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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 271-300 of 324 questions

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271 Medium
271 of 324
Analyzing +4 | -1
If ∑r=025{50Cr⋅50−rC25−r}=K(50C25)\sum\limits_{r = 0}^{25} {\left\{ {{}^{50}{C_r} \cdot {}^{50 - r}{C_{25 - r}}} \right\} = K\left( {{}^{50}{C_{25}}} \right)}. Then, KK is equal to
View answer and explanation
Correct option: C
225{2^{25}}
Option C (225{2^{25}}) is correct because the given summation simplifies to 225{2^{25}} after applying combinatorial identities.
272 Medium
272 of 324
Application +4 | -1
In a shop there are five types of ice-creams (each type is available unlimited). A child buys six ice-creams.
Statement-1: The number of different ways the child can buy the six ice-creams is 10C5{ }^{10} C_5.
Statement-2: The number of different ways the child can buy the six ice-creams is equal to the number of different ways of arranging 6A′s6 {A}^{\prime} \mathrm{s} and 4B4 {B} 's in a row.
View answer and explanation
Correct option: C
Statement - 1 is false, Statement - 2 is true.
Statement - 1 is false, Statement - 2 is true. Statement 1 incorrectly calculates the number of ways to choose the ice creams, while Statement 2 correctly relates it to arranging A's and B's.
273 Easy
273 of 324
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.
274 Easy
274 of 324
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.
275 Easy
275 of 324
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.
276 Easy
276 of 324
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.
277 Easy
277 of 324
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.
278 Easy
278 of 324
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.
279 Easy
279 of 324
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.
280 Easy
280 of 324
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.
281 Medium
281 of 324
Application +4 | -1
A dice is rolled 4 times, the numbers appearing on the dice in different throws are listed, such that the largest number appearing in the list is not 4 , in how many ways this can be done
View answer and explanation
Correct option: D
1121
Option D (1121) is correct because it's calculated by subtracting the cases where the largest number is 4 from the total possible outcomes.
282 Easy
282 of 324
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}}.
283 Medium
283 of 324
Analyzing +4 | -1
The sum ∑r=110(r2+1)×(r!)\sum_{r=1}^{10}\left(r^{2}+1\right) \times(r !) is equal to :
View answer and explanation
Correct option: B
10×(1110 \times(11 !)
The correct answer is B (10 ×(11 !)) because the sum telescopes to 10 × 11! after simplification and summation.
284 Easy
284 of 324
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.
285 Easy
285 of 324
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.
286 Easy
286 of 324
Application +4 | -1
There are four balls of different colours and four boxes of colours same as those of the balls. The number of ways in which the balls, one in each box, could be placed such that a ball does not go to box of its own colour, is
View answer and explanation
Correct option: C
9
Option C (9) is correct because it represents the number of derangements of 4 items, which can be calculated using the formula or by listing out the possibilities.
287 Easy
287 of 324
Application +4 | -1
If nPr=nPr+1     and nCr=nCr−1,{}^n{P_r} = {}^n{P_{r + 1\,\,\,}}\,\,\mathrm{and}\,\mathrm{ }{}^n{C_r} = {}^n{C_{r - 1}}, then the value of rr is equal to
View answer and explanation
Correct option: C
2
Option C (2) is correct because solving the permutation and combination equations yields r = 2.
288 Easy
288 of 324
Analyzing +4 | -1
The number of ways can a team of six horses be selected out of a stud of 16 , so that there shall always be three out of A,B,C,D,E,FA, B, C, D, E, F but never AD,BEA D, B E or CFC F together, is
View answer and explanation
Correct option: C
960
Option C (960) is correct because it calculates the number of ways to select the team of horses with the given constraints, considering the combinations of the special horses A, B, C, D, E, F and the remaining horses.
289 Medium
289 of 324
Application +4 | -1
If the number of 5-element subsets of the set A={a1,a2,…,a20}A=\left\{a_{1}, a_{2}, \ldots, a_{20}\right\} of 20 distinct elements is kk times the number of 5 -element subsets containing a4a_{4}, then kk is
View answer and explanation
Correct option: C
4
Option C (4) is correct because the number of 5-element subsets of A is 4 times the number of 5-element subsets containing a4.
