datavorous/entrance-exam-dataset huggingface dataset:
The number of values of $\mathrm{k}$ for which the linear equations $4 x+k y+2 z=0 ; k x+4 y+z=0 ; 2 x+2 y+z=0$ possess a non-zero solution is

Options:
A
2

B
1

C
zero

D
3

The correct answer is:
2

Solution:
$\left|\begin{array}{lll}4 & k & 2 \\ k & 4 & 1 \\ 2 & 2 & 1\end{array}\right|=0 \Rightarrow k^2-6 k+8=0 \Rightarrow k=4,2$

Topic: Mathematics, Determinants, JEE Main, JEE Main 2011
The number of values of k, for which the system of equations :

k+1x+8y=4k

kx+k+3y=3k-1

has no solution, is :

Options:
A 2
B 3
C Infinite
D 1

The correct answer is:
1

Solution:
For a1x+b1y=c1anda2x+b2y=c2 to have no solution,

a1a2=b1b2≠c1c2

∴k+1k=8k+3≠4k3k-1

k+1k+3=8k⇒k2-4k+3=0

⇒k=1,3

but, for k=1,8k+3=4k3k-1

∴k=3

Hence, only one value

Topic: Mathematics, Determinants, JEE Main, JEE Main 2013 (07 Apr)
The number of values of k for which the system of linear equations k+2x+10y=k & kx+k+3y=k-1 has no solution is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
1

Solution:
For no solution

k+2k=10k+3=kk-1

⇒k+2k+3=10k

⇒k2-5k+6=0⇒k=2, 3

k≠2 for k=2 both lines are identical.

So, only k=3 is possible.

So the number of values of k is 1.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2018 (16 Apr Online)
The number of values of \( k \) for which the system of linear equations \( (k+2) x+10 y=k \& k x+(k+3) y=k-1 \) has no solution is

Options:
A \( 1 \)
B \( 2 \)
C \( 3 \)
D \( 4 \)

The correct answer is:
\( 1 \)

Solution:
For no solution

k+2k=10k+3=kk-1

⇒k+2k+3=10k

⇒k2-5k+6=0⇒k=2, 3

k≠2 for k=2 both lines are identical.

So, only k=3 is possible.

So the number of values of k is 1.

Topic: Mathematics, Determinants, JEE Main
The number of values of p for which the lines x+y-1=0, px+2y+1=0 and 4x+2py+7=0 are concurrent is equal to

Options:
A 0
B 2
C 1
D Infinite

The correct answer is:
1

Solution:
11-1p2142p7=0
â2p2+9p-26=0
âp=-132,2
p=2 is rejected because at p=2 all 3 lines becomes parallel
â Hence, p=-132

Topic: Mathematics, Straight Lines, JEE Main
The number of values of the parameter Î±â0,2Ï for which the quadratic function sinâ¡Î±x2+2cosâ¡Î±x+12cosâ¡Î±+sinâ¡Î± is the square of a linear function is

Options:
A 2
B 4
C 6
D 8

The correct answer is:
4

Solution:
For the quadratic function to be square of a linear function

D=0
⇒4cos2⁡α-4·12sin⁡αcos⁡α+sin⁡α=0⇒2cos⁡α+sin⁡αcos⁡α-sin⁡α=0
⇒tan⁡α=1 or -2
For α∈0,2π there are two solutions for each.
⇒ total four values of α.

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of values of xâ-2Ï,2Ï which satisfy the equation cosecx=1+cotâ¡x is equal to

Options:
A 0
B 2
C 4
D 6

The correct answer is:
2

Solution:
cosecx=1+cotâ¡x
1sinâ¡x=sinâ¡x+cosâ¡xsinâ¡x
âsinâ¡x+cosâ¡x=1 and sinâ¡xâ 0
âcosâ¡x-Ï4=12
âx-Ï4=Â±Ï4,Â±7Ï4
âx=0,Ï2,2Ï,-3Ï2
But, sinâ¡xâ 0âx=Ï2,-3Ï2

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of values of x in -4,-1, for which the matrix 3-1+x23-1x+2x+3-12 is singular, is

