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Exponent of a prime in n! - class-XI

Description: exponent of a prime in n!
Number of Questions: 61
Created by:
Tags: maths permutations and combinations combinatorics and mathematical induction
Attempted 0/61 Correct 0 Score 0

The number of words that can be formed by using the letter of the word "MATHEMATICS", taken all at a time is

  1. $11!$

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

  3. $\dfrac{11!}{(2!)^{3}}$

  4. none of these


Correct Option: C

Let $P _m$ stand for $^m P _m$, then,
$1 + P _1 + 2P _2 + 3 P _3 + ... + n.P _n$ is equal to 

  1. $(n - 1)!$

  2. $n !$

  3. $(n + 1)! - 1$

  4. $(n + 1)!$


Correct Option: A

The given relation is  $1.P(1,)+2.P(2,2)+3.P(3,3)+......+n.P(n,n)=P)(n+1,n+1)-3$.

  1. True

  2. False


Correct Option: B

If $^{ 56 }{ { P } _{ r+6 } }:^{ 54 }{ { P } _{ r+3 }}=30800$, then $r$ is

  1. $39$

  2. $41$

  3. $28$

  4. $43$


Correct Option: B
Explanation:
$^{ 56 }{ P } _{ r+6 }:^{ 54 }{ P } _{ r+3 }=30800$
$\cfrac { \cfrac { 56! }{ \left( 50-r \right) ! }  }{ \cfrac { 54! }{ \left( 51-r \right) ! }  } =30800$
$\cfrac { 56!\times \left( 51-r \right) ! }{ 54!\left( 50-r \right) ! } =30800$
$56\times 55\times \left( 51-r \right) =30800$
$\left( 51-r \right) =\cfrac { 30800 }{ 56\times 55 }$
$\left( 51-r \right) =10$
$r=41$

In how many ways unique can arrange the  letters  in the word "SUCCESSFUL" 

  1. $\dfrac{10!}{2!2!3!}$

  2. $\dfrac{7!}{2!2!}. ^8P _3$

  3. $^8P _3$

  4. $\dfrac{7!}{2!2!}. \dfrac{^8P _3}{3!}$


Correct Option: A
Explanation:
In the word 'SUCCESSFUL',
We have to arrange $10$ letters, out of which S occurs thrice, U and C occurs twice.
$\therefore$ Number of ways of arranging these letters $= \cfrac{10!}{\left( 3! \right) \left( 2! \right) \left( 2! \right)}$

Hence the correct answer is $\cfrac{10!}{\left( 3! \right) \left( 2! \right) \left( 2! \right)}$.

The given relation is  $1.P(1,1)+2.P(2,2)+3.P(3,3)++n.P(n,n)=P(n+1,n+1)-3$.

  1. True

  2. False


Correct Option: B

How many $4$-letter words, with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?

  1. $5040$

  2. $1000$

  3. $2500$

  4. $2060$


Correct Option: A
Explanation:

There are $10$ letters in the word 'LOGARITHMS'.
So, the number of $4$-letter word$=$Number of arrangements of $10$ letters, taken $4$ at a time
$=$ $^{10}P _4=5040$.

If $^{10}P _r\,= 5040$, then find the value of $r$.

  1. $2$

  2. $4$

  3. $6$

  4. $8$


Correct Option: B
Explanation:

Given, $^{10}P _r\,= 5040$
$\Rightarrow \cfrac{10!}{(10-r)!} =5040 $

$\Rightarrow (10-r)! =\cfrac{9!}{504}=\cfrac{9\times 8\times 7\times 6!}{504} = 6!$
$\therefore r=4$

If $ {^n}P _r $ $=$ 5040, then $(n, r)$ $= $

  1. (9,4)

  2. (10,4)

  3. (11,3)

  4. (11,4)


Correct Option: B
Explanation:

$:^{n}P _{r}$ $=\dfrac{n!}{(n-r)!}$.
Now $5040$ $=10(9)(8)(7)$
$=\dfrac{10!}{(6)!}$
$=\dfrac{10!}{(10-4)!}$
$=:^{10}P _{4}$
Hence, $n=10$ and $r=4$

If the last four letters of the word 'CONCENTRATION' are written in reverse order followed by next two in the reverse order and next three in the reverse order and then followed by the first four in the reverse order counting from the end which letter would be eighth in the new arrangement? 

