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JEE • MATHEMATICS • PERMUTATION AND COMBINATION

Permutation and combination 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 31-60 of 485 questions
31 Easy
31 of 485
Application +4 | -1
The sum of all 3- digited numbers that are divisible by either 3 or 4 but not by 48 is
View answer and explanation
Correct option: D
2,37,1142,37,114
Option D (2,37,114) is correct because it is the sum of all 3-digit numbers divisible by 3 or 4 but not by 48.
32 Easy
32 of 485
Application +4 | -1
The figure 4, 5, 6, 7, 8 are written in every possible order. The number of numbers greater than 56000 is
View answer and explanation
Correct option: C
90
Option C (90) is correct because it is the sum of numbers greater than 56000 when arranging the digits 4, 5, 6, 7, 8.
33 Easy
33 of 485
Application +4 | -1
Find the remainder when 1!+2!+3!+…+n1 !+2 !+3 !+\ldots+n! is divided by 15, if n≥5n \geq 5.
View answer and explanation
Correct option: C
3
Option C (3) is correct because for n ≥ 5, all terms from 5! onwards are divisible by 15, so the remainder is determined by the sum of the first four terms.
34 Easy
34 of 485
Application +4 | -1
No.of ways of seating 20 persons at two round tables. Seating 10 to each table is
View answer and explanation
Correct option: A
20!100\frac{20 !}{100}
Option A (20!100\frac{{20 !}}{{100}}) is correct because it represents the number of ways to divide 20 people into two groups of 10 and arrange them around two round tables.
35 Easy
35 of 485
Application +4 | -1
Number of ways in which the letters of the word ARRANGE can be arranged such that both R do not come together is
View answer and explanation
Correct option: B
900
Option B (900) is correct because it calculates the total arrangements minus the arrangements where both R's are together.
36 Easy
36 of 485
Application +4 | -1
The number of words that can be formed out of the letters of the words ‘ARTICE’ so that the vowels occupy even places, is
View answer and explanation
Correct option: D
144
Option D (144) is correct because the number of ways to arrange the letters with vowels in even places is calculated by multiplying the arrangements of vowels and consonants.
37 Easy
37 of 485
Application +4 | -1
The number of arrangements of the letters of the word "INDEPENDENCE" in which all the vowels always occur together is
View answer and explanation
Correct option: A
16800
Option A (16800) is correct because it accounts for arranging the consonants and the block of vowels, and then arranging the vowels within the block.
38 Easy
38 of 485
Application +4 | -1
The number of arrangements which can be made using all the letters of the word LAUGH, if the vowels are adjacent, is
View answer and explanation
Correct option: C
48
Option C (48) is correct because it accounts for arranging the consonants and the vowel group, and then arranging the vowels within the group.
39 Easy
39 of 485
Application +4 | -1
The number of ways in which the letters of 'ARRANGE' be arranged so that "Two R's come together but Two A's do not come together is
View answer and explanation
Correct option: C
240
Option C (240) is correct because it subtracts the arrangements where both R's and A's are together from the arrangements where only R's are together.
40 Easy
40 of 485
Application +4 | -1
The total number of three-digit numbers, the sum of whose digits is even, is equal to
View answer and explanation
Correct option: A
450
Option A (450) is correct because it correctly calculates the total number of three-digit numbers where the sum of the digits is even.
41 Easy
41 of 485
Application +4 | -1
If a man and his wife enter in a bus, in which five seats are vacant, then the number of different ways in which they can be seated, is
View answer and explanation
Correct option: C
20
Option C (20) is correct because the man can choose from 5 seats, and then the wife can choose from the remaining 4 seats, giving 5*4 = 20 possibilities.
42 Easy
42 of 485
Application +4 | -1
The number of ways in which 5 boys and 3 girls can be seated in a row so that each girl is between two boys is
View answer and explanation
Correct option: A
2880
Option A (2880) is correct because it considers the arrangement of boys and the placement of girls between them.
