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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 . Then, is equal to
View answer and explanation
Correct option: C
Option C () is correct because the given summation simplifies to 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 .
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 and 's in a row.
Statement-1: The number of different ways the child can buy the six ice-creams is .
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 and '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 .
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 simplifies to , which gives .
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 be the set of 4 digit number then 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 is the number of triangles formed by joining these points, then
View answer and explanation
Correct option: B
Option B () 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 on the side , excluding and , of a triangle . Similarly there are 6 points on the side and 7 points on the side of the triangle. The number of triangles, that can be formed using the points 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
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
Option D () 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, , is correct because the sum of even-indexed binomial coefficients equals .
283
Medium
283 of 324
Analyzing
+4 | -1
The sum is equal to :
View answer and explanation
Correct option: B
!)
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 , then the value of 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
Option C () is correct because after including 2 and excluding 4 players, 9 more players must be selected from the remaining 16.
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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 then the value of 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 but never or 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 of 20 distinct elements is times the number of 5 -element subsets containing , then 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 is a convex quadrilateral with and 6 points, marked on sides and 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 denotes the number of combinations of things taken things at a time, then the expression is
View answer and explanation
Correct option: A
Option A () is correct because it simplifies the given expression using the combination identity .
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 and , then 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
is equal to
View answer and explanation
Correct option: B
Option B () is correct because the summation of binomial coefficients can be simplified using the identity .
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 .
297
Easy
297 of 324
Application
+4 | -1
A question paper is divided into two parts and 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 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
Option A (4) is correct because it counts the number of positive odd divisors of 216, which are the divisors of its odd factor.
Revision notes and key takeaways
Review COMBINATIONS in your textbook or class notes, then use this set to check your understanding.
Showing 271-300 of 324 questions