  1. N

  2. T

  3. E

  4. R


Correct Option: D
Explanation:

The new letter sequence is 
NOITARTNECNOC
The eighth letter from the end is R

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! $


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

  1. $7$

  2. $8$

  3. $6$

  4. $9$


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

  1. $100$

  2. $101$

  3. $99$

  4. $86$


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

  1. $180$

  2. $120$

  3. $720$

  4. $1024$


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 $MATHEMATICS$ 

  1. $2234$

  2. $2542$

  3. $2346$

  4. $2454$


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

  1. $18!$

  2. $17!\times 2!$

  3. $18!\times 2!$

  4. $20!$


Correct Option: A

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


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

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

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

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

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


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]

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

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

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

  4. none


Correct Option: A

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

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

  2. $14$

  3. $6$

  4. $8$


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 

  1. $7$

  2. $5$

  3. $10$

  4. $8$


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

  1. $6!$

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

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

  4. none of these


Correct Option: C

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


Correct Option: C

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

  1. $10$

  2. $14$

  3. $20$

  4. $15$


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 

  1. $2\times 10!$

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

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

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


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

  1. $2x10!$

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

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

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


Correct Option: A

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

  1. $n = 2$

  2. $n = 3$

  3. $n = 4$

  4. $n = 5$


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 

  1. $2\times (7!)$

  2. $3\times 7!$

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

  4. $4\times 7!$


Correct Option: A

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

  1. 420

  2. 840

  3. 2520

  4. 5040


Correct Option: A

There are m apples and n oranges to be placed in a line such that the two extreme fruits being both oranges. Let P denotes the number of arrangements if the fruits of the same species are different and Q the corresponding figure when the fruits of the same species are alike, then the ratio P/Q has the value equal to :

  1. $^{ n }{ P } _{ { 2 }^{ - } }\quad ^{ m }{ P } _{ { m }^{ - } }\quad (n-2)!$

  2. $^{ m }{ P } _{ { 2 }^{ - } }\quad ^{ n }{ P } _{ { n }^{ - } }\quad (n-2)!$

  3. $^{ n }{ P } _{ { 2 }^{ - } }\quad ^{ n }{ P } _{ { n }^{ - } }\quad (m-2)!$

  4. none


Correct Option: A

Exponent of $4$ in $80\ !$ is

  1. $26$

  2. $77$

  3. $39$

  4. $38$


Correct Option: A
Explanation:

Exponent of $4$ is $80!$ is

$[\cfrac{80}{4}]+[\cfrac{80}{4^2}]+[\cfrac{80}{4^3}]=20+5+1=26$

If $^{n}P _{5}=9 \times ^{n-1}P _{4}$, then the value of $n$ is 

  1. $6$

  2. $8$

  3. $5$

  4. $9$


Correct Option: A

In the word $ENGINEERIGNG if all $Es$ are not together and $Ns$ come together then number of permutations is

  1. $\dfrac{9!}{2!2!}-\dfrac{7!}{2!2!}$

  2. $\dfrac{9!}{3!2!}-\dfrac{7!}{2!2!}$

  3. $\dfrac{9!}{3!2!2!}-\dfrac{7!}{2!2!2!}$

  4. $\dfrac{9!}{3!2!2!}-\dfrac{7!}{2!2!}$


Correct Option: A

There are m apples and n oranges to be placed in a line such that the two extreme fruits being both oranges. Let P denotes the number of arrangements if the fruits of the same species are different and Q the corresponding figure when the fruits of the same species are alike, then the ratio P/Q has the value equal to :

  1. $^{ n }{ P } _{ 2^{ . } }\quad ^{ m }{ P } _{ { m }^{ . } }(n-2)!$

  2. $^{ m }{ P } _{ 2^{ . } }\quad ^{ n }{ P } _{ { n }^{ . } }(n-2)!$

  3. $^{ m }{ P } _{ 2^{ . } }\quad ^{ n }{ P } _{ { n }^{ . } }(n-2)!$

  4. none


Correct Option: A

If $3.^{n _{1}-n _{2}}P _{2}=^{n _{1}+n _{2}}P _{2}=90$, then the ordered $(n _{1},n _{2})$ is:

