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JEE • MATHEMATICS • COMBINATIONS

Combinations practice · Medium

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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 31-60 of 83 questions
31 Medium
31 of 83
Application +4 | -1
A five digit number divisible by 6 is to be formed by using the digits 0,1,2,3,40,1,2,3,4 and 88 without repetition. The total number of ways in which this can be done is
View answer and explanation
Correct option: B
150
Option B (150) is correct because it accounts for the divisibility rule of 6 (divisible by both 2 and 3) and the arrangements of the digits 0, 1, 2, 3, 4, and 8 without repetition.
32 Medium
32 of 83
Application +4 | -1
The value of ∑k=0651−kC3\sum_{k=0}^{6}{ }^{51-k} C_{3} is
View answer and explanation
Correct option: B
52C4−45C4{ }^{52} C_{4}-{ }^{45} C_{4}
Option B ({ }^{52} C_{4}-{ }^{45} C_{4}) is correct because it represents the sum of combinations using the Hockey Stick Identity.
33 Medium
33 of 83
Application +4 | -1
Seven people leave their bags outside temple and while returning after worshiping the deity, picked one bag each at random. In how many ways at least one and at most three of them get their correct bags?
View answer and explanation
Correct option: A
7C3⋅9+7C5⋅44+7C1⋅265{ }^7 C_3 \cdot 9+{ }^7 C_5 \cdot 44+{ }^7 C_1 \cdot 265
The correct answer is A (7C3⋅9+7C5⋅44+7C1⋅265{}^7 C_3 \cdot 9+{}^7 C_5 \cdot 44+{}^7 C_1 \cdot 265) because it correctly calculates the number of ways for at least one and at most three people to get their correct bags using derangements.
34 Medium
34 of 83
Application +4 | -1
If the total number of mm elements subsets of the set A={a1,a2,a3,…an}A = \left\{ {{a_1},{a_2},{a_3}, \ldots {a_n}} \right\} is λ\lambda times the number of 3 elements subsets containing a4{a_4}, then nn is
View answer and explanation
Correct option: B
mλm\lambda
Option D (mλm\lambda) is correct because it is derived from the given relationship between the total number of m-element subsets and the number of 3-element subsets containing a specific element.
35 Medium
35 of 83
Application +4 | -1
If nn is the number of ways five different employees can sit into four indistinguishable offices where any office may have any number of persons including zero, then nn is equal to:
View answer and explanation
Correct option: D
51
Option D (51) is correct because it is the total number of ways to distribute the employees into the indistinguishable offices, considering all possible distributions.
36 Medium
36 of 83
Application +4 | -1
A man XX has 7 friends, 4 of them are ladies and 3 are men. His wife YY also has 7 friends, 3 of them are ladies and 4 are men. Assume and YY have no common friends. Then, the total number of ways in which XX and YY together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of XX and YY are in this party, is
View answer and explanation
Correct option: A
485
Option A (485) is correct because it is the sum of all possible combinations of selecting 3 ladies and 3 men from X and Y's friends.
37 Medium
37 of 83
Application +4 | -1
If n∈Nn \in N, then value of the expression n−2Cr+2.n−2Cr−1+n−2Cr−2={ }^{n-2} C_r+2 .^{n-2} C_{r-1}+{ }^{n-2} C_{r-2}=
View answer and explanation
Correct option: D
nCr{ }^n C_r
Option D (nCr{ }^n C_r) is correct because the given expression simplifies to nCr{ }^n C_r using the properties of combinations.
38 Medium
38 of 83
Application +4 | -1
Number of ways in which 6 different toys can be distributed among two brothers in 1:21: 2, is
View answer and explanation
Correct option: A
30
Option A (30) is correct because one brother gets 2 toys and the other gets 4. We choose 2 toys for the first brother and then decide which brother gets the smaller share.
39 Medium
39 of 83
Application +4 | -1
Let AA be the set of 4 -digit numbers a1a2a3a4a_{1} a_{2} a_{3} a_{4} where a1>a2>a3>a_{1} > a_{2} > a_{3} > a4a_{4}. Then n(A)n(A) is equal to
View answer and explanation
Correct option: C
210
Option C (210) is correct because any selection of four digits from the ten digits gives one such number.
40 Medium
40 of 83
Application +4 | -1
If a,  b,  c∈N,a,\;b,\;c \in N, The anumber of points having position vector ai^+bj^+ck^a\mathbf{\hat i} + b\mathbf{\hat j} + c\mathbf{\hat k} such that 6≤a+b+c≤106 \le a + b + c \le 10, is
View answer and explanation
Correct option: A
110
Option A (110) is correct because it represents the total number of possible combinations of a, b, and c that satisfy the given condition using combinations with repetition.
