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JEE • MATHEMATICS • COMBINATIONS

Combinations practice · Easy

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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 181-210 of 236 questions

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181 Easy
181 of 236
Application +4 | -1
If a polygon has 44 diagonals, then the number of its sides are
View answer and explanation
Correct option: A
11
The correct answer is A (11) because it is the number of sides of a polygon with 44 diagonals, calculated using the formula for the number of diagonals.
182 Easy
182 of 236
Application +4 | -1
14C7+∑i=1317−iC6={ }^{14} {C}_7+\sum\limits_ {{i}=1}^3{ }^{17-{i}} {C}_6=
View answer and explanation
Correct option: B
17C7{ }^{17} {C}_7
Option B (17C7{ }^{17} {C}_7) is correct because it is the simplified form of the given expression after applying the identity nCr+nCr−1=n+1Cr{ }^{n} {C}_r + { }^{n} {C}_{r-1} = { }^{n+1} {C}_r multiple times.
183 Easy
183 of 236
Application +4 | -1
Six points in a plane be joined in all possible ways by indefinite straight lines and if no two of them be coincident or parallel, and no three pass through the same point (with the exception of the original 6 points). The number of distinct points or intersection is equal to
View answer and explanation
Correct option: C
51
Option C (51) is correct because it accounts for the intersections of lines formed by joining the points, subtracting overcounted intersections at the original points, and adding back the original points.
184 Easy
184 of 236
Analyzing +4 | -1
The value of 11!50!+13!48!+15!46!+….+149!2!+15!1!\frac{1}{1 ! 50 !}+\frac{1}{3 ! 48 !}+\frac{1}{5 ! 46 !}+\ldots .+\frac{1}{49 ! 2 !}+\frac{1}{5 ! 1 !} is :
View answer and explanation
Correct option: A
25051!\frac{2^{50}}{51 !}
Option A (25051!\frac{{2^{{50}}}}{{51 !}}) is correct because the given expression can be transformed into a binomial coefficient sum, which simplifies to this value.
185 Easy
185 of 236
Application +4 | -1
From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. Then, the number of such arrangements is
View answer and explanation
Correct option: A
atleast 1000
Option A (atleast 1000) is correct because the total number of arrangements is 1080, which is greater than 1000.
186 Easy
186 of 236
Application +4 | -1
There are 3 sections in a question paper and each section contains 5 questions. A candidate has to answer a total of 5 questions, choosing at least one question from each section. Then the number of ways, in which the candidate can choose the questions, is
View answer and explanation
Correct option: D
2250
Option D (2250) is correct because it correctly counts the number of ways to select 5 questions with at least one from each section, considering the possible distributions of questions across the three sections.
187 Easy
187 of 236
Application +4 | -1
The number of integers which lie between 1 and 106{10^6} and which have the sum of the digits equal to 12, is
View answer and explanation
Correct option: C
6062
Option C (6062) is correct because it is the coefficient of x12{x^{12}} in the expansion of a generating function representing the sum of digits.
188 Easy
188 of 236
Application +4 | -1
Consider three boxes, each containing 10 balls labelled 1, 2, …, 10. Suppose one ball is randomly drawn from each of the boxes. Denote by ni,{n_i}, the label of the ball drawn from ii th box, (i = 1, 2, 3). Then, the number of ways in which the balls can be chosen such that n1<n2<n3{n_1} < {n_2} < {n_3} is
View answer and explanation
Correct option: B
120
Option B (120) is correct because it's the number of ways to choose 3 distinct numbers from 1 to 10, which is a combination problem.
189 Easy
189 of 236
Application +4 | -1
Find nC21{ }^n C_{21}, if nC10=nC11{ }^n C_{10}={ }^n C_{11}.
View answer and explanation
Correct option: A
1
Option A (1) is correct because given nC10=nC11{ }^n C_{10}={ }^n C_{11}, we find n=21n=21, and 21C21=1{ }^{21} C_{21}=1.
