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JEE • MATHEMATICS • COMBINATIONS

Combinations practice · Mixed difficulty

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 151-180 of 324 questions

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151 Easy
151 of 324
Application +4 | -1
From 12 books, the difference between number of ways a selection of 5 books when one specified book is always excluded and one specified book is always included, is
View answer and explanation
Correct option: C
132
The correct answer is 132 because it is the difference between the number of ways to select 5 books with one specified book excluded and one specified book included.
152 Easy
152 of 324
Application +4 | -1
If   35Cn+7=  35C4  n−2{\;^{35}}{C_{n + 7}} = {\;^{35}}{C_{4\;n - 2}}, then all the values of nn are given by
View answer and explanation
Correct option: B
3,6
Option B (3,6) is correct because it solves for n using the property that if nCr = nCs, then r = s or r + s = n.
153 Easy
153 of 324
Application +4 | -1
The number of ways in which two teams AA and BB of 11 players each can be made up from 22 players so that two particular players are on the opposite sides is
View answer and explanation
Correct option: A
369512
Option A (369512) is correct because it calculates the number of ways to form two teams with two specific players on opposite sides by choosing the remaining players from the remaining pool.
154 Easy
154 of 324
Application +4 | -1
The interior angles of a regular polygon measure 1600{160^0} each. The number of diagonals of the polygon are
View answer and explanation
Correct option: C
135
Option C (135) is correct because it is the number of diagonals calculated using the formula after finding the number of sides of the polygon.
155 Easy
155 of 324
Application +4 | -1
The number of integral points on the hyperbola x2−y2=(2000)2x^{2}-y^{2}=(2000)^{2} is (an integral point is a point both of whose co-ordinates are integer)
View answer and explanation
Correct option: A
98
Option A (98) is correct because it represents the number of integral points on the given hyperbola.
156 Easy
156 of 324
Application +4 | -1
The straight lines I1,I2,I3{I_1},{I_2},{I_3} are parallel and lie in the same plane. A total numbers of mm points are taken on I1,n{I_1},n points on I2,k{I_2},k points on I3{I_3}. The maximum number of triangles formed with vertices at these points is
View answer and explanation
Correct option: B
  m+n+kC3−mC3−nC3−kC3{\;^{m + n + k}}{C_3}{ - ^m}{C_3}{ - ^n}{C_3}{ - ^k}{C_3}
Option B (  m+n+kC3−mC3−nC3−kC3{\;^{m + n + k}}{C_3}{ - ^m}{C_3}{ - ^n}{C_3}{ - ^k}{C_3}) is correct because it subtracts the number of ways to choose 3 points on the same line from the total number of ways to choose 3 points.
157 Easy
157 of 324
Analyzing +4 | -1
The number of times the digit 3 will be written when listing the integers from 1 to 1000 , is
View answer and explanation
Correct option: C
300
The correct answer is C (300) because when listing integers from 1 to 1000, the digit 3 appears 300 times in the ones, tens, and hundreds places.
158 Medium
158 of 324
Analyzing +4 | -1
If ∑r=1n[log⁡2r]=1898\sum\limits_{{r=1}}^n\left[\log _2 r\right]=1898 (where [.] denotes greatest integer function), then number of zeroes at the end of n\mathrm{n} is/are
View answer and explanation
Correct option: A
74
Option A (74) is correct because the sum of the greatest integer function of log base 2 of r from 1 to n equals 1898, leading to n=300, and the number of zeroes at the end of 300! is 74.
159 Medium
159 of 324
Analyzing +4 | -1
(nn!)! is divisible by
View answer and explanation
Correct option: D
(n!)(n−1)!(n !)^{(n-1) !}
Option D is correct because (n!)! is divisible by (n!)^((n-1)!) as shown by breaking down the factorial product into smaller groups.
160 Medium
160 of 324
Understanding +4 | -1
The number of ways in which 12 balls can be divided between two friends, one receiving 8 an the other 4, is
View answer and explanation
Correct option: B
12  !2  !8  !4  !\frac{{12\;!2\;!}}{{8\;!4\;!}}
Option A (12  !2  !8  !4  !\frac{{12\;!2\;!}}{{8\;!4\;!}}) is correct because we need to consider the two possible distributions: one friend gets 8 and the other gets 4, or vice versa.