Options:
A 1
B 3
C 4
D 5

The correct answer is:
1

Solution:
For the matrix to be singular, its determinant value should be zero
3-1+x23-1x+2x+3-12=0
â3-2+x+2-x-16-x+2x+3+2-3+x+3=0
â3x+x-1xx+5+2x=0
âx5+x2+4x-5=0âx=0,-4

But, x=0 is rejected

Hence, only one value in -4,-1

Topic: Mathematics, Determinants, JEE Main
The number of values of $x$ in the interval $[0,3 \pi]$ satisfying the equation $2 \sin ^2 x+5 \sin x-3=0$ is

Options:
A
4

B
6

C
1

D
2

The correct answer is:
4

Solution:
$2 \sin ^2 x+5 \sin x-3=0$
$\Rightarrow(\sin x+3)(2 \sin x-1)=0$
$\Rightarrow \sin x=\frac{1}{2} \quad \therefore \ln (0,3 \pi), x$ has 4 values

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2006
The number of values of x in the interval 0,5π satisfying the equation 3sin2⁡x-7sin⁡x+2=0 is

Options:
A 0
B 5
C 6
D 10

The correct answer is:
6

Solution:
3sin2⁡x-7sin⁡x+2=0
3sin2⁡x-6sin⁡x-sin⁡x+2=0
3sin⁡x-1sin⁡x-2=0
⇒sin⁡x=13,sin⁡x=2 not possible
Fro the figure, 6 roots in 0,5π

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of values of x in the interval π4,7π4 for which 14cosec2x-2sin2x=21-4cos2x holds, is ______.

Options:

The correct answer is:
4

Solution:
Given x∈π4,7π4

14cosec2x-2sin2x=21-4cos2x

=21-41-sin2x

=17+4sin2x

14sin2x-6sin2x=17

Let sin2x=p

14p-6p=17⇒14-6p2=17p

6p2+17p-14=0

p=-72,23⇒sin2x=23

⇒sinx=±23

∴ There are 4 solutions

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2022 (25 Jun Shift 1)
The number of values of x lying in the interval -2π,2π satisfying the equation 1+cos⁡10xcos⁡6x=2cos28x+sin28x is equal to

Options:

The correct answer is:
7

Solution:
We have
1+cos⁡10xcos⁡6x=2cos28x+sin28x
=cos28x+1
So, cos⁡10xcos⁡6x=cos28x=1+cos⁡16x2
⇒cos⁡16x+cos⁡4x=1+cos⁡16x
⇒cos⁡4x=1
⇒4x=2nπ,n∈Z
⇒x=nπ2,n∈Z
in -2π,2π, x=±3π2,±π,±π2,0
So, the number of values of x is 7.

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of values of x such that x, x and x are in arithmetic progression is equal to (where, ⋅ denotes the greatest integer function and ⋅ denotes the fractional part function)

Options:
A 0
B 1
C 2
D 4

The correct answer is:
2

Solution:
We have,
2x=x+x=x+2x
⇒x=x2.
Now, 0≤x<1
⇒0≤x2<1
⇒0≤x<2⇒x=0,1
for x=0, x=0⇒x=0
for x=1, x=12⇒x=32
Thus, x=0 and 32.

Topic: Mathematics, Sequences and Series, JEE Main
The number of water molecules in gypsum, dead burnt plaster and plaster of Paris, respectively are:

Options:
A 2, 0 and 0.5
B 0.5, 0 and 2
C 2, 0 and 1
D 5, 0 and 0.5

The correct answer is:
2, 0 and 0.5

Solution:
Gypsum =CaSO4·2H2O
Plaster of paris (POP)=CaSO4·12H2O
Dead burnt of plaster =CaSO4

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2021 (27 Aug Shift 1)
The number of water molecules not coordinated to copper ion directly in CuSO4·5H2O, is

Options:
A 4
B 2
C 1
D 3

The correct answer is:
1

Solution:
In CuSO4·5H2O, 4 H2O molecules are directly attached to the Cu atom while one is present as the water of crystallisation.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2019 (09 Apr Shift 1)
The number of waves contained in unit length of the medium is called

Options:
A Elastic wave
B Wave number
C Wave pulse
D Electromagnetic wave

The correct answer is:
Wave number

Solution:
Wave number is the reciprocal of wavelength and is written as $n=\frac{1}{\lambda}$

Topic: Physics, Waves and Sound, JEE Main
The number of ways, 16 identical cubes, of which 11 are blue and rest are red, can be placed in a row so that between any two red cubes there should be at least 2 blue cubes, is ______.

