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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 91-120 of 324 questions
91 Easy
91 of 324
Application +4 | -1
The number of triplets (x,y,z)(x, y, z) where x,y,zx, y, z are distinct non negative integers satisfying x+y+z=15x+y+z=15, is
View answer and explanation
Correct option: C
114
Option C (114) is correct because it represents the number of distinct non-negative integer triplets that sum to 15, accounting for the condition that x, y, and z must be distinct.
92 Medium
92 of 324
Application +4 | -1
A student is to answer 10out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is
View answer and explanation
Correct option: C
196
Option C (196) is correct because it calculates the sum of combinations where at least 4 questions are chosen from the first five.
93 Easy
93 of 324
Understanding +4 | -1
Consider the following statements:

Statement I: The number of diagonals of n−sidedn-\text{sided} polygon is nC2−n{ }^n C_2-n.

Statement II: A polygon has 44 diagonals. The number of its sides are 10.

Choose the correct option from the choices given below.
View answer and explanation
Correct option: A
Only I is true
Option A (6) is correct because there are 6 ways to choose 2 objects from 4 distinct objects without regard to order.
94 Medium
94 of 324
Application +4 | -1
In an steamer there are stalls for 12 animals and there are hours, cows and calves (not less then 12 each) ready to be shipped in how many ways can the ship load be made?
View answer and explanation
Correct option: B
312{3^{12}}
Option B (312{3^{12}}) is correct because each stall can be filled with any of the 3 types of animals, leading to 3 options for each of the 12 stalls.
95 Easy
95 of 324
Application +4 | -1
If 10Cx=10Cx+4{ }^{10} {C}_x={ }^{10} {C}_{x+4}, then the value of xx is
View answer and explanation
Correct option: A
3
Option A (3) is correct because using the property of combinations, either x = x+4 (which is not possible) or x + x + 4 = 10, leading to x = 3.
96 Easy
96 of 324
Application +4 | -1
These are nn distinct points on the circumference of a circle. The number of pentagons that can be formed with these points as vertices is equal to the number of possible triangles. Then, the value of nn is
View answer and explanation
Correct option: B
8
The correct answer is 8 because it is the value of n for which the number of pentagons equals the number of triangles formed from n points on a circle.
97 Easy
97 of 324
Application +4 | -1
In a chess tournament where the participants were to play one game with one another, two players fell ill having played 6 games each, without playing among themselves. If the total number of games is 117, then the number of participants at the beginning was
View answer and explanation
Correct option: A
15
Option A (15) is correct because solving the equation derived from the total number of games played gives the number of participants.
98 Easy
98 of 324
Analyzing +4 | -1
If Sn=∑r=0n1  nCr{S_n} = \sum \limits_{r = 0}^n \frac{1}{{{\;^n}{C_r}}} and tn=∑r=0nr  nCr{t_n} = \sum \limits_{r = 0}^n \frac{r}{{{\;^n}{C_r}}} then tnsn\frac{{{t_n}}}{{{s_n}}} is equal to
View answer and explanation
Correct option: A
n2\frac{n}{2}
Option A (n2\frac{n}{2}) is correct because it is derived by manipulating the summation formulas and using properties of binomial coefficients.
99 Easy
99 of 324
Analyzing +4 | -1
The coefficient of x18{x^{18}} in the product (1+x)(1−x)10(1+x+x2)9\left( {1 + x} \right){\left( {1 - x} \right)^{10}}{\left( {1 + x + {x^2}} \right)^9} is
View answer and explanation
Correct option: A
84
Option A (84) is correct because it is the coefficient of x18 after simplifying the product using the difference of cubes factorization.
100 Easy
100 of 324
Application +4 | -1
Let y be element of the set A={1,2,3,5,6,10,15A=\{1,2,3,5,6,10,15, 30 }\} and x1,x2,x3x_1, x_2, x_3 be integers such that x1x2x3=yx_1 x_2 x_3=y, the number of positive integral solutions of x1x2x3=yx_1 x_2 x_3=y, is
View answer and explanation
Correct option: B
64
Option B (64) is correct because the number of positive integral solutions is found by distributing the prime factors of y among the three variables.
