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JEE • MATHEMATICS • COMBINATIONS

Combinations practice · Easy

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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 121-150 of 236 questions

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121 Easy
121 of 236
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.
122 Easy
122 of 236
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.
123 Easy
123 of 236
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.
124 Easy
124 of 236
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.
125 Easy
125 of 236
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.
126 Easy
126 of 236
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.
127 Easy
127 of 236
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.
128 Easy
128 of 236
Application +4 | -1
If 20%20 \% of three subsets (i.e., subsets containing exactly three elements ) of the set A={a1,a2⋯an}A=\left\{a_1, a_2 \cdots a_n\right\} contain a1a_1, then the value of nn is
View answer and explanation
Correct option: A
15
Option A (15) is correct because it satisfies the condition that 20% of the three-element subsets of A contain a1.
129 Easy
129 of 236
Application +4 | -1
A lady gives a dinner party to 5 guests to be selected from nine friends. The number of ways of forming the party of 5 , given that two of the friends will not attend the party together, is
View answer and explanation
Correct option: C
91
Option C (91) is correct because it calculates the total number of ways to form the party minus the number of ways where the two friends both attend.
130 Easy
130 of 236
Application +4 | -1
Pyramid has a square base. Vertices are to be selected from the set {1,2,3,4,5}\{1,2,3,4,5\} such that no two adjacent vertices take same number, the number of ways is
View answer and explanation
Correct option: C
420
The correct answer is C (420) because it accounts for all possible arrangements of vertices with the given constraint of no two adjacent vertices having the same number.
131 Easy
131 of 236
Understanding +4 | -1
The number of 2×22 \times 2 matrices having elements 0 and 1, is
View answer and explanation
Correct option: B
16
Option B (16) is correct because each of the 4 elements in the matrix can be either 0 or 1, giving 2 options for each element, resulting in 24=162^4 = 16 possible matrices.
132 Easy
132 of 236
Understanding +4 | -1
Statement-1 The number of ways distributing 10 identical balls in 4 distinct boxes such that no box is empty, is 9C3{ }^9 C_3.
Statement-2 The number of ways of choosing any 3 places from 9 different places is 9C3{ }^9 C_3.
View answer and explanation
Correct option: A
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1 because distributing 10 identical balls into 4 distinct boxes with no box empty is equivalent to placing 3 dividers among 9 spaces between the balls, which is 9C3{ }^9 C_3.
133 Easy
133 of 236
Application +4 | -1
Three straight lines L1,  L2,  L3{L_1},\;{L_2},\;{L_3} are parallel and lie in the same plane. A total of m points are taken on L1,  n{L_1},\;n points on L2,  k{L_2},\;k points on L3{L_3}. The maximum number of triangles formed with vertices at these points are
View answer and explanation
Correct option: D
None of the above
The correct answer is None of the above because the correct expression for the maximum number of triangles is   m+n+kC3−  mC3−  nC3−  kC3{\;^{m + n + k}}{C_3} - {\;^m}{C_3} - {\;^n}{C_3} - {\;^k}{C_3}.
134 Easy
134 of 236
Application +4 | -1
If a,b,ca, b, c are odd positive integers, then number of integral solutions of a+b+c=13a+b+c=13, is
View answer and explanation
Correct option: B
21
Option B (21) is correct because the number of integral solutions is found by transforming the equation and using combinations.
135 Easy
135 of 236
Application +4 | -1
At an election there are five candidates and three members are to be elected, and a voter may vote for any number of candidates not greater than the number to be elected. The number of ways in which the person can vote is
View answer and explanation
Correct option: A
25
Option A (25) is correct because it sums the number of ways to vote for 1, 2, or 3 candidates out of 5.
136 Easy
136 of 236
Application +4 | -1
The number of diagonals that can be drawn by joining the vertices of an octagon is
View answer and explanation
Correct option: C
20
Option C (20) is correct because it subtracts the number of sides (8) from the total number of lines that can be formed by joining the vertices (28).
137 Easy
137 of 236
Analyzing +4 | -1
The number of positive integer solutions of a+b+c=60a+b+c=60, where aa is a factor of bb and cc, is
View answer and explanation
Correct option: C
144
The correct answer is C (144) because by considering all possible values of 'a' that are factors of 60 and then finding the number of solutions for 'b' and 'c', we arrive at a total of 144 solutions.
138 Easy
138 of 236
Application +4 | -1
The number of ways of dividing 52 cards amongst four players so that three players have 17 cards each and the fourth players just one card, is
View answer and explanation
Correct option: A
52!(17!)3\frac{{52!}}{{{{\left( {17!} \right)}^3}}}