Options:

The correct answer is:
56

Solution:
Given, 4 blue & 5 red spherical ball

Now according to question there should be minimum 2 blue ball in between any two red ball.

So assume a,b,c,d,e,f are number of blue balls,

So, a+b+c+d+e+f=11 ......(i)

Where b,c,d,e⩾2

Now let b=b'+2

c=c'+2

d=d'+2

e=e'+2

So equation (i) becomes

a+b'+2+c'+2+d'+2+e'+2+f=11

⇒ a+b'+c'+d'+e'+f=3

Now total ways will be

= 3+6-1C6-1=8C5=8C3=56

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2022 (27 Jun Shift 1)
The number of ways in which 2n distinct letters (addressed) can be distributed in N distinct mail boxes such that there are exactly K letters n<K≤2n in one of the mail boxes is

Options:
A C2nK⋅N
B CK2n·NN-12n-K
C C2nKN-12n-K
D C2nK2n-KN-1⋅N

The correct answer is:
CK2n·NN-12n-K

Solution:
Select K letters from 2n in C2nK ways.
Select 1 of the mail boxes in CN1 way.
Remaining 2n-K letters each have N-1 options and can be allotted in N-12n-K ways.
Hence, the required number of ways =CK2nC1NN-12n-K.

Topic: Mathematics, Permutation Combination, JEE Main
The number of ways in which \( 4 \) boys and \( 4 \) girls can sit alternatively in a row if a particular boy and a particular girl are never adjacent to each other, are

Options:
A \( 1152 \)
B \( 504 \)
C \( 1656 \)
D \( 648 \)

The correct answer is:
\( 648 \)

Solution:
Total number of ways in which boys and girls can seat alternatively =4!×4!×2=1152.
When particular boy and particular girl are always together, then number of ways =3!×3!×7×2=504.
∴ Required number of ways =1152-504= 648.

Topic: Mathematics, Permutation Combination, JEE Main
The number of ways in which 5 beads of different colours form a necklace is

Options:
A $12$
B $24$
C $120$
D $60$

The correct answer is:
$12$

Solution:
The number of ways in which 5 beads of different colours can be arranged in a circle to form a necklace are $(5-1)!=4$ !.
But the clockwise and anticlockwise arrangement are not different (because when the necklace is turned over one gives rise to another)
Hence the total number of ways of arranging the beads $=\frac{1}{2}(4!)=12$.

Topic: Mathematics, Permutation Combination, JEE Main
The number of ways in which 5 boys and 3 girls can be seated on a round table if a particular boy B1 and a particular girl G1 never sit adjacent to each other, is:

Options:
A 7!
B 5×6!
C 6×6!
D 5×7!

The correct answer is:
5×6!

Solution:
Number of ways = Total - when B1 & G1 sit together

Total ways in which 5 boys+3 girls=8 people can be arranged is 7!

Now, when B1 & G1 sit together the number of ways (other 6and B1G1 grouped together) is 6!×2!
Hence, the required numer of ways is 7!-6!2!=5×6!

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2017 (09 Apr Online)
The number of ways in which 5 boys and 4 girls can be arranged on a circular table such that no two girls sit together and two particular boys are always together is

Options:
A 288
B 44
C 720
D 540

The correct answer is:
288

Solution:
Let two particular boys as one boy, we have only four boys which can be seated at a round table in 3! ways.
The two boys together can be arranged in 2 ways.
So, boys can be seated in 2Ã3! ways.
B1B2 are together. Now 4 girls can be seated at four places (marked Ã) in 4! ways.
â Required number of ways =3!Ã2Ã4!
=6Ã2Ã24=288 ways

Topic: Mathematics, Permutation Combination, JEE Main
The number of ways in which 5 boys and 4 girls sit around a circular tables. So, that no two girls sit together is

