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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 241-270 of 324 questions

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241 Easy
241 of 324
Application +4 | -1
In a football championship, there were played 153 matches. Every team played one match with each other. The number of teams participating in the championship is
View answer and explanation
Correct option: B
18
Option B (18) is correct because the number of teams is found by solving the combination equation   nC2=153{{\;^n}{C_2}} = 153, which results in n=18n = 18.
242 Easy
242 of 324
Application +4 | -1
The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets atleast one ball, is
View answer and explanation
Correct option: B
150
Option B (150) is correct because it represents the total number of ways to distribute the balls such that each person gets at least one ball, calculated using inclusion-exclusion principle.
243 Easy
243 of 324
Application +4 | -1
The number of five digit numbers that contain 7 exactly once is
View answer and explanation
Correct option: A
(41)(93)\left( {41} \right)\left( {{9^3}} \right)
Option A ((41)(93)\left( {41} \right)\left( {{9^3}} \right)) is correct because it accounts for the placement of the digit 7 in different positions within the five-digit number.
244 Easy
244 of 324
Analyzing +4 | -1
AA is a set containing nn elements. A subset PP of AA is chosen. The set AA is reconstructed by replacing the elements of PP. A subset QQ of AA is again chosen, the number of ways of choosing so that (P∪Q)(P \cup Q) is a proper subset of AA, is
View answer and explanation
Correct option: D
4n−3n4^n-3^n
The correct answer is D (4n−3n4^n - 3^n) because by considering all possible combinations of elements in subsets P and Q and subtracting the cases where their union is not a proper subset of A, we find the number of ways is 4n−3n4^n - 3^n.
245 Medium
245 of 324
Analyzing +4 | -1
If (1+x)n=∑r=0narxr,br=1+arar−1(1+x)^n=\sum\limits_{{r=0}}^n a_r x^r, b_r=1+\frac{a_r}{a_{r-1}} and ∏r=1nbr=(101)100100!\prod\limits_{{r=1}}^n b_r=\frac{(101)^{100}}{100 !}, then value of nn is
View answer and explanation
Correct option: B
100
Option B (100) is correct because the value of n is 100, derived from the given binomial expansion and product.
246 Easy
246 of 324
Application +4 | -1
On the sides AB,BC,CA\mathrm{AB}, \mathrm{BC}, \mathrm{CA} of a △ABC,3,4,5\triangle \mathrm{ABC}, 3,4,5 distinct points (excluding vertices A, B, C) are respectively chosen. The number of triangles that can be constructed using these chosen points as vertices are :
View answer and explanation
Correct option: B
205
Option B (205) is correct because the total number of triangles that can be formed is calculated by subtracting the number of collinear points from the total possible combinations.
247 Easy
247 of 324
Application +4 | -1
The expression   nCr+4.  nCr−1+6.nCr−2+4.  nCr−3+  nCr−4{\;^n}{C_r} + 4.{\;^n}{C_{r - 1}} + {6.^n}{C_{r - 2}} + 4.{\;^n}{C_{r - 3}} + {\;^n}{C_{r - 4}} equals
View answer and explanation
Correct option: A
  n+4Cr{\;^{n + 4}}{C_r}
Option A (  n+4Cr{{\;^{n + 4}}{C_r}}) is correct because the given expression simplifies to this using the identity nCr + nCr-1 = n+1Cr repeatedly.
248 Easy
248 of 324
Application +4 | -1
All possible two factors products are formed from numbers 1, 2, 3, 4,…,200. The number of factors out of the total obtained which are multiples of 5, is
View answer and explanation
Correct option: B
7180
Option B (7180) is correct because it subtracts the number of two-factor products not divisible by 5 from the total number of two-factor products.
249 Easy
249 of 324
Application +4 | -1
A parallelogram is cut by two sets of mm lines parallel to its sides. The number of parallelograms thus formed is
View answer and explanation
Correct option: C
(  m+2C2)2{\left( {{\;^{m + 2}}{C_2}} \right)^2}
Option C ((  m+2C2)2{\left( {{\;^{m + 2}}{C_2}} \right)^2}) is correct because each parallelogram is formed by choosing two lines from each of the two sets of (m+2) parallel lines.
250 Easy
250 of 324
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.
251 Easy
251 of 324
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.
252 Easy
252 of 324
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.
253 Easy
253 of 324
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.
254 Easy
254 of 324
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.
255 Easy
255 of 324
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.
256 Easy
256 of 324
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.
257 Medium
257 of 324
Application +4 | -1
A shopkeeper sells three varieties of perfumes and he has a large number of bottles of the same size of each variety in his stock. There are 5 places in a row in his showcase. The number of different ways of displaying the three varieties of perfumes in the show case is
View answer and explanation
Correct option: C
150
Option C (150) is correct because it accounts for arranging the three perfume varieties in the showcase, considering both cases: three of one type and two distinct, and two of one type, two of another, and one of the remaining.
258 Easy
258 of 324
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.
259 Easy
259 of 324
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.
260 Easy
260 of 324
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}}.
261 Medium
261 of 324
Application +4 | -1
From an unlimited number of red, white, blue and green balls, a selection of 18 balls is to be made so that there are at least two of each colour. If the number of selection is kk, then kk is equal to:
View answer and explanation
Correct option: B
286
Option B (286) is correct because it represents the number of ways to select 18 balls with at least two of each color, calculated using combinations.
262 Easy
262 of 324
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.
263 Easy
263 of 324
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.
264 Easy
264 of 324
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.
265 Easy
265 of 324
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.
266 Easy
266 of 324
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.
267 Easy
267 of 324
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.
268 Easy
268 of 324
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.
269 Easy
269 of 324
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.
270 Easy
270 of 324
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.
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