101 Easy
101 of 324
Application +4 | -1
Let l1{l_1} and l2{l_2} be two lines intersecting at PP. If A1,  B1,  C1{A_1},\;{B_1},\;{C_1} are points on l1{l_1} and A2,  B2,  C2,  D2,  E2{A_2},\;{B_2},\;{C_2},\;{D_2},\;{E_2} are points on l2{l_2} and if none of these coincides with PP, then the number of triangles formed by these eight points, is
View answer and explanation
Correct option: D
45
Option D (45) is correct because it accurately counts the number of triangles formed by selecting three points from the given set, accounting for the collinearity of points on each line.
102 Easy
102 of 324
Application +4 | -1
A committee of at least three members is to be formed from a group of 6 boys and 6 girls such that it always has a boy and a girl. The number of ways to form such a committee is
View answer and explanation
Correct option: B
212−27−352^{12}-2^7-35
Option B (212−27−352^{12}-2^7-35) is correct because it accurately calculates the number of ways to form the committee by subtracting the invalid cases (no boys or no girls) from the total possible committees.
103 Medium
103 of 324
Analyzing +4 | -1
Let N\mathbf{N} be the total number of six digits number formed which consists the digit {0,1,2,3}\{0,1,2,3\} at least once and whose sum of digits is 10 , then the sum of even divisors of N\mathrm{N} is
View answer and explanation
Correct option: A
640
Option A (640) is correct because it represents the total number of six-digit numbers formed with the given conditions, calculated using combinations and permutations.
104 Medium
104 of 324
Application +4 | -1
The number of square matrices of order 5 with
entries from the set {0, 1}, such that the sum of all
the elements in each row is 1 and the sum of all the
elements in each column is also 1, is
View answer and explanation
Correct option: A
120
Option D (120) is correct because it represents the number of permutations of placing 5 ones in a 5x5 matrix such that each row and column has exactly one 1.
105 Easy
105 of 324
Application +4 | -1
The maximum number of points of intersection of 8 circles is
View answer and explanation
Correct option: D
56
Option D (56) is correct because each pair of circles can intersect at most at two points, and there are 8C2 ways to choose a pair of circles.
106 Easy
106 of 324
Application +4 | -1
The number of ways in which any four letters can be selected from the word ‘CORGOO’ is
View answer and explanation
Correct option: C
7
Option C (7) is correct because we consider all possible selections of four letters from the word 'CORGOO'.
107 Medium
107 of 324
Application +4 | -1
A committee of 11 members is to be formed from 8 males and 5 females. If mm is the number of ways the committee is formed with at least 6 males and nn is the number of ways the committee is formed with atleast 3 females, then
View answer and explanation
Correct option: C
m=n=78m = n = 78
Option C (m=n=78m = n = 78) is correct because both the number of ways with at least 6 males and at least 3 females are equal to 78.
108 Easy
108 of 324
Application +4 | -1
The total number of integral solutions of xyz=24x y z=24 is
View answer and explanation
Correct option: D
120
Option D (120) is correct because it accounts for both positive and negative integral solutions of the equation xyz=24.
109 Hard
109 of 324
Application +4 | -1
There are two bags each of which contains ' nn ' balls. A man has to select an equal number of balls from both the bags. The number of ways in which a man can choose at least one ball from each bag is
View answer and explanation
Correct option: D
2nCn−1{ }^{2 n} C_n-1
Option D (2nCn−1{}^{2 n} C_n-1) is correct because it represents the number of ways to select an equal number of balls from two bags, each containing 'n' balls, with at least one ball selected from each bag.
110 Hard
110 of 324
Application +4 | -1
The number of solutions of x1+x2+x3=51(x1,x2,x3x_{1}+x_{2}+x_{3}=51\left(x_{1}, x_{2}, x_{3}\right. being odd natural numbers) is
View answer and explanation
Correct option: B
325
Option B (325) is correct because it represents the number of solutions to the equation with odd natural numbers, transformed into a simpler equation.