Options:
A $5 ! 4 !$
B $3 ! 3 !$
C $5 !$
D $4 !$

The correct answer is:
$5 ! 4 !$

Solution:
Number of ways in which the boys can be arranged $=4$ !
Number of ways in which girls alternately arranged $=5$ !
Total number of ways $=5 ! 4 !$

Topic: Mathematics, Permutation Combination, JEE Main
The number of ways, in which 5 girls and 7 boys can be seated at a round table so that no two girls sit together is

Options:
A 720
B 126(5!)2
C 73602
D 77202

The correct answer is:
126(5!)2

Solution:
Given,

7 boys and 5 girls are to be seated around a circular such that no two girls to be seated together,

Now we know that n objects can be arranged in a circle in (n-1)! ways.

Let us first arrange 7 boys in circular arrangement in 7-1! ways.

Now there will be 7 gaps.

So let us select any 5 gaps out of 7 gaps and arrange 5 girls in the chosen gaps. This can be done in C57×5! ways.

Hence, required arrangements are 6!×C57×5!

=6×5!×7×62×5!

=1265!2.

Therefore, required arrangements are 1265!2

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2023 (08 Apr Shift 1)
The number of ways in which 6 men and 5 women can dine at a found table if no two women are to sit together is given by

Options:
A
$7 \times 5$

B
$6 \times 5$

C
30

D
$5 \times 4$

The correct answer is:
$7 \times 5$

Solution:
No.of ways in which $6 \mathrm{~mm}$ can be arranged at a round table $=(6-1)$ ! Now women can be arranged in 6 ! ways.
Total number of ways $=6 ! \times 5$ !

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2003
The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by

Options:
A $6!\times 5$ !
B $30$
C $5!\times 4$ !
D $7!\times 5$ !

The correct answer is:
$6!\times 5$ !

Solution:
No. of ways in which 6 men can be arranged at a round table $=(6-1)!$
Now women can be arranged in 6! ways.
Total Number of ways $=6!\times 5!$

Topic: Mathematics, Permutation Combination, JEE Main
The number of ways in which 9 persons can be divided into three equal groups

Options:
A $1680$
B $840$
C $560$
D $280$

The correct answer is:
$1680$

Solution:
Total ways $=\frac{9!}{(3!)^3}=\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4}{3 \times 2 \times 3 \times 2}=1680$.

Topic: Mathematics, Permutation Combination, JEE Main
The number of ways in which 10 balls can be selected from 10 identical green balls, 10 identical blue balls and 9 identical red balls are

Options:
A 63
B 64
C 65
D 66

The correct answer is:
65

Solution:
Required number of ways = coefficient of x10 in
x0+x1+x2+....+x102x0+x1+x2+....+x9
= coeff. of x10 in 1-x111-x21-x101-x
= coeff. of x10 in 1–x–31–x1121–x10
= coeff. of x10 in 1–x–31+x22–2x111–x10
= coeff. of x10 in 1–x–31+x22–2x11–x10–x32+2x21
= coeff. of x10 in 1–x–31–x10–2x11+2x21+x22–x32
= coeff. of x10 in 1–x–3– coeff. of x0 in 1–x–3
=10+3-1C10-1=12C10-1=12×112–1=65

Topic: Mathematics, Permutation Combination, JEE Main
The number of ways in which 10 boys can be divided into 2 groups of 5, such that two tallest boys are in two different groups, is equal to

Options:
A 70
B 35
C 252
D 126

The correct answer is:
70

Solution:
Excluding the 2 tallest boys, we can divide the 8 boys into 2 groups of 4 each by
8!4!4!2! ways.
The tallest boys can be assigned to the groups in 2 ways.
The desired number of ways are
8!4!4!2!×2!=8×7×6×54×3×2=70

Topic: Mathematics, Permutation Combination, JEE Main
The number of ways in which 10 boys can take positions around a circular round table, if two particular boys must not be seated side by side, is

Options:
A 2×48!
B 12
C 360
D 7×8!

The correct answer is:
7×8!