41 Medium
41 of 83
Application +4 | -1
A gentlemen hosts a party of (m+n)(m+n) guests and places ' mm ' at one round table and the remaining ' nn ' at the other round table. Number of ways the guests can be arranged is
View answer and explanation
Correct option: A
(m+n)!m⋅n\frac{(m+n) !}{m \cdot n}
Option A ((m+n)!m⋅n\frac{{(m+n) !}}{{m \cdot n}}) is correct because it represents the number of ways to arrange (m+n) guests into two groups of m and n and then arrange them around two round tables.
42 Medium
42 of 83
Application +4 | -1
If xk≥0,k=1,2,3,4,5{x}_{{k}} \geq 0, {k}=1,2,3,4,5 and x1+x2+x3+x4+x5=20{x}_1+{x}_2+{x}_3+{x}_4+{x}_5=20 and x1+x2+x3=5{x}_1+{x}_2+{x}_3=5 then the number of integral solution of (x1,x2,x3,x4,x5)\left(x_1, x_2, x_3, x_4, x_5\right)
View answer and explanation
Correct option: D
336
The correct answer is D (336) because it is the product of the number of solutions for the two independent equations derived from the given conditions.
43 Medium
43 of 83
Analyzing +4 | -1
If ∑r=1n[log⁡2r]=1898\sum\limits_{{r=1}}^n\left[\log _2 r\right]=1898 (where [.] denotes greatest integer function), then number of zeroes at the end of n\mathrm{n} is/are
View answer and explanation
Correct option: A
74
Option A (74) is correct because the sum of the greatest integer function of log base 2 of r from 1 to n equals 1898, leading to n=300, and the number of zeroes at the end of 300! is 74.
44 Medium
44 of 83
Analyzing +4 | -1
(nn!)! is divisible by
View answer and explanation
Correct option: D
(n!)(n−1)!(n !)^{(n-1) !}
Option D is correct because (n!)! is divisible by (n!)^((n-1)!) as shown by breaking down the factorial product into smaller groups.
45 Medium
45 of 83
Understanding +4 | -1
The number of ways in which 12 balls can be divided between two friends, one receiving 8 an the other 4, is
View answer and explanation
Correct option: B
12  !2  !8  !4  !\frac{{12\;!2\;!}}{{8\;!4\;!}}
Option A (12  !2  !8  !4  !\frac{{12\;!2\;!}}{{8\;!4\;!}}) is correct because we need to consider the two possible distributions: one friend gets 8 and the other gets 4, or vice versa.
46 Medium
46 of 83
Understanding +4 | -1
Statement 1: If AA and BB be two sets having pp and qq elements respectively, where q>pq>p. Then the total number of functions from set AA to set BB is qpq^{p}.
[Online May 12, 2012]
Statement 2: The total number of selections of pp different objects out of qq objects is qCp{ }^{q} \mathrm{C}_{p}.
View answer and explanation
Correct option: D
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation of Statement 1 .
Option D. Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation of Statement 1 is correct because statement 1 correctly states the number of functions from set A to set B, and statement 2 correctly states the number of selections of p objects from q objects.
47 Medium
47 of 83
Application +4 | -1
Given 5 different green dyes, 4 different blue dyes and 3 different red dyes. The number of combinations of dyes that can be chosen by taking at least one green and one blue dye is
View answer and explanation
Correct option: C
3720
Option C (3720) is correct because it calculates the total number of combinations by considering at least one green and one blue dye, along with any combination of red dyes.
48 Medium
48 of 83
Application +4 | -1
Find the number of three digit numbers which can be formed using the digits 0,1,2,2,3,3,3.0,1,2,2,3, 3,3 .
View answer and explanation
Correct option: A
35
Option A (35) is correct because it's derived from considering all possible combinations of digits (0, 1, 2, 2, 3, 3, 3) to form three-digit numbers, accounting for repetitions and the restriction that the first digit cannot be zero.
49 Medium
49 of 83
Application +4 | -1
Out of 10 consonants and 4 vowels, words with 6 consonants and 3 vowels are formed. The number of such words is
View answer and explanation
Correct option: A
711(12!)\frac{7}{11}(12 !)
Option A (711(12!)\frac{7}{11}(12 !)) is correct because it calculates the number of ways to choose 6 consonants and 3 vowels and then arranges them to form words.