111 Easy
111 of 324
Application +4 | -1
There are five different green dyes, four different blue dyes and three different red dyes. The total number of combinations of dyes that can be chosen taking at least one green and one blue dye is
View answer and explanation
Correct option: C
3720
The correct answer is C (3720) because it represents the total number of dye combinations possible with at least one green and one blue dye.
112 Easy
112 of 324
Application +4 | -1
Let Tn{T_n} be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If Tn+1−Tn=10,{T_{n + 1}} - {T_n} = 10, then the value of nn is
View answer and explanation
Correct option: B
5
Option B (5) is correct because solving the equation Tn+1−Tn=10{T_{n + 1}} - {T_n} = 10 using the formula for the number of triangles formed by vertices of a polygon yields n = 5.
113 Easy
113 of 324
Application +4 | -1
If 2nC3:nC3=10:1{ }^{2 n} C_{3}:{ }^{n} C_{3}=10: 1, then the ratio (n2+3n):(n2−\left(n^{2}+3 n\right):\left(n^{2}-\right. 3n+4)3 n+4) is
View answer and explanation
Correct option: D
2:12: 1
Option D (2:12: 1) is correct because solving the given ratio of combinations leads to n=8, and substituting this value into the expression yields 2.
114 Easy
114 of 324
Application +4 | -1
The total number of ways in which a person can put 8 different rings in the fingers of his right hand is equal to
View answer and explanation
Correct option: B
11P8{ }^{11} P_{8}
Option B (11P8{ }^{11} P_{8}) is correct because it considers the number of ways to distribute the rings among the fingers and then arrange them.
115 Medium
115 of 324
Application +4 | -1
A five digit number divisible by 6 is to be formed by using the digits 0,1,2,3,40,1,2,3,4 and 88 without repetition. The total number of ways in which this can be done is
View answer and explanation
Correct option: B
150
Option B (150) is correct because it accounts for the divisibility rule of 6 (divisible by both 2 and 3) and the arrangements of the digits 0, 1, 2, 3, 4, and 8 without repetition.
116 Easy
116 of 324
Application +4 | -1
If 1  4Cn=1  5Cn+1  6Cn\frac{1}{{{\;^4}{C_n}}} = \frac{1}{{{\;^5}{C_n}}} + \frac{1}{{{\;^6}{C_n}}}, then nn is equal to
View answer and explanation
Correct option: B
2
Option B (2) is correct because solving the given equation involving combinations results in n = 2.
117 Easy
117 of 324
Application +4 | -1
There are 18 points in a plane such that no three of them are in the same line except five points which are collinear. The number of triangles formed by these points is
View answer and explanation
Correct option: B
806
Option B (806) is correct because we subtract the number of degenerate triangles formed by the collinear points from the total possible triangles.
118 Easy
118 of 324
Analyzing +4 | -1
There are nn-points (n>2)(n > 2) in each of two parallel lines. Every point on one line is joined to every point on the other line by a line segment drawn within the lines. The number of points (between the lines) in which these segments intersect is
View answer and explanation
Correct option: C
  nC2×  nC2{\;^n}{C_2} \times {\;^n}{C_2}
Option C (  nC2×  nC2{\;^n}{C_2} \times {\;^n}{C_2}) is correct because each intersection point is uniquely determined by choosing two points from each of the two parallel lines.
119 Easy
119 of 324
Application +4 | -1
The number of diagonals that can be drawn in a polygon of 15 sides, is
View answer and explanation
Correct option: C
90
The correct answer is 90 because it is calculated by subtracting the number of sides from the total number of lines that can be formed from the vertices.
120 Easy
120 of 324
Application +4 | -1
If nCr:nCr+1:nCr+2=1:2:3{ }^{{n}} C_r:{ }^{{n}} C_{r+1}:{ }^n C_{r+2}=1: 2: 3, then rr is equal to
View answer and explanation
Correct option: B
4
Option A (7/11(12!)) is correct because it is the result of multiplying the number of ways to choose the consonants and vowels by the number of ways to arrange them.
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