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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 211-236 of 236 questions
211 Easy
211 of 236
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.
212 Easy
212 of 236
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.
213 Easy
213 of 236
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.
214 Easy
214 of 236
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.
215 Easy
215 of 236
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.
216 Easy
216 of 236
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}.
217 Easy
217 of 236
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.
218 Easy
218 of 236
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).
219 Easy
219 of 236
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.
220 Easy
220 of 236
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.
221 Easy
221 of 236
Understanding +4 | -1
If nn is even and nC0<nC1<nC2<....<nCr>nCr+1>nCr+2>....>nCn′^n{C_0}{ < ^n}{C_1}{ < ^n}{C_2} < ....{ < ^n}{C_r}{ > ^n}{C_{r + 1}}{ > ^n}{C_{r + 2}} > ....{ > ^n}{C_{n'}} then, r=r =
View answer and explanation
Correct option: A
n2\frac{n}{2}
Option A (n2\frac{n}{2}) is correct because the binomial coefficients nCr^{n}C_r are maximized when rr is closest to n2\frac{n}{2}, and since nn is even, r=n2r = \frac{n}{2} gives the maximum value.
222 Easy
222 of 236
Application +4 | -1
Sum of all the odd divisors of 720 is
View answer and explanation
Correct option: B
78
Option B (78) is correct because the sum of odd divisors is calculated by considering the powers of odd prime factors in the prime factorization of 720.
223 Easy
223 of 236
Application +4 | -1
The number of ways of distributing 8 identical balls in 3 distinct boxes, so that none of the boxes is empty, is
View answer and explanation
Correct option: B
21
Option B (21) is correct because it represents the number of ways to distribute 8 identical balls into 3 distinct boxes such that no box is empty, which is calculated using the formula   n−1Cr−1{\;^{n-1}}{C_{r-1}}.
224 Easy
224 of 236
Analyzing +4 | -1
A rectangle with sides 2m−12m - 1 and 2n−12n - 1 divided into squere of unit length. The number of rectangle which can be formed with sides of odd length is
View answer and explanation
Correct option: A
m2n2{m^2}{n^2}
The correct answer is A (m2n2{m^2}{n^2}) because it represents the total number of rectangles with odd length sides that can be formed within the given rectangle.
225 Easy
225 of 236
Application +4 | -1
In a conference of 8 persons, if each person shake hand with the other one only, then the total number of shake hands shall be
View answer and explanation
Correct option: D
28
Option D (28) is correct because the total number of handshakes is calculated using combinations, specifically 8C2^8C_2, which equals 28.
226 Easy
226 of 236
Application +4 | -1
If an=∑r=0m1  nCr,  then  ∑r=0nr  nCr{a_n} = \sum \limits_{r = 0}^m \frac{1}{{{\;^n}{C_r}}},\;{\text{then}}\;\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 (12\frac{1}{2}n{a_n}) is correct because by using the property of combinations and summing the series in reverse, we can find the required sum.
227 Easy
227 of 236
Application +4 | -1
The number of ways in which a committee can be formed of 5 members from 6 men and 4 women if the committee has at least one woman, is
View answer and explanation
Correct option: B
246
Option B (246) is correct because it represents the total number of ways to form the committee with at least one woman, calculated by considering all possible combinations of men and women.
228 Easy
228 of 236
Application +4 | -1
The value of ∑r=115r2(15Cr15Cr−1)\sum_{\mathrm{r}=1}^{15} \mathrm{r}^{2}\left(\frac{{ }^{15} \mathrm{C}_{\mathrm{r}}}{{ }^{15} \mathrm{C}_{\mathrm{r}-1}}\right) is equal to :
View answer and explanation
Correct option: D
680
Option D (680) is correct because after simplifying the summation expression and applying the formulas for the sum of first n natural numbers and the sum of squares of first n natural numbers, the result is 680.
229 Easy
229 of 236
Application +4 | -1
The sides AB,  BC,  CAAB,\;BC,\;CA of a triangle ABCABC have 3,4 and 5 interior points respectively on them. The total number of triangles that can be constructed by using these points as vertices is
View answer and explanation
Correct option: C
205
Option C (205) is correct because it calculates the total number of triangles formed by subtracting the number of collinear point combinations from the total possible combinations.
230 Easy
230 of 236
Understanding +4 | -1
There is a set of mm parallel lines intersecting a set of another nn parallel lines in a plane. The number of parallelograms formed, is
View answer and explanation
Correct option: B
  mC2.  nC2{\;^m}{C_2}.{\;^n}{C_2}
Option B (  mC2.  nC2{\;^m}{C_2}.{\;^n}{C_2}) is correct because to form a parallelogram, we need to choose 2 lines from the set of m parallel lines and 2 lines from the set of n parallel lines.
231 Easy
231 of 236
Application +4 | -1
The value of ∑r=110r⋅nCrnCr−1\sum_{r=1}^{10} r \cdot \frac{{ }^n C_r}{{ }^n C_{r-1}} is equal to
View answer and explanation
Correct option: A
5(2n−9)5(2 n-9)
Option A (5(2n−9)5(2n-9)) is correct because the summation can be simplified using the properties of combinations and then evaluated.
232 Easy
232 of 236
Application +4 | -1
If in a regular polygon the number of diagonals is 54 , then the number of sides of this polygon is
View answer and explanation
Correct option: A
12
Option A (12) is correct because solving the equation for the number of diagonals in a polygon yields n=12.
233 Easy
233 of 236
Application +4 | -1
If n+2C6n−2P2=11\frac{{ }^{n+2} C_{6}}{{ }^{n-2} P_{2}}=11, then n\mathrm{n} satisfies the equation :
View answer and explanation
Correct option: C
n2+3n−108=0n^{2}+3 n-108=0
Option C (n2+3n−108=0n^2 + 3n - 108 = 0) is correct because solving the given equation for n yields n=9, which satisfies the equation n2+3n−108=0n^2 + 3n - 108 = 0.
234 Easy
234 of 236
Application +4 | -1
The number of ways of selecting two numbers a and b,a∈{2,4,6,….100}b, \quad a \in\{2,4,6, \ldots .100\} and b∈(1,3,5,…,99)b \in(1,3,5, \ldots, 99) such that 2 is the remainder when a+ba+b is divided by 23 is
View answer and explanation
Correct option: A
108
Option A (108) is correct because by considering the possible values of λ\lambda in the equation a+b=23λ+2a+b = 23\lambda + 2, and counting the number of pairs (a, b) that satisfy the conditions, the total number of ways is 108.
235 Easy
235 of 236
Application +4 | -1
The number of ways in which a committee of 3 women and 4 men be chosen from 8 women and 7 men if Mr. XX refuses to serve on the committee if Mr\mathrm{Mr}. YY is a member of the committee is
View answer and explanation
Correct option: D
1400
Option D (1400) is correct because it considers the cases where Mr. Y is on the committee and Mr. X is not, and the case where Mr. Y is not on the committee.
236 Easy
236 of 236
Application +4 | -1
If a,b,ca, b, c and dd are odd natural numbers such that a+b+c+d=20a+b+c+d=20, the number of values of the ordered quadruplet (a,b,c,d)(a, b, c, d) is
View answer and explanation
Correct option: A
165
Option A (165) is correct because it calculates the number of ordered quadruplets of odd natural numbers that sum to 20.
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