Option A (52!(17!)3\frac{{52!}}{{{{\left( {17!} \right)}^3}}}) is correct because it represents the distribution of cards among the players with the specified number of cards each.
139 Easy
139 of 236
Application +4 | -1
Let Tn{T_n} denote the number of triangles which can be formed using the vertices of a regular polygon of nn sides. If Tn+1−Tn=21{T_{n + 1}} - {T_n} = 21, then nn equals
View answer and explanation
Correct option: B
7
Option B (7) is correct because solving the equation Tn+1−Tn=21{T_{n + 1}} - {T_n} = 21 with Tn=  nC3{T_n} = {\;^n}{C_3} leads to n = 7.
140 Easy
140 of 236
Application +4 | -1
The number of subsets of {1, 2, 3, …, 9}\left\{ {1,{\text{ }}2,{\text{ }}3,{\text{ }} \ldots ,{\text{ }}9} \right\} containing at least one odd number is
View answer and explanation
Correct option: C
496
Option C (496) is correct because it is calculated by subtracting the number of subsets containing only even numbers and the null set from the total number of subsets.
141 Easy
141 of 236
Understanding +4 | -1
In a shop, there are five types of ice-creams available. A child buys six ice-creams.
Statement-1 The number of different ways the child can buy the six ice-creams is 10C4{ }^{10} \mathrm{C}_4.
Statement-2 The number of different ways the child can buy six ice-creams is equal to the number of different ways to arranging 6A′s6 A^{\prime} s and 4B′s4 B^{\prime} s in a row.
View answer and explanation
Correct option: A
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1 because the problem is equivalent to distributing 6 identical items into 5 distinct boxes, which can be solved using stars and bars method, and is equivalent to arranging 6 A's and 4 B's.
142 Easy
142 of 236
Application +4 | -1
Out of 11 consecutive natural numbers if three numbers are selected at random (without repetition), then the probability that they are in AP with positive common difference, is
View answer and explanation
Correct option: C
533\frac{5}{{33}}
Option C (533\frac{5}{{33}}) is correct because it accurately calculates the probability of selecting three numbers in AP from 11 consecutive numbers by considering the possible combinations of even and odd numbers.
143 Easy
143 of 236
Application +4 | -1
Number of triangles that can be formed joining the angular points of decagon is
View answer and explanation
Correct option: D
120
Option D (120) is correct because the number of triangles formed by joining the vertices of a decagon is a combination problem.
144 Easy
144 of 236
Application +4 | -1
If   nCr{\;^n}{C_r} denotes the number of combinations of nn things takes rr at a time, then the expression   nCr+1+  nCr−1+2×  nCr{\;^n}{C_{r + 1}} + {\;^n}{C_{r - 1}} + 2 \times {\;^n}{C_r}, equals
View answer and explanation
Correct option: B
  n+2Cr+1{\;^{n + 2}}{C_{r + 1}}
Option B (  n+2Cr+1{\;^{n + 2}}{C_{r + 1}}) is correct because the expression simplifies to this using the combination identity   nCr+  nCr+1=  n+1Cr+1{\;^n}{C_r} + {\;^n}{C_{r + 1}} = {\;^{n + 1}}{C_{r + 1}}.
145 Easy
145 of 236
Application +4 | -1
On the occasion of Diwali festival each student of a class sends greeting cards to the others. If there are 20 students in the lass, then the total number of greeting cards exchanged by the students is
View answer and explanation
Correct option: B
2.  20C22.{\;^{20}}{C_2}
Option B (2.  20C22.{\;^{20}}{C_2}) is correct because each student sends a card to every other student, and each exchange involves two students.
146 Easy
146 of 236
Application +4 | -1
The number of divisors of the number of 38808 (excluding 1 and the number itself)is
View answer and explanation
Correct option: A
70
Option A (70) is correct because it calculates the total number of divisors, excluding 1 and the number itself, based on the prime factorization.
147 Easy
147 of 236
Application +4 | -1
If an=∑r=0n1nCra_n=\sum\limits_{{r=0}}^n \frac{1}{{ }^n C_r}, then ∑r=0nrnCr\sum\limits_{{r=0}}^n \frac{r}{{ }^n C_r} equals
View answer and explanation
Correct option: C
12nan\frac{1}{2} n a_n
Option C is correct because it is derived from the properties of binomial coefficients and summation.
148 Easy
148 of 236
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. Then the number of intersection points the lines have is equal to
View answer and explanation
Correct option: A
535
Option A (535) is correct because it accurately calculates the number of intersection points considering the lines passing through points A and B.
149 Easy
149 of 236
Application +4 | -1
If   n−1C6+  n−1C7>  nC6,{\;^{n - 1}}{C_6} + {\;^{n - 1}}{C_7} > {\;^n}{C_6}, then
View answer and explanation
Correct option: D
n>13n > 13
Option D (n>13n > 13) is correct because simplifying the inequality leads to the condition n>13n > 13.
150 Easy
150 of 236
Application +4 | -1
A person is permitted to select at least one and at most nn coins from a collection of 2n+12n + 1 (distinct) coins. If the total number of ways in which he can select coins is 255, then nn equals
View answer and explanation
Correct option: A
4
Option A (4) is correct because solving the binomial equation derived from the problem statement leads to n=4.
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