161 Easy
161 of 324
Application +4 | -1
A class consists of 4 boys and gg girls. Every Sunday five students, including at least three boys go for a picnic to Appu Ghar, a different group being sent every week. During, the picnic, the class teacher gives each girl in the group a doll. If the total number of dolls distributed was 85, then value of gg is
View answer and explanation
Correct option: D
5
Option D (5) is correct because it satisfies the equation derived from the total number of dolls distributed based on the given conditions.
162 Easy
162 of 324
Application +4 | -1
Out of 6 boys and 4 girls, a group of 7 is to be formed. In how many ways can this be done, if the group is to have a majority of boys?
View answer and explanation
Correct option: D
100
Option D (100) is correct because the total number of ways is the sum of the combinations where boys are in majority.
163 Easy
163 of 324
Application +4 | -1
The number of group that can be made from5 different green balls, 4 different blue balls and 3 different red balls, if at latest 1 green and 1 blue ball is to be included, is
View answer and explanation
Correct option: B
3720
Option B (3720) is correct because it calculates the number of ways to select at least one green and one blue ball, along with any number of red balls.
164 Easy
164 of 324
Application +4 | -1
Sita has 5 coins each of the different denomination. The number different sums of money she can form is
View answer and explanation
Correct option: C
31
Option C (31) is correct because it represents all possible non-empty subsets of the 5 coins.
165 Medium
165 of 324
Understanding +4 | -1
Statement 1: If AA and BB be two sets having pp and qq elements respectively, where q>pq>p. Then the total number of functions from set AA to set BB is qpq^{p}.
[Online May 12, 2012]
Statement 2: The total number of selections of pp different objects out of qq objects is qCp{ }^{q} \mathrm{C}_{p}.
View answer and explanation
Correct option: D
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation of Statement 1 .
Option D. Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation of Statement 1 is correct because statement 1 correctly states the number of functions from set A to set B, and statement 2 correctly states the number of selections of p objects from q objects.
166 Easy
166 of 324
Analyzing +4 | -1
A person wishes to make up as many different parties as he can out of his 20 friends such that each party consists of the same number of persons. The number of friends he should invite at a time, is
View answer and explanation
Correct option: B
10
Option B (10) is correct because the maximum number of parties is formed when the number of friends invited is half the total number of friends, which maximizes the binomial coefficient.
167 Easy
167 of 324
Application +4 | -1
The value of 50C4+∑r=1656−rC3{ }^{50} C_4+\sum\limits_{{r=1}}^6{ }^{56-r} C_3 is
View answer and explanation
Correct option: B
56C4{ }^{56} C_4
The correct answer is 56C4{ }^{56} C_4 because it is the result of applying the identity nCr+nCr−1=n+1Cr{ }^n C_r + { }^n C_{r-1} = { }^{n+1} C_r repeatedly to simplify the summation.
168 Easy
168 of 324
Application +4 | -1
  nCr+2  nCr−1+  nCr−2{\;^n}{C_r} + 2{\;^n}{C_{r - 1}} + {\;^n}{C_{r - 2}} is equal to
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the simplified expression is   n+2Cr{\;^{n + 2}}{C_r}, which is not among the given options.
169 Easy
169 of 324
Application +4 | -1
Number of diagonals of a polygon are 65 , then the number of sides of polygon are
View answer and explanation
Correct option: C
13
Option C (13) is correct because the number of sides of a polygon with 65 diagonals can be found by solving the equation n(n-3)/2 = 65.
170 Easy
170 of 324
Application +4 | -1
The value of 50C4+∑r=1656−rC3{ }^{50} C_{4}+\sum_{r=1}^{6}{ }^{56-r} C_{3} is
View answer and explanation
Correct option: D
56C4{ }^{56} C_{4}
Option D (56C4{}^{56}C_4) is correct because applying the identity nCr+nCr−1=n+1Cr{}^{n}C_r + {}^{n}C_{r-1} = {}^{n+1}C_r repeatedly simplifies the expression to 56C4{}^{56}C_4.
171 Medium
171 of 324
Application +4 | -1
Given 5 different green dyes, 4 different blue dyes and 3 different red dyes. The number of combinations of dyes that can be chosen by taking at least one green and one blue dye is
View answer and explanation
Correct option: C
3720
Option C (3720) is correct because it calculates the total number of combinations by considering at least one green and one blue dye, along with any combination of red dyes.