190 Easy
190 of 236
Application +4 | -1
The value of   50C4+∑r=16  56−rC3{\;^{50}}{C_4} + \sum \limits_{r = 1}^6 {\;^{56 - r}}{C_3} is
View answer and explanation
Correct option: A
  56C4{\;^{56}}{C_4}
Option A (  56C4{\;^{56}}{C_4}) is correct because the summation simplifies to a single combination using the identity   nCr+  nCr+1=  n+1Cr+1{\;^n}{C_r} + {\;^n}{C_{r + 1}} = {\;^{n + 1}}{C_{r + 1}}.
191 Easy
191 of 236
Application +4 | -1
The results of 21 football matches (win, lose or draw) are to be predicted. The number of forecasts that contain exactly 18 correct results is
View answer and explanation
Correct option: B
21C1823{ }^{21} C_{18} 2^3
Option B (21C1823{ }^{21} C_{18} 2^3) is correct because it calculates the number of ways to choose 18 correct results and then the number of ways to have the remaining 3 results be incorrect.
192 Easy
192 of 236
Application +4 | -1
There are 10 points in a plane, out of these 6 are collinear. The number of triangles formed by joining these points is
View answer and explanation
Correct option: A
100
Option A (100) is correct because it is the result of subtracting the number of degenerate triangles (formed by collinear points) from the total possible number of triangles.
193 Easy
193 of 236
Application +4 | -1
The total number of different combinations of letters which can be made from the letters of the word MISSISSIPPI is
View answer and explanation
Correct option: C
149
Option C (149) is correct because it accounts for all possible combinations of the letters in the word MISSISSIPPI, including the option of selecting no letters.
194 Easy
194 of 236
Application +4 | -1
A person goes for an examination in which there are four papers with a maximum of m marks from each paper. The number of ways in which one can get 2m2m marks, is
View answer and explanation
Correct option: C
13(m+1)(2m2+4m+3)\frac{1}{3}\left( {m + 1} \right)\left( {2{m^2} + 4m + 3} \right)
Option C (13(m+1)(2m2+4m+3)\frac{1}{3}(m + 1)(2m^2 + 4m + 3)) is correct because it represents the number of ways to obtain 2m marks from four papers, each with a maximum of m marks.
195 Easy
195 of 236
Analyzing +4 | -1
If the number of ways of selecting nn cards out of unlimited number of cards bearing the number 0,9,30,9,3, so that they cannot be used to write the number 903 is 93 , then nn is equal to
View answer and explanation
Correct option: C
5
The correct answer is C (5) because by analyzing the conditions under which the cards cannot be used to write the number 903, we find that n = 5.
196 Easy
196 of 236
Application +4 | -1
A lady gives a dinner party for six guests. The number of ways in which they may be selected from among ten friends, if two of the friends will not attent the party together, is
View answer and explanation
Correct option: B
140
Option B (140) is correct because it considers the cases where neither, or only one, of the two friends are invited.
197 Easy
197 of 236
Application +4 | -1
In a cricket championship there are 36 matches. The number of terms, if each plays 1 match with other are
View answer and explanation
Correct option: A
9
Option A (9) is correct because solving the combination equation nC2 = 36 yields n = 9.
198 Easy
198 of 236
Application +4 | -1
At an election, a voter may vote for any number of candidates, not greater than number to be elected. There are 10 candidates and 4 are to be selected. If a voter votes for atleast one candidate, then number of ways in which he can vote, is
View answer and explanation
Correct option: C
385
Option C (385) is correct because it sums the number of ways to vote for 1, 2, 3, or 4 candidates out of 10.
199 Easy
199 of 236
Application +4 | -1
The number of ordered pairs of non-negative integers {x,y}\{x, y\} having sum 7596 without any carries is equal to
View answer and explanation
Correct option: B
3360
Option B (3360) is correct because it represents the number of ordered pairs that sum to 7596 without any carries.
200 Easy
200 of 236
Application +4 | -1
The number of diagonals of a polygon of 20 sides is
View answer and explanation
Correct option: D
170
Option D (170) is correct because it uses the formula for calculating the number of diagonals in a polygon, which is nC2−n^nC_2 - n.
201 Easy
201 of 236
Application +4 | -1
In a club election the number of contestants is one more than the number of maximum candidates for which a voter can vote. If the total number of ways in which a voter can vote be 126, then the number of contestants is
View answer and explanation
Correct option: D
7
Option D (7) is correct because the equation   nC1+…+  nCn−1=126{\;^n}{C_1} + \ldots + {\;^n}{C_{n - 1}} = 126 simplifies to 2n−2=1262^n - 2 = 126, which gives n=7n = 7.