Solution:
Total number of ways without any restriction =9!
Considering two of the particular boys sit together and then assuming them as one, the total no. of ways =8!
They them selves can be arranged in 2 ways.
So, the required number of ways
=9!-2×8!=7×8!

Topic: Mathematics, Permutation Combination, JEE Main
The number of ways in which a cricket team of 11 players be chosen out of a batch of 15 players so that the captain of the team is always included, is

Options:
A 165
B 364
C 1001
D 1365

The correct answer is:
1001

Solution:
If captain is always included then we can choose 10 more players out of the remaining 14 players. So
${ }^{14} \mathrm{C}_{10}=\frac{14 !}{10 ! 4 !}=1001$

Topic: Mathematics, Permutation Combination, NDA, NDA 2015 (Phase 2)
The number of ways in which an examiner can assign 30 marks to 8 questions, giving not less than 2 marks to any question, is :

Options:
A
${ }^{30} C_7$

B
${ }^{21} C_8$

C
${ }^{21} C_7$

D
${ }^{30} C_8$

The correct answer is:
${ }^{21} C_7$

Solution:
30 marks to be alloted to 8 questions. Each question has to be given $\geq 2$ marks Let questions be $a, b, c, d, e, f, g, h$ and $a+b+c+d+e+f+g+h=30$
Let $a=a_1+2$ so, $a_1 \geq 0$ $b=a_2+2$ so, $a_2 \geq 0, \ldots \ldots . a_8 \geq 0$
So, $\left.\begin{array}{c}a_1+a_2+\ldots \ldots+a_8 \\ +2+2+\ldots . .+2\end{array}\right\}=30$ $\Rightarrow a_1+a_2+\ldots \ldots+a_8=30-16=14$
So, this is a problem of distributing 14 articles in 8 groups.
Number of ways $={ }^{14+8-1} \mathrm{C}_{8-1}={ }^{21} \mathrm{C}_7$

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2013 (22 Apr Online)
The number of ways in which four different letters can be put in their four addressed envelopes such that atleast two of them are in the wrong envelopes are

Options:
A 23
B 25
C 6
D 7

The correct answer is:
23

Solution:
Two in wrong envelops
=4C2×2!12!=6
Three in wrong envelops
=4C1×3!12!-13!=43-1=8
Four in wrong envelops
=4!12!-13!+14!=12-4+1=9
The total ways =6+8+9=23

Topic: Mathematics, Permutation Combination, JEE Main
The number of ways in which four different toys and five indistinguishable marbles can be distributed between 3 boys, if each boy receives at least one toy and at least one marble, is

Options:
A 42
B 100
C 150
D 216

The correct answer is:
216

Solution:
5 identical marbles can be distributed to 3 boys, each receiving at least one in 5-1C3-1=C42 ways
4 different toys can be grouped as 2,1,1 in 4!2!1!1!=4!2! ways
Two boys will get 1 each and they can be selected in C23 ways.
Required number of ways =C42×4!2!×C32=6×12×3=216

Topic: Mathematics, Permutation Combination, JEE Main
The number of ways in which letters of the word ‘ARRANGE’ can be arranged, such that no two R’s are together, is

Options:
A 160
B 200
C 360
D 900

The correct answer is:
900

Solution:
AANGE can be arranged in 5!2!=60 ways
Six gaps will be created, two can be selected in C26 ways where two R’s can be arranged in one way
So, required number of ways
=6×52×60=15×60=900

Topic: Mathematics, Permutation Combination, JEE Main
The number of ways in which the letters of the word ‘ARRANGE’ can be arranged so that two A’s are together is

Options:
A 160
B 200
C 360
D 900

The correct answer is:
360

Solution:
Let us tie the two A’s.
AA RRNGE → Total 6 objects, two identical
6!2!=6×5×4×3
=360

Topic: Mathematics, Permutation Combination, JEE Main
The number of ways in which the letters of the word MACHINE can be arranged such that the vowels may occupy only odd positions, is

Options:
A 288
B 625
C 576
D 1152

The correct answer is:
576

Solution:
A, I, E (3 vowels)
M, C, H, N (4 consonants)
There are 4 odd places. So, three vowels can be arranged in four places in $4_{\mathrm{P}_3}$ ways and 4 consonants can be arranged in remaining four places in $4_{\mathrm{P}_4}$ ways.
Hence total number of ways $=4_{\mathrm{P}_3} \times 4_{\mathrm{P}_4}=576$