50 Medium
50 of 83
Application +4 | -1
Let there are 6 shirts of different colours and 6 trousers of same colours as that of shirts. If the number of ways in which these can be put on by 5 men such that no men wear the shirt and the trouser of the same colour is (k)6!,(k) 6!, then find the value of kk.
View answer and explanation
Correct option: A
309
Option A (309) is correct because it's derived from using derangements and combinations to calculate the number of ways 5 men can wear shirts and trousers of different colors, then solving for k.
51 Medium
51 of 83
Application +4 | -1
The number of triangles which can be formed by using the vertices of a regular polygon of (n+3)\left( {n + 3} \right) sides is 220. Then nn is equal to
View answer and explanation
Correct option: B
9
Option B (9) is correct because solving the combination equation yields n=9.
52 Medium
52 of 83
Application +4 | -1
A student is allowed to select at most nn books from a collection of (2n+1)\left( {2n + 1} \right) books. If the total number of ways in which he can select one book is 63, then the value of nn is equal to
View answer and explanation
Correct option: B
3
Option B (3) is correct because solving the binomial equation derived from the problem statement results in n = 3.
53 Medium
53 of 83
Application +4 | -1
The number of ways in which 13 gold coins can be distributed among three persons such that each one gets at least 2 gold coins is
View answer and explanation
Correct option: A
36
Option A (36) is correct because it represents the number of ways to distribute 13 gold coins among three persons such that each one gets at least 2 gold coins.
54 Medium
54 of 83
Application +4 | -1
Team A consists of 7 boys and n girls and team B has 4 boys and 6 girls. If a total of 52 single matches can be arranged between these two teams when a boy plays against a boy and a girl plays against a girl, then n=
View answer and explanation
Correct option: D
4
Option D (4) is correct because the total number of matches is the sum of boy-boy matches and girl-girl matches, which can be set up as an equation and solved for n.
55 Medium
55 of 83
Application +4 | -1
There are two parallel lines one having 10 point and other having 5 points. The number of triangle formed with vertice as these points
View answer and explanation
Correct option: C
325
The provided answer 20C3−19C3{ }^{20} {C}_3-{ }^{19} {C}_3 is incorrect. The correct approach involves stars and bars. The correct answer should be the sum of solutions to x+y+z = k, where k ranges from 20 to 50. This requires a different calculation than the options provided.
56 Medium
56 of 83
Application +4 | -1
The number of ways in which four letters can be selected from the word degree is
View answer and explanation
Correct option: A
7
Option A (7) is correct because it accounts for the different combinations of letters that can be selected from the word 'degree'.
57 Medium
57 of 83
Application +4 | -1
The number of positive integers satisfying the inequality n+1Cn−2−n+1Cn−1≤50n + {1_{{C_{n - 2}}}} - n + {1_{{C_{n - 1}}}} \le 50 is
View answer and explanation
Correct option: C
7
Option C (7) is correct because it represents the number of positive integers that satisfy the given inequality.
58 Medium
58 of 83
Application +4 | -1
In how many ways can Rs 16 be divided into 4 persons when none of them get less than Rs 3?
View answer and explanation
Correct option: B
35
Option B (35) is correct because it represents the number of ways to distribute the remaining amount after giving each person the minimum required amount.
59 Medium
59 of 83
Analyzing +4 | -1
The packs of 52 cards are shuffled together. The number of ways in which a man can be dealt 26 cards so that he does not get two cards of the same suit and same denomination, is
View answer and explanation
Correct option: A
  52C26.226{\;^{52}}{C_{26}}{.2^{26}}
The correct answer is A (  52C26.226{\;^{52}}{C_{26}}{.2^{26}}) because it accounts for choosing 26 cards out of 52 and then considering that each card can come from either of the two packs.
60 Medium
60 of 83
Application +4 | -1
The set S={1,  2,  3,  …,  12}S = \left\{ {1,\;2,\;3,\; \ldots ,\;12} \right\} is to be partitioned into three sets A,  B,  CA,\;B,\;C of equal size
Thus, A∪B∪C=SA \cup B \cup C = S
A∪B=B∪C=A∪C=∅A \cup B = B \cup C = A \cup C = \emptyset
The number of ways to partition SS is
View answer and explanation
Correct option: C
12!/(4!)312!/{\left( {4!} \right)^3}
Option C (12!/(4!)312!/{\left( {4!} \right)^3}) is correct because it represents the number of ways to divide the set into three equal subsets without considering the order of the subsets.
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