43 Easy
43 of 485
Application +4 | -1
The number of different words that can be formed from the letters of the word ‘PENCIL’ so that no two vowels are together, is
View answer and explanation
Correct option: D
480
Option D (480) is correct because the consonants can be arranged in 4! ways, and the vowels can be arranged in the remaining spaces in   5P2{{\;^5}{P_2}} ways.
44 Easy
44 of 485
Application +4 | -1
There are 10 persons among whom two are brother. The total number of ways in which these persons can be seated around a round table so that exactly one person sit between the brothers, is equal to
View answer and explanation
Correct option: B
(2!)(8!)(2!) (8!)
Option B ((2!)(8!)(2!) (8!)) is correct because the two brothers can be arranged in 2! ways, and the remaining 8 people (including the block of brothers) can be arranged around the table in 8! ways.
45 Easy
45 of 485
Application +4 | -1
If there are 12 persons in a party, and if each of them shakes hands with each other, then number of handshakes happen in the party is
View answer and explanation
Correct option: A
66
Option A (66) is correct because it represents the number of ways to choose 2 people out of 12 for a handshake, calculated using combinations.
46 Easy
46 of 485
Application +4 | -1
The number of words that can be formed from the letters of the word DAUGHTER so that the vowels always come together, is
View answer and explanation
Correct option: A
4320
Option A (4320) is correct because it calculates the arrangements of the word DAUGHTER treating the vowels as a single unit.
47 Easy
47 of 485
Application +4 | -1
The sum of all the four-digit numbers that can be formed using all the digits 2,1,2,32,1,2,3 is equal to____
View answer and explanation
Correct option: A
26664
Option A (26664) is correct because it is the sum of all possible four-digit numbers formed using the digits 2, 1, 2, 3, calculated by considering the place values and the frequency of each digit.
48 Easy
48 of 485
Application +4 | -1
The number of ways of painting the faces of a cube with six different colours is
View answer and explanation
Correct option: A
1
Option A (1) is correct because when painting the faces of a cube with different colors, the orientation doesn't matter, so there's only one unique way.
49 Easy
49 of 485
Analyzing +4 | -1
The number of ways in which the letters of the word 'SUCCESSFUL' be arranged such that no two S's will come together is
View answer and explanation
Correct option: D
7!2!2!8P33!\frac{7 !}{2 ! 2 !} \frac{{ }^8 P_3}{3 !}
The correct answer is D (7!2!2!8P33!\frac{7 !}{2 ! 2 !} \frac{{ }^8 P_3}{3 !}) because it accounts for arranging the non-S letters and then placing the S's in the gaps.
50 Easy
50 of 485
Application +4 | -1
X{X} has four friends A,B,C,D{A}, {B}, {C}, {D} he wants to invite one or more of them at dinner. If the invited persons(s) & X\& \,X are sitting around a circular table then the arrangement in which C & DC\,\&\,D (if both are invited) are sitting together is
View answer and explanation
Correct option: B
48
Option B (848^4) is correct because each of the four prizes can be given to any of the 8 students, independently. This results in 8×8×8×8=848 \times 8 \times 8 \times 8 = 8^4 possible ways.
51 Easy
51 of 485
Application +4 | -1
The total number of functions ' ff ' from the set {1,2,3}\{1,2,3\} into the set {1,2,3,4,5}\{1,2,3,4,5\} such that f(i)⩽f(j),∀i<jf(i) \leqslant f(j),\forall i < j, is
View answer and explanation
Correct option: A
35
Option A (35) is correct because it counts the number of non-decreasing functions from a set of size 3 to a set of size 5.
52 Easy
52 of 485
Application +4 | -1
Let aa be a factor of 120, then the number of positive integral solutions of x1x2x3=ax_1 x_2 x_3=a is
View answer and explanation
Correct option: B
320
Option B (320) is correct because it accurately calculates the number of positive integral solutions by considering the distribution of prime factors of 120 among the three variables, given that 'a' is a factor of 120.