Topic: Mathematics, Permutation Combination, MHT CET, MHT CET 2022 (05 Aug Shift 2)
The number of ways in which the letters of the word 'VERTICAL' can be arranged without changing the order of the vowels is

Options:
A \(6 ! \times 3 !\)
B \(\frac{8 !}{3}\)
C \(6 ! \times 3\)
D \(\frac{8 !}{3 !}\)

The correct answer is:
\(\frac{8 !}{3 !}\)

Solution:
Hint : Vowels 'EIA' to be kept in order, so out of 8 places 3 places can be chosen in \({ }^8 \mathrm{C}_3\) ways. Remaining 5 letters can be arranged in 5 ! ways.
\(\therefore\) Total number of ways \(={ }^8 \mathrm{C}_3 \times 5 !=\frac{8 !}{3 !}\)

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2023
The number of ways in which three numbers in arithmetic progression can be selected from 1,2,3,……,50 is

Options:
A 276
B 600
C 840
D 640

The correct answer is:
600

Solution:
a+c=2b
So, both a and c will be even or both will be odd as their sum is even.
In first 50 natural number 25 are even and 25 are odd.
Required number of ways 2C52+ 2C52=2×25×242=600

Topic: Mathematics, Permutation Combination, JEE Main
The number of ways in which we can put 5 different balls in 5 different boxes such that atmost three boxes are empty, is equal to

Options:
A 55+5
B 55-10
C 55-5
D 55-45

The correct answer is:
55-5

Solution:
5 different balls can be put in 5 different boxes in 55 ways.
At most 3 empty boxes are allowed
Number of ways in which 4 boxes are empty =5C1×15=5.
Number of ways in which 5 boxes are empty =0, as balls has to be placed in one of the boxes.
⇒ Required number of ways =55-5.

Topic: Mathematics, Permutation Combination, JEE Main
The number of ways of arranging 8 men and 4 women around a circular table such that no two women can sit together, is

Options:
A 8!
B 4!
C 8!4!
D 7! 8P4

The correct answer is:
7! 8P4

Solution:
The number of ways of arranging 8 men =7!
The number of ways of arranging 4 women such that no two women can sit together = 8P4
â´ Required number of ways =7! 8P4

Topic: Mathematics, Permutation Combination, JEE Main
The number of ways of arranging 18 boys such that 3 particular boys are always separate is equal to

Options:
A 18!-16!3!
B 16!P173
C 15!P163
D 17!P173

The correct answer is:
15!P163

Solution:
First arrange 15 boys in 15! ways.
16 gaps will be created.
Select any 3 gaps out of 16 gaps in C163 ways.
Arrange 3 boys to be separated in these selected 3 gaps in 3! ways.
Number of ways =15!ÃC163Ã3!
=15!ÃP163

Topic: Mathematics, Permutation Combination, JEE Main
The number of ways of arranging the letters AAAAA, BBB, CCC, D, EE and F in a row, if the letters B are separated from one another, is equal to

Options:
A 13!5!3!3!2!
B 14!3!3!2!
C 15!3!22!5!
D C313×12!5!3!2!

The correct answer is:
C313×12!5!3!2!

Solution:
All AAAAA, CCC, D, EE, F can be arranged in 12!5!3!2! ways.
Between the gaps B can be arranged in C313 ways
Hence, the total number of ways =C133Ã12!5!3!2!