290 Easy
290 of 324
Application +4 | -1
If ABCDA B C D is a convex quadrilateral with 3,4,53,4,5 and 6 points, marked on sides AB,BC,CDA B, B C, C D and DAD A respectively, then the number of triangles with vertices on different sides, is
View answer and explanation
Correct option: D
342
Option D (342) is correct because it sums the number of triangles formed by choosing one vertex from each of the three sides in all possible combinations.
291 Medium
291 of 324
Application +4 | -1
If nCr{ }^n C_r denotes the number of combinations of nn things taken rr things at a time, then the expression nCr+1+nCr−1+2nCr{ }^n C_{r+1}+{ }^n C_{r-1}+2{ }^n C_r is
View answer and explanation
Correct option: A
n+2Cr+1{ }^{n+2} C_{r+1}
Option A (n+2Cr+1{ }^{n+2} C_{r+1}) is correct because it simplifies the given expression using the combination identity nCr+nCr−1=(n+1)Cr{ }^n C_r+{ }^n C_{r-1}={ }^{(n+1)} C_r.
292 Medium
292 of 324
Application +4 | -1
The number of all the odd divisors of 3600 is
View answer and explanation
Correct option: D
9
Option D (9) is correct because the number of odd divisors is found by considering only the odd prime factors in the prime factorization.
293 Medium
293 of 324
Application +4 | -1
In an examination a candidate has to pass in each of the papers to be successful. It the total number of ways to fail is 63 , how many papers are there in the examination?
View answer and explanation
Correct option: A
6
Option A (6) is correct because the total number of ways to fail or pass is 2^n, and subtracting the one way to pass (all papers passed) gives 63, leading to n=6.
294 Easy
294 of 324
Application +4 | -1
If P(n  ,r)=1680P\left( {n\;,r} \right) = 1680 and C(n,  r)=70C\left( {n,\;r} \right) = 70, then 69n+r!69n + r! is equal to
View answer and explanation
Correct option: B
576
Option B (576) is correct because solving the permutation and combination equations leads to n=8 and r=4, and substituting these values into 69n + r! gives 576.
295 Easy
295 of 324
Application +4 | -1
∑k=0651−kC3\sum\limits_{{k=0}}^{6}{ }^{51-k} C_{3} is equal to
View answer and explanation
Correct option: B
52C4−45C4{ }^{52} C_{4}-{ }^{45} C_{4}
Option B (52C4−45C4{ }^{52} C_{4}-{ }^{45} C_{4}) is correct because the summation of binomial coefficients can be simplified using the identity nCr+nCr+1=n+1Cr+1{ }^{n}C_{r} + { }^{n}C_{r+1} = { }^{n+1}C_{r+1}.
296 Easy
296 of 324
Application +4 | -1
Ramesh has 6 friends. In how many ways can he invite one or more of them at a dinner?
View answer and explanation
Correct option: C
63
Option C (63) is correct because it represents the total number of non-empty subsets of a set with 6 elements, which is 26−1=632^6 - 1 = 63.
297 Easy
297 of 324
Application +4 | -1
A question paper is divided into two parts AA and BB and each part contain 5 questions. The number of ways in which a candidate can answer 6 questions selecting at least two questions from each part is
View answer and explanation
Correct option: C
200
Option C (200) is correct because it sums the number of ways to select 6 questions with at least 2 from each (2 from A, 4 from B), (3 from A, 3 from B), and (4 from A, 2 from B).
298 Easy
298 of 324
Application +4 | -1
Total number of ways of selecting five letters from letters of the word INDEPENDENT is
View answer and explanation
Correct option: A
72
Option A (72) is correct because it accounts for all possible combinations of selecting five letters from the word INDEPENDENT, considering repeated letters.
299 Easy
299 of 324
Application +4 | -1
The total number of proper divisors of 38808 is
View answer and explanation
Correct option: B
70
Option B (70) is correct because it calculates the total number of divisors and subtracts 1 for the number itself to find the proper divisors.
300 Medium
300 of 324
Application +4 | -1
There are nn straight lines in a plane, no two of which are parallel and no three pass through the same point. Their points of intersection are joined. Then, the number of fresh lines thus obtained is
View answer and explanation
Correct option: C
n(n−1)(n−2)(n−3)8\frac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)}}{8}
Option A (4) is correct because it counts the number of positive odd divisors of 216, which are the divisors of its odd factor.
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