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JEE • MATHEMATICS • COMBINATIONS

Combinations practice · Mixed difficulty

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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 181-210 of 324 questions

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181 Easy
181 of 324
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.
182 Easy
182 of 324
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.
183 Easy
183 of 324
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.
184 Medium
184 of 324
Application +4 | -1
Let there are 6 shirts of different colours and 6 trousers of same colours as that of shirts. If the number of ways in which these can be put on by 5 men such that no men wear the shirt and the trouser of the same colour is (k)6!,(k) 6!, then find the value of kk.
View answer and explanation
Correct option: A
309
Option A (309) is correct because it's derived from using derangements and combinations to calculate the number of ways 5 men can wear shirts and trousers of different colors, then solving for k.
185 Easy
185 of 324
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.
186 Easy
186 of 324
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.
187 Easy
187 of 324
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}.
188 Easy
188 of 324
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.
189 Medium
189 of 324
Application +4 | -1
The number of triangles which can be formed by using the vertices of a regular polygon of (n+3)\left( {n + 3} \right) sides is 220. Then nn is equal to
View answer and explanation
Correct option: B
9
Option B (9) is correct because solving the combination equation yields n=9.
190 Easy
190 of 324
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.
191 Easy
191 of 324
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).
192 Easy
192 of 324
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.
193 Hard
193 of 324
Analyzing +4 | -1
If ∑i=120(20Ci−120Ci+20Ci−1)3=k21\sum_{i=1}^{20}\left(\frac{{ }^{20} \mathrm{C}_{i-1}}{{ }^{20} \mathrm{C}_{i}+{ }^{20} \mathrm{C}_{i-1}}\right)^{3}=\frac{k}{21}, then kk equals:
View answer and explanation
Correct option: D
100
The correct answer is D (100) because after simplifying the expression inside the summation and using the formula for the sum of cubes of first n natural numbers, we get k = 100.
194 Easy
194 of 324
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.
195 Easy
195 of 324
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.
196 Easy
196 of 324
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.
197 Medium
197 of 324
Application +4 | -1
A student is allowed to select at most nn books from a collection of (2n+1)\left( {2n + 1} \right) books. If the total number of ways in which he can select one book is 63, then the value of nn is equal to
View answer and explanation
Correct option: B
3
Option B (3) is correct because solving the binomial equation derived from the problem statement results in n = 3.
198 Easy
198 of 324
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.
199 Easy
199 of 324
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.
200 Easy
200 of 324
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.
201 Medium
201 of 324
Application +4 | -1
The number of ways in which 13 gold coins can be distributed among three persons such that each one gets at least 2 gold coins is
View answer and explanation
Correct option: A
36
Option A (36) is correct because it represents the number of ways to distribute 13 gold coins among three persons such that each one gets at least 2 gold coins.
202 Easy
202 of 324
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}}.
203 Easy
203 of 324
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.
204 Medium
204 of 324
Application +4 | -1
Team A consists of 7 boys and n girls and team B has 4 boys and 6 girls. If a total of 52 single matches can be arranged between these two teams when a boy plays against a boy and a girl plays against a girl, then n=
View answer and explanation
Correct option: D
4
Option D (4) is correct because the total number of matches is the sum of boy-boy matches and girl-girl matches, which can be set up as an equation and solved for n.
205 Medium
205 of 324
Application +4 | -1
There are two parallel lines one having 10 point and other having 5 points. The number of triangle formed with vertice as these points
View answer and explanation
Correct option: C
325
The provided answer 20C3−19C3{ }^{20} {C}_3-{ }^{19} {C}_3 is incorrect. The correct approach involves stars and bars. The correct answer should be the sum of solutions to x+y+z = k, where k ranges from 20 to 50. This requires a different calculation than the options provided.
206 Easy
206 of 324
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.
207 Medium
207 of 324
Application +4 | -1
The number of ways in which four letters can be selected from the word degree is
View answer and explanation
Correct option: A
7
Option A (7) is correct because it accounts for the different combinations of letters that can be selected from the word 'degree'.
208 Easy
208 of 324
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.
209 Easy
209 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. 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.
210 Easy
210 of 324
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.
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