Tag: factorial notation

Questions Related to factorial notation

If $ {^1}{^2}P _r  =$ 1320, then r $=$

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

  1. 2

  2. 3

  3. 4

  4. 5

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Correct Option: B

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

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

  1. $11!$

  2. $\dfrac{11!}{2!.4!}$

  3. $\dfrac{11!}{2!.4!.4!}$

  4. $ 2!.4!.4! $

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Correct Option: C
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!} $

If $n$ books can be arranged in a linear shelf in $5040$ different ways 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. $8$

  3. $6$

  4. $9$

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Correct Option: A
Explanation:

If $n$ books are to be arranged, the no. of ways is equal to $=n!$.
$\Rightarrow$ $n!$ $=5040$
$\Rightarrow$ $n$ $=7$

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

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

  1. $100$

  2. $101$

  3. $99$

  4. $86$

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Correct Option: A
Explanation:

Given  $ ^nP _{100} = ^nP _{99} $ 


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

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


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

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

$\Rightarrow {n}-99=1$

$\Rightarrow {n}=100$

How many numbers can be formed by using all the given digits $1,2,8,9,3,5$ when repetition is not allowed

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

  1. $180$

  2. $120$

  3. $720$

  4. $1024$

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Correct Option: C
Explanation:

Number of ways of permuting $6$ different digits with no repetition $= 6! = 720$

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

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

  1. $2234$

  2. $2542$

  3. $2346$

  4. $2454$

Show answer
Correct Option: D
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 $=1680+756+18$
                             $=2456$

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

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

  1. $18!$

  2. $17!\times 2!$

  3. $18!\times 2!$

  4. $20!$

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Correct Option: A

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

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

  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. $P _ { 3 } \times ^ { n - 3 } P _ { r- 3 }$

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Correct Option: A
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
$\\\therefore\>Permutation=^{n-3}C _{r-3}(r-2)!\>3!$

$2^n P _n$ is .equal to

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

  1. $(n + 1) (n + 2) ....... (2n)$

  2. $2^n[1.3.5 .....(2n - 1)]$

  3. $(2).(6).(10) .... (4n - 2)$

  4. $n!(2 ^nC _n )$

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Correct Option: A,B,C,D
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.

12 normal dice are thrown once. The number of ways in which each of the values 2,3,4,5 and 6 occurs exactly twice is : [1,1, 2,2, 3,3, 4,4, 5,5, 6,6 can come in any order]

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

  1. $\frac{(12)!}{6}$

  2. $\frac{(12)!}{2^{6}.6!}$

  3. $\frac{(12)!}{2^{6}}$

  4. none

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Correct Option: A