Topic: Mathematics, Permutation Combination, JEE Main
The Number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remaining 21 are distinct, is:

Options:
A 220
B 221
C 220+1
D 220-1

The correct answer is:
220

Solution:
There is only one way in which we can choose one, two or more objects from 10 identical objects.
Number of ways of selecting 10 objects
= 21C10+ 21C9+ 21C8+ 21C7+ 21C6+ 21C5+ 21C4+ 21C3+ 21C2+ 21C1+ 21C0
We know nCr= nCn-r and
⇒221= 21C0+ 21C1+ 21C2+ 21C3+ 21C4+.........+ 21C21            ∵ 21C21= 21C0 21C20= 21C1 21C19= 21C2
=2 21C0 + 21C1 + 21C2 +……………+ 21C10
⇒ 21C0+ 21C1 +21C2 +.........+ 21C10 = 220

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2019 (12 Apr Shift 1)
The Number of ways of choosing \( 10 \) objects out of \( 31 \) objects of which \( 10 \) are identical and the remaining \( 21 \) are distinct, is:

Options:
A \( 2^{20} \)
B \( 2^{21} \)
C \( 2^{20}+1 \)
D \( 2^{20}-1 \)

The correct answer is:
\( 2^{20} \)

Solution:
There is only one way in which we can choose one, two or more objects from 10 identical objects.
Number of ways of selecting 10 objects
= 21C10+ 21C9+ 21C8+ 21C7+ 21C6+ 21C5+ 21C4+ 21C3+ 21C2+ 21C1+ 21C0
We know nCr= nCn-r and
⇒221= 21C0+ 21C1+ 21C2+ 21C3+ 21C4+.........+ 21C21            ∵ 21C21= 21C0 21C20= 21C1 21C19= 21C2
=2 21C0 + 21C1 + 21C2 +……………+ 21C10
⇒ 21C0+ 21C1 +21C2 +.........+ 21C10 = 220

Topic: Mathematics, Permutation Combination, JEE Main
The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is

Options:
A
5

B
${ }^8 C_3$

C
$3^8$

D
21

The correct answer is:
21

Solution:
Number of balls $=8$
number of boxes $=3$
Hence number of ways $={ }^7 C_2=21$.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2004
The number of ways of dividing 52 cards amongst four players equally, are

Options:
A $\frac{52!}{(13!)^4}$
B $\frac{52!}{(13!)^2 4!}$
C $\frac{52!}{(12!)^4(4!)}$
D None of these

The correct answer is:
$\frac{52!}{(13!)^4}$

Solution:
Number of ways
\(\begin{aligned}
&={ }^{52} \mathrm{C}_{13} \times{ }^{39} \mathrm{C}_{13} \times{ }^{26} \mathrm{C}_{13} \times{ }^{13} \mathrm{C}_{13} \\
&=\frac{52!}{13!(52-13)!} \times \frac{39!}{13!(39-13)!} \times \frac{26!}{13!(26-13)!} \times \frac{13!}{13!(13-13)!} \quad\left({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=\frac{\mathrm{n !}}{\mathrm{r !}(\mathrm{n}-\mathrm{r})!}\right) \\
&=\frac{52!}{13!\times 39!} \times \frac{39!}{13!\times 26!} \times \frac{26!}{13!\times 13!}(\because 0!=1) \\
&=\frac{52!}{(13!)^4}
\end{aligned}\)

Topic: Mathematics, Permutation Combination, JEE Main
The number of ways of giving 20 distinct oranges to 3 children such that each child gets at least one orange is _____

Options:

The correct answer is:
171

Solution:
Given,

We have to distribute 20 identical oranges to 3 children,

Let children A get OA oranges, children B gets OB oranges and children C gets OC oranges,

So, OA+OB+OC=20

Now this can be distributed using the multinomial theorem,

So, Number of ways will be,

=coefficient of x20 in x+x2+….+x183

=coefficient of x20 in x31+x+x2+….x173

=coefficient of x17 in 1-x181-x3

=coefficient of x17 in (1-x)-3

=C219=171

As we know that expansion of (1-x)-3 is 1+C13x+C24x2+C35x3+......+C1719x17........∞

Note: This is bonus question as here 20 oranges are considered identical but in question it is considered distinct.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2023 (06 Apr Shift 1)
The number of ways of selecting 15 teams from 15 men and 15 women, such that each team consists of a man and a woman is

Options:
A 1960
B 15!
C 15!2
D 14!

The correct answer is:
15!

Solution:
First team can be selected in (15C1.15C1) way, second in (14C1.14C1) way and so on. So, number of ways of selecting 15 teams =15!215!=15!

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2015 (10 Apr Online)