  1. $(8,2)$

  2. $(7,3)$

  3. $(16,8)$

  4. $(9,2)$


Correct Option: A

If $^{2n+1}P _{n-1}:^{2n-1}P _n=7:10$, then $^nP _3$ equals

  1. 60

  2. 24

  3. 120

  4. 6


Correct Option: A

There are m apples and n oranges to be placed in a line such that the two extreme fruits being both oranges. Let P denotes the number of arrangements if the fruits of the same species are different and Q the corresponding figure when the fruits of the same species are alike, then the ratio P/Q has the value equal to :

  1. $^{ n }{ P } _{ { 2 }^{ . } }\quad ^{ m }{ P } _{ { m }^{ . } }(n-2)!$

  2. $^{ m }{ P } _{ { 2 }^{ . } }\quad ^{ n }{ P } _{ { n }^{ . } }(n-2)!$

  3. $^{ n }{ P } _{ { 2 }^{ . } }\quad ^{ n }{ P } _{ { n }^{ . } }(m-2)!$

  4. none


Correct Option: A

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

  1. 4

  2. 6

  3. 8

  4. 3


Correct Option: A

Number of ways in which these $16$ players can be divided into equal groups, such that when the best player is selected from each group, ${P} _{6}$ is one among them, is $(k)\dfrac{12!}{{4!}^{3}}$. The value of $k$ is:

  1. $36$

  2. $24$

  3. $18$

  4. $20$


Correct Option: A

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. ${3}^{5}$


Correct Option: A

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


Correct Option: A

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

  1. $10$

  2. $4$

  3. $5$

  4. $9$


Correct Option: A
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 } $
$n-3=7$
$n=10$
$\therefore n=10$

Total number of $6-$digit numbers in which all the odd digits and only odd digits appears, is

  1. $\dfrac {5}{2}(6\ !)$

  2. $6!$

  3. $\dfrac {1}{2}(6\ !)$

  4. $\dfrac {3}{2}(6\ !)$


Correct Option: A

Total number of $6-$ digit numbers in which all the odd digits and only odd digits appear, is

  1. $\dfrac{5}{2}\left(6!\right)$

  2. $6!$

  3. $\dfrac{1}{2}\left(6!\right)$

  4. $none$


Correct Option: A

The number of many one functions from $A=}1,2,3}$ to $B={a,b,c,d}$ is 

  1. $64$

  2. $24$

  3. $40$

  4. $0$


Correct Option: A

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


Correct Option: A,C
Explanation:

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

Find the value of $n$ when:

  1. $^{n-1}P _{3}:^{n+1}P _{3}=5:12$

  2. $^{n}P _{6}=10.^{n}P _{5}$

  3. $^{56}P _{n+6}:^{54}P _{n+3}=30800$

  4. $^{6+n}P _{2}:^{6+n}P _{2}=56:12$


Correct Option: A

If P(n, n) denotes the number of permutations of n different things taken all at a time then P(n, n) is also identical to

  1. n.P(n 1, n 1)

  2. P(n, n 1)

  3. n! 

  4. (n r) . P(n, r)


Correct Option: C
Explanation:
$P(n,n)=\dfrac{n!}{0!}=n!$
No one correct, answer is $n.$

$\displaystyle ^{n}P {n}=$___.

  1. $n!$

  2. $(n-1)!(3)$

  3. $1$

  4. $(n+1)6$


Correct Option: A
Explanation:

$^nP _r = \dfrac { n! }{ (n-r)! } $

So, $ ^nP _n = \dfrac { n! }{ (n-n)! } =\dfrac { n! }{ 0! }  = n ! $ (Since $ 0 ! = 1 )$

In an examination hall, there are four rows of chairs. Each row has $8$ chairs one behind the other. There are two classes sitting for the examination with $16$ students in each class. It is desired that in each row all students belong to the same class and that no two adjacent rows are allotted to the same class. In how many ways can these $32$ students be seated?