172 Easy
172 of 324
Analyzing +4 | -1
There were two women participating in a chess tournament. Every participant played two games with the other participants. The number of games that the men played between themselves proved to exceed by 66 the number of games that the men played with the women. The number of participants is
View answer and explanation
Correct option: C
13
The correct answer is 13 because solving the equation derived from the problem statement gives 11 men, and adding the 2 women results in 13 participants.
173 Easy
173 of 324
Application +4 | -1
Let TnT_n denote the number of triangles which can be formed using the vertices of a regular polygon of nn sides. If Tn+1−Tn=21T_{n+1}-T_n=21, then nn equals
View answer and explanation
Correct option: A
7
Option A (7) is correct because the equation n+1C3−nC3=21{ }^{n+1} C_3-{ }^n C_3=21 simplifies to nC2=21{ }^n C_2=21, which yields n=7n=7.
174 Easy
174 of 324
Application +4 | -1
In a plane there are 37 straight lines, of which 13 pass through the point AA and 11 pass through the point BB. Besides, no three lines pass through one point, no line passes through both points AA and BB and no two are parallel, the number of intersection points the lines have, is
View answer and explanation
Correct option: A
535
Option A (535) is correct because it calculates the number of intersection points of the lines, accounting for the lines passing through points A and B.
175 Easy
175 of 324
Application +4 | -1
The straight lines I1,I2,I3I_1, I_2, I_3 are parallel and lie in the same plane. A total number of mm points on I1;nI_1 ; n points on I2;kI_2 ; k points on I3I_3, the maximum number of triangles formed with vertices at these points is
View answer and explanation
Correct option: B
m+n+kC3−mC3−nC3−kC3{ }^{m+n+k} C_3-{ }^m C_3-{ }^n C_3-{ }^k C_3
Option B (m+n+kC3−mC3−nC3−kC3{ }^{m+n+k} C_3-{ }^m C_3-{ }^n C_3-{ }^k C_3) is correct because it subtracts the number of invalid triangles (formed by collinear points) from the total possible combinations of points.
176 Easy
176 of 324
Application +4 | -1
There are nn points in a plane, no three of which are collinear except ' pp ' points all of which are on a line. How many
(a) straight lines can be formed and
(b) triangles can be formed out of these nn points?
View answer and explanation
Correct option: A
(i) nC2− pC2+1^n{C_2} - {\,^p}{C_2} + 1 , (ii) nC3− pC3^n{C_3} - {\,^p}{C_3}
Option A ((i) nC2− pC2+1^n{C_2} - {\,^p}{C_2} + 1 , (ii) nC3− pC3^n{C_3} - {\,^p}{C_3}) is correct because it accurately calculates the number of straight lines and triangles that can be formed.
177 Medium
177 of 324
Application +4 | -1
Find the number of three digit numbers which can be formed using the digits 0,1,2,2,3,3,3.0,1,2,2,3, 3,3 .
View answer and explanation
Correct option: A
35
Option A (35) is correct because it's derived from considering all possible combinations of digits (0, 1, 2, 2, 3, 3, 3) to form three-digit numbers, accounting for repetitions and the restriction that the first digit cannot be zero.
178 Medium
178 of 324
Application +4 | -1
Out of 10 consonants and 4 vowels, words with 6 consonants and 3 vowels are formed. The number of such words is
View answer and explanation
Correct option: A
711(12!)\frac{7}{11}(12 !)
Option A (711(12!)\frac{7}{11}(12 !)) is correct because it calculates the number of ways to choose 6 consonants and 3 vowels and then arranges them to form words.
179 Easy
179 of 324
Application +4 | -1
The number of ways of selecting 4 cards of an ordinary pack of playing cards so that exactly 3 of them are of the same denomination is
View answer and explanation
Correct option: A
2496
Option A (2496) is correct because it accurately calculates the number of ways to select 4 cards with exactly 3 of the same denomination.
180 Easy
180 of 324
Application +4 | -1
The number of diagonals of a polygon of 15 sides is
View answer and explanation
Correct option: B
90
Option B (90) is correct because the number of diagonals in a polygon is calculated by subtracting the number of sides from the total number of lines that can be formed by joining any two vertices.
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