53 Easy
53 of 485
Application +4 | -1
The sum 1(1!)+2(2  !)+3(3!)+...+n(n!)1\left( {1!} \right) + 2\left( {2\;!} \right) + 3\left( {3!} \right) + ... + n\left( {n!} \right) equal to
View answer and explanation
Correct option: C
(n+1)!−1!\left( {n + 1} \right)! - 1!
Option C ((n+1)!−1!\left( {n + 1} \right)! - 1!) is correct because the sum can be rewritten as a telescoping series, which simplifies to the given expression.
54 Easy
54 of 485
Application +4 | -1
In a group of boys, two boys are brothers and in this group, 6 more boys are there. In how many ways, they can sit if the brothers are not to sit alongwith each other :
View answer and explanation
Correct option: D
None of these
The correct answer is None of these because the calculated number of ways is 30240, which is not among the options.
55 Easy
55 of 485
Application +4 | -1
A library has a copies of one book, bb copies of each of two books, cc copies of each of three books and single copies of d books. The total number of ways in which these book can be distributed, is
View answer and explanation
Correct option: B
(a+2b+3c+d)!a!(b!)2(c!)3\frac{{\left( {a + 2b + 3c + d} \right)!}}{{a!{{\left( {b!} \right)}^2}{{\left( {c!} \right)}^3}}}
Option B is correct because it accounts for the repetitions of identical books within each group (two books with b copies each and three books with c copies each) when calculating the total number of arrangements.
56 Easy
56 of 485
Application +4 | -1
Let Sn=∑r=1nr!S_n=\sum\limits_{{r=1}}^n r!. If Tn=Sn−7[Sn7]T_n=S_n-7\left[\frac{S_n}{7}\right], where [x][x] denotes greatest integer ≤x\leq x, then T100T_{100} is equal to
View answer and explanation
Correct option: B
5
The correct answer is 5 because for n >= 6, n! is divisible by 7, so the sum modulo 7 stabilizes, and we can calculate the sum of the first few terms modulo 7.
57 Easy
57 of 485
Application +4 | -1
If k\mathrm{k} be the number of 3 digit natural numbers, having sum of their digits at least 10 , then find k−35100\frac{k-35}{100}.
View answer and explanation
Correct option: A
7
Option A (7) is correct because the number of 3-digit numbers with a digit sum of at least 10 is 735, and (735-35)/100 = 7.
58 Easy
58 of 485
Application +4 | -1
If   2n+1Pn−1:  2n−1Pn:3:5{\;^{2n + 1}}{P_{n - 1}}:{\;^{2n - 1}}{P_n}:3:5, then the value of nn is equal to
View answer and explanation
Correct option: A
4
The value of n is 4 (Option A). This is found by setting up the ratio of permutations, simplifying the equation, and solving the resulting quadratic equation for n.
59 Easy
59 of 485
Application +4 | -1
If 15Cr+1:15C3r=3:11{ }^{15} C_{{r}+1}:{ }^{15} C_{3 {r}}=3: 11, then value of rr is:
View answer and explanation
Correct option: B
3
Option B (3) is correct because substituting r=3 into the ratio 15Cr+1:15C3r=3:11{}^{15} C_{{r}+1}:{}^{15} C_{3 {r}}=3: 11 satisfies the equation after simplifying the combinations.
60 Easy
60 of 485
Application +4 | -1
There are 10 points in a plane, of which no 3 points are collinear and 4 points are concyclic. The number of different circles that can be drawn through atleast 3 of these points, is
View answer and explanation
Correct option: C
117
Option C (117) is correct because it calculates the total number of circles possible minus the circles formed by the 4 concyclic points, plus 1 (for the circle formed by the 4 concyclic points).
Revision notes and key takeaways

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