  1. $2 \times 16! \times 16!$

  2. $4 \times 16! \times 16!$

  3. $2 \times 8! \times 16!$

  4. None of these


Correct Option: A
Explanation:

Since there are 4 rows, let us label the rows as row 1,2,3,4.

each row has 8 chairs. since all the students of the same class sit in the same row. and no adjacent row is alloted to the same class.

therefore one class can be alloted either in 1 and 3 rows or 2 and 4 rows. therefore there are 2 ways to allot the rows to the class.

now 16 students of this class can be arranged in 16 seats, the number of ways to arrange 16 students in 16 seats=16!

similarly 16 students of other class can be arranged in 16! ways.

therefore total number of ways=2$\times$16!$\times$16! ways

$\displaystyle ^{5}P {4}=$___.

  1. $720$

  2. $120$

  3. $60$

  4. $360$


Correct Option: B
Explanation:

$ { { n } _{ P } } _{ r } = \frac { n! }{ (n-r)! } $

So, $ { { 5 } _{ P } } _{ 4 } = \frac { 5! }{ (5-4)! } =\frac { 5! }{ 1! }  = 5 ! = 5 \times 4 \times 3 \times 2 \times 1 = 120 $ (Since $ 1 ! = 1 )$

'$X$' completes a job in $2$ days and '$Y$' completes it in $3$ days and '$Z$' takes $4$ days to complete it. If they work together and get Rs. $3,900$ for the job, then how much amount does '$Y$' get?

  1. Rs. $1,800$

  2. Rs. $ 1,200$

  3. Rs. $ 900$

  4. Rs. $ 800$


Correct Option: B
Explanation:

$X$ do job in $2$ days, $Y$ completes it in $3$ days, $Z$ takes $4$ days.
If $X, Y, Z$ together can do in $1$ day, then 
$= \dfrac {1}{2} + \dfrac {1}{3} + \dfrac {1}{4} = \dfrac {13}{12}$ of work
Therefore, the whole work is done in $\dfrac {12}{13}$ of a day.
Daily wages of $Y = \dfrac {1}{3}\times $ Rs. $ 3,900 =$ Rs. $ 1,300$
$\therefore$ Amount of $Y = \dfrac {12}{13} \times$ Rs. $ 1,300 =$ Rs. $1,200$

In how many ways the letters of the word $'LEADER '$ can be arranged?

  1. $5!\times2!$

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

  3. $6!$

  4. $4!\times2!$


Correct Option: B
Explanation:

Number of letters in the word $= 6$, $E$ is repeated twice
So, total number of ways of arranging $=\dfrac{6!}{2!} $

Find the number of words, with meaning or without meaning, that can be formed by arranging the letters of the word $'EIGHT'$ in all possible ways 

  1. $180$

  2. $120$

  3. $24$

  4. $720$


Correct Option: B
Explanation:

Number of ways of permuting $5$ distinct elements = $5!$
$\therefore$ The answer $=5! = 5\times4\times 3\times2 \times1=120$
Hence option B

Find the number of ways in which the letters of the word $'AEROPLANE'$ can be arranged such that the vowels are always together.

  1. $\dfrac{5!}{2!}$

  2. $\dfrac{5!}{2!^2}$

  3. $\dfrac{5!^2}{2!^2}$

  4. $\dfrac{9!}{2!^2}$


Correct Option: C
Explanation:

$'AEROPLANE'$ Total number of letters $=9$
Vowels - $ 2A,2E,1O$
Considering all vowels as a single letter, we are left with $5$ letters
They can be arranged in $5!$ ways 
Now vowels can be arranged among themselves in $\dfrac{5!}{2!.2!}$
So required number of ways $=\dfrac{5!^2}{2!^2} = 3600 $

In how many ways can you partition $6$ into ordered summands? (For example, $3$ can be partitioned in $3$ ways as : $1 + 2, \,2 + 1, \,1 + 1 + 1$)

  1. $27$

  2. $29$

  3. $31$

  4. $33$


Correct Option: C
Explanation:

$1+1+1+1+1+1$      $1$ ways

$2+1+1+1+1$             $5$ ways
$3+1+1+1$                    $4$ ways
$4+1+1$                           $3$ ways
$5+1$                                  $2$ ways
$2+3+1$                           $6$ ways
$2+4$                                  $2$ ways
$3+3$                                  $1$ ways
$2+2+2$                           $1$ ways
$2+2+1+1$                    $6$ ways
Total number of ways of partitioning $6=1+5+4+3+2+6+2+1+1+6=31$.
Hence, option C is correct.

