$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given words $MATHEMATICS$
for our simplicity we can write it as
$M\ A\ T\ H\ E\ I\ C\ S\ $
$M\ A\ T\ $
a) words of type $ABCD$
(All four letters are different)
(4 letters are chosen from $MATHEICS$ and are arranged)
$={ 8 } _{ { c } _{ 4 } }\times 4!=1680$

b) words of type $AABC$
(2 are alike, 2 are different)
(1 pair of letters is selected from M, A, T and 2 letters
are chosen from the remaining 7 letters and are arranged)
$={3} _{{c} _{1}} \times {7} _{{c} _{2}}\times \dfrac{4!}{2!}=756$

c) words of type $AABB$
(2 are alike of 1 kind, 2 are alike of another kind)
(2 pair of letter are chosen from M, A, T and arranged) 
$={3} _{c _2} \times \dfrac{4!}{2!2!}=18$
Total no. of word $=1680+756+18$
                             $=2456$
Multiple choice exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

There are $20$ persons among whom are two brothers. The number of ways in which we can arrange them around a circle so that there is exactly one person between the two brothers, is

  1. $18!$
  2. $17!\times 2!$
  3. $18!\times 2!$
  4. $20!$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

To have exactly one person between the two specific brothers in a circle of 20 people, we can treat the two brothers and the person between them as a single fixed block or use combinatorial placement. Specifically, choose the person between them in 18 ways, place the two brothers in 2! ways relative to that person, and arrange the remaining 17 entities around the circle in (18 - 1)! = 17! ways, leading to 18 * 2! * 17! = 18! * 2!.

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

The number of permutations of $n$ distinct objects taken $r$ together in which include $3$ particular things must occur together

  1. $^ { n-3 } C _ { r -3}(r-2)! \times 3 !$
  2. $^ { n } C _ { r - 3 } \times 3 !$
  3. $^ { n - 3 } p _ { r - 3 } \times 3 !$
  4. $P _ { 3 } \times ^ { n - 3 } P _ { r- 3 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Number of ways of selection of (r-3) objects out of (n-3) objects
$\\=^{n-3}C _{r-3}$
When we bundle 3 things together,
$\\no. of \> objects=(r-3)+1$
$\\=(r-2)$
which can be arranged in (r-2)! ways also 3 objects
in the bundle can be arranged among themselves in 3! ways
$\\\therefore\>Permutation=^{n-3}C _{r-3}(r-2)!\>3!$
Multiple choice exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

$2^n P _n$ is .equal to

  1. $(n + 1) (n + 2) ....... (2n)$
  2. $2^n[1.3.5 .....(2n - 1)]$
  3. $(2).(6).(10) .... (4n - 2)$
  4. $n!(2 ^nC _n )$
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation
$2^n{P _n}=\dfrac{2n!}{(2n-n)!}$
as ${^nP _r}=\dfrac{n!}{n-n!}$
             $=\dfrac{(2n)!}{n!}$
We can write it as,
$n^n[1.3.5.......(2n-1)]$
or
$2.6.10........(4n-2)$
also we $2^n{P _n}=(2n)(2n-2)........(n+1)$
and checking from option (d)
$n!(2{^nC _n})=\dfrac{n!(2n)!}{n!n!}$
                  $=2^n{P _n}$
$\therefore$ $(a),(b),(c),(d)$ all are correct.
Hence, all the answers are correct.