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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 301-324 of 324 questions

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301 Medium
301 of 324
Application +4 | -1
Determine nn if 2nC3:nC2=12:1{ }^{2 n} {C}_3:{ }^n {C}_2=12: 1
View answer and explanation
Correct option: A
5
Option A (5) is correct because the ratio 2nC3:nC2=12:1{ }^{2 n} {C}_3:{ }^n {C}_2=12: 1 simplifies to 2n−1=92n-1 = 9, which gives n=5n=5.
302 Easy
302 of 324
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.
303 Easy
303 of 324
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.
304 Medium
304 of 324
Understanding +4 | -1
Consider the following statements.

I. nCr+2 nCr−1+nCr−2=n+2Cr{ }^n C_r+2 \, { }^n C_{r-1}+{ }^n C_{r-2}={ }^{n+2} C_r

II. nCp=nCq⇒p=q{ }^n C_p={ }^n C_q \Rightarrow p=q or p+q=np+q=n

Choose the correct option.
View answer and explanation
Correct option: C
Both are true
Option C is correct because both statements are true. Statement I is a combinatorial identity, and Statement II describes the condition for equality of combinations.
305 Medium
305 of 324
Application +4 | -1
The number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remaining 21 are distinct, is
View answer and explanation
Correct option: C
220{2^{20}}
Option C (220{2^{20}}) is correct because it is the total number of ways to choose 10 objects with the given constraints.
306 Easy
306 of 324
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}}.
307 Easy
307 of 324
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.
308 Easy
308 of 324
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.
309 Medium
309 of 324
Application +4 | -1
If (30C1)2+2(30C2)2+3(30C3)2+…+30(30C30)2\left({ }^{30} C_{1}\right)^{2}+2\left({ }^{30} C_{2}\right)^{2}+3\left({ }^{30} C_{3}\right)^{2}+\ldots+30\left({ }^{30} C_{30}\right)^{2} =α60!(30!)2=\frac{\alpha 60 !}{(30 !)^{2}} then α\alpha is equal to :
View answer and explanation
Correct option: C
15
Option C (15) is correct because the given expression can be simplified using combinatorial identities to find the value of alpha.
310 Easy
310 of 324
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.
311 Easy
311 of 324
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.
312 Easy
312 of 324
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.
313 Easy
313 of 324
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.
314 Easy
314 of 324
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.
315 Medium
315 of 324
Application +4 | -1
mCr+1+∑k=mnkCr{ }^m C_{r+1}+\sum\limits_{{k=m}}^n{ }^k C_r is equal to
View answer and explanation
Correct option: B
n+1Cr+1{ }^{n+1} C_{r+1}
Option B (n+1Cr+1{ }^{n+1} C_{r+1}) is correct because the given expression simplifies to n+1Cr+1{ }^{n+1} C_{r+1} using the properties of combinations.
316 Medium
316 of 324
Understanding +4 | -1
Statement-1: The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is empty is 9C3{ }^{9} C_{3}.
Statement-2: The number of ways of choosing any 3 places from 9 different places is 9C3{ }^{9} C_{3}.
View answer and explanation
Correct option: A
Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
Option A is correct. Statement-1 is true as the number of ways to distribute identical balls is calculated using combinations, and Statement-2 is also true, but it doesn't explain Statement-1.
317 Medium
317 of 324
Application +4 | -1
If mm denotes the number of 5 digit number if each successive digits are in their descending order magnitude and n{n} is the corresponding figure when the digit are in their ascending order of magnitude
View answer and explanation
Correct option: B
9C5{ }^9 {C}_5
Option A (325) is correct because it correctly calculates the number of triangles that can be formed by choosing vertices from the two parallel lines.
318 Easy
318 of 324
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.
319 Easy
319 of 324
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.
320 Easy
320 of 324
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.
321 Medium
321 of 324
Application +4 | -1
The number of different ways in which five alike dashes and eight alike dots can be arranged using only seven of these 'dashes' and 'dots' is equal to
View answer and explanation
Correct option: C
120
Option C (120) is correct because it represents the sum of combinations of arranging dashes and dots, considering the constraints on their numbers.
322 Easy
322 of 324
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.
323 Easy
323 of 324
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.
324 Easy
324 of 324
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.
Revision notes and key takeaways

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