If $^{56}P _{r+6} : ^{54}P _{r+3} = 30800:1$ find $r$.

  1. $1280$

  2. $1440$

  3. $1520$

  4. $1640$


Correct Option: D
Explanation:

We have:
$\dfrac { 56! }{ (50-r)! } .\dfrac { (51-r)! }{ 54! } =30800\ =>56.55.(51-r)=30800\ =>51-r=10=>r=41$

Therefore, $ _{ 2 }^{ r }{ P }= _{ 2 }^{ 41 }{ P }=41.40=1640$
Hence, (D) is correct.

A bag contains  $4$ red,  $3$ black, and  $2$ white balls. If  $2$  balls are selected at random, the probability of selecting atleast one white ball is

  1. $\dfrac { 7 } { 12 }$

  2. $\dfrac { 5 } { 12 }$

  3. $\dfrac { 1 } { 3 }$

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


Correct Option: C
Explanation:
Bag contains $4$ red, $3$ black and, $2$ white balls
Two balls are selected at random
The total no. of ways of doing that is ${ 10 } _{ { C } _{ 2 } }=\dfrac { 10! }{ 8!\times 2! } =\dfrac { 10\times 9\times 8! }{ 8!\times 2! } =\dfrac { 10\times 9 }{ 2 } =45$
Now in the selection we need to ensure that at least on white ball is selected.
Case $1$ :  $1$ white ball $+$ $1$ ball of any other color
This can be done in ${ 2 } _{ { C } _{ 1 } }\times { 7 } _{ { C } _{ 1 } }=2\times 7=14$ ways
Case $2$ :  $2$ white ball
This can be done in ${ 2 } _{ { C } _{ 2 } }=1$ way
$\therefore$   probability of selecting atleast one white ball is $=\dfrac { 14+1 }{ 45 } =\dfrac { 15 }{ 45 } =\dfrac { 1 }{ 3 } $
Answer : Option C.

Which of the following is true ?

  1. $^nP _r = ^{n-1}P _r + r\times ^{n-1}P _{r-1}$

  2. $^nP _r = ^{n-1}P _{r-1} + r\times ^{n-1}P _{r-1}$

  3. $^nC _r = ^{n-1}C _{r-1} + r \times^{n-1}C _{r-1}$

  4. None of these


Correct Option: A
Explanation:

$\mbox{R.H.S. =}  ^{n-1}P _r + r ^{n-1}P _{r-1}$
$\quad = \displaystyle\frac{(n-1)!}{(n-1-r)!} + r\displaystyle\frac{(n-1)!}{(n-1-r+1)!} = \displaystyle\frac{(n-1)!}{(n-r-1)!} + \displaystyle\frac{r(n-1)!}{(n-r)!}$

$\quad = \displaystyle\frac{(n-r)(n-1)!}{(n-r)(n-1-r)!} + \displaystyle\frac{r(n-1)!}{(n-1)!} = \displaystyle\frac{(n-r)(n-1)!}{(n-r)!} + \displaystyle\frac{r(n-1)!}{(n-1)!} \quad [\therefore \space\alpha(\alpha - 1)! = \alpha!, \alpha\in N]$

$\quad = \displaystyle\frac{(n-1)!}{(n-r)!}[n - r + r] = \displaystyle\frac{n(n-1)!}{(n-1)!} = \displaystyle\frac{n!}{(n-r)!}$

$\quad = \space ^nP _r$

$\quad = \space \mbox{L.H.S}$

If $\displaystyle \frac{^{n}P _{r-1}}{a}=\frac{^{n}P _{r}}{b}=\frac{^{n}P _{r+1}}{c}$,then which of the following holds good 

  1. $c^{2}=a(b+c)$

  2. $a^{2}=c(a+b)$

  3. $b^{2}=a(b+c)$

  4. $\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$


Correct Option: C
Explanation:

$\displaystyle \frac{^{n}p _{r-1}}{a}=\frac{^{n}P _{r}}{b}$

$\displaystyle \Rightarrow n-r=\frac{b}{a}-1$

and $\displaystyle \frac{^{n}P _{r}}{b}=\frac{^{n}P _{r+1}}{c}$

$\displaystyle \Rightarrow n-r=\frac{c}{b}$
On dividing (i) and (ii) we get
$b^{2}=a(b+c)$

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