202 Easy
202 of 236
Application +4 | -1
At an electron, a voter may vote for any number of candidates not greater than the number to be elected. There are 10 candidates and 4 are to be elected. If a voter votes for at least one candidate, then the number of ways in which he can vote, is
View answer and explanation
Correct option: B
385
Option B (385) is correct because it sums the number of ways to vote for 1, 2, 3, or 4 candidates out of 10.
203 Easy
203 of 236
Application +4 | -1
Let AA be the set of 4 digit number a1a2a3a4,  where  a1<a2<a3<a4,{a_1}{a_2}{a_3}{a_4},\;\mathrm{where}\;{a_1} < {a_2} < {a_3} < {a_4}, then n(A)n\left( A \right) is equal to
View answer and explanation
Correct option: B
126
Option B (126) is correct because it calculates the number of ways to choose 4 distinct digits from the set {1, 2, 3, 4, 5, 6, 7, 8, 9} where the order is strictly increasing.
204 Easy
204 of 236
Application +4 | -1
There are 10 points in a plane, out of these 6 are collinear. If NN is the number of triangles formed by joining these points, then
View answer and explanation
Correct option: B
N⩽100N \leqslant 100
Option B (N⩽100N \leqslant 100) is correct because the number of triangles formed is calculated by subtracting the number of collinear point combinations from the total possible combinations.
205 Easy
205 of 236
Application +4 | -1
Given 6 different toys of green colour, 5 different toys of blue co- lour and 4 different toys of red colour. Combination of toys that can be chosen taking at least one green and one blue toys are
View answer and explanation
Correct option: B
31248
Option B (31248) is correct because it accurately calculates the number of combinations with at least one green and one blue toy.
206 Easy
206 of 236
Application +4 | -1
There are 5 points P1,P2,P3,P4,P5{P}_1, {P}_2, {P}_3, {P}_4, {P}_5 on the side AB{AB}, excluding AA and BB, of a triangle ABCA B C. Similarly there are 6 points P6,P7,…,P11{P}_6, {P}_7, \ldots, {P}_{11} on the side BC{BC} and 7 points P12,P13,…,P18{P}_{12}, {P}_{13}, \ldots, {P}_{18} on the side CA{CA} of the triangle. The number of triangles, that can be formed using the points P1,P2,…,P18{P}_1, {P}_2, \ldots, {P}_{18} as vertices, is:
View answer and explanation
Correct option: B
751
Option B (751) is correct because it calculates the total number of possible triangles from 18 points, subtracting those formed only by points on the same side of the triangle.
207 Easy
207 of 236
Application +4 | -1
pp points are chosen on each of the three coplanar lines. The maximum number of triangles formed with vertices at these points is
View answer and explanation
Correct option: D
p2(4p−3){p^2}\left( {4p - 3} \right)
Option D (p2(4p−3){p^2}\left( {4p - 3} \right)) is correct because it is the correct expression for the maximum number of triangles formed by the points.
208 Easy
208 of 236
Application +4 | -1
The number of diagonals in a octagon will be
View answer and explanation
Correct option: B
20
Option A, 2n−1{2^{n - 1}}, is correct because the sum of even-indexed binomial coefficients equals 2n−1{2^{n - 1}}.
209 Easy
209 of 236
Analyzing +4 | -1
If   nCr=  nCr−1  and    nPr=  nPr+1{\;^n}{C_r} = {\;^n}{C_{r - 1}}{\text{\;and}}\;{\;^n}{P_r} = {\;^n}{P_{r + 1}}, then the value of nn is
View answer and explanation
Correct option: A
3
The correct answer is 3. The problem involves solving two equations derived from combinations and permutations, leading to n=3.
210 Easy
210 of 236
Application +4 | -1
The number of ways in which a team of eleven players can be selected from 22 players including 2 of them and excluding 4 of them is
View answer and explanation
Correct option: C
  16C9{\;^{16}}{C_9}
Option C (  16C9{\;^{16}}{C_9}) is correct because after including 2 and excluding 4 players, 9 more players must be selected from the remaining 16.
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