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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 241-270 of 485 questions

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241 Easy
241 of 485
Application +4 | -1
How many numbers with no more than three digits can be formed using only the digits 1 through 7 with no digit used more than once in a given number?
View answer and explanation
Correct option: A
259
Option A (259) is correct because it sums the number of 1-digit, 2-digit, and 3-digit numbers that can be formed using digits 1 through 7 without repetition.
242 Easy
242 of 485
Application +4 | -1
The number of integral solutions of x1+x2+x3=0x_1+x_2+x_3=0, with xi≥−5x_i \geq-5, is
View answer and explanation
Correct option: C
17C2{ }^{17} C_2
Option C (17C2{ }^{17} C_2) is correct because it represents the number of integral solutions of the given equation with the specified constraints, after transforming the variables to non-negative integers.
243 Easy
243 of 485
Analyzing +4 | -1
The number of different pairs of words (□□□□,□□□)(\square \square \square \square, \square \square \square) that can be made with the letters of the word 'STATICS' is
View answer and explanation
Correct option: B
1260
Option B (10 or 11) is correct because the number of groups is maximized when the group size is closest to half the total number of students.
244 Easy
244 of 485
Application +4 | -1
In how many ways can 5 boys and 5 girls sit in a circle so that no two boys sit together?
View answer and explanation
Correct option: B
4!×5!4! \times 5!
Option B (4!×5!4! \times 5!) is correct because it represents the arrangement of girls in a circle and boys in the spaces between them.
245 Easy
245 of 485
Application +4 | -1
The largest integer nn such that 33 ! is divisible by 2n2^n is
View answer and explanation
Correct option: B
31
Option B (31) is correct because it represents the largest integer n such that 33! is divisible by 2^n, calculated using Legendre's formula.
246 Easy
246 of 485
Application +4 | -1
If m1m_1 and m2m_2 satisfy the relation m+5Pm+1=112(m−1)(m+3Pm),{ }^{m+5} P_{m+1}=\frac{11}{2}(m-1)\left({ }^{m+3} P_m\right), then m1+m2m_1+m_2 is equal to
View answer and explanation
Correct option: C
13
Option C (13) is correct because solving the permutation equation leads to a quadratic equation with roots 6 and 7, and their sum is 13.
247 Easy
247 of 485
Application +4 | -1
If eight persons are to address a meeting, then the number of ways in which a specified speaker is to speak before another specified speaker is
View answer and explanation
Correct option: B
20160
Option B (20160) is correct because in half of the total arrangements, the specified speaker will speak before the other specified speaker.
248 Easy
248 of 485
Application +4 | -1
A bag contains 2 apples, 3 oranges and 4 bananas. The number of ways in which 3 fruits can be selected, if atleast one banana is always in the combination (assume fruit of same species to be alike) is
View answer and explanation
Correct option: A
6
Option A (6) is correct because it represents the number of ways to select 3 fruits with at least one banana, considering the fruits of the same species to be alike.
249 Easy
249 of 485
Application +4 | -1
Total number of non-negative integral solutions of x1+x2+x3=10x_1+x_2+x_3=10 is equal to
View answer and explanation
Correct option: C
12C2{ }^{12} C_2
Option C (12C2{ }^{12} C_2) is correct because it represents the number of non-negative integral solutions of the given equation using stars and bars method.
250 Easy
250 of 485
Application +4 | -1
Words with or without meaning are to be formed using all the letters of the word EXAMINATION. The probability that the letter M appears at the fourth position in any such word is
View answer and explanation
Correct option: B
1/111/11
Option B (1/11) is correct because it represents the ratio of favorable arrangements (M at the fourth position) to the total possible arrangements of the letters in EXAMINATION.
251 Easy
251 of 485
Application +4 | -1
The number of positive integral solutions of the equation x1x2x3x4x5=420x_1 x_2 x_3 x_4 x_5=420 is
View answer and explanation
Correct option: A
1875
Option A (1875) is correct because it calculates the number of ways to distribute the prime factors of 420 among the five variables.
252 Easy
252 of 485
Application +4 | -1
A number of 4 different digits is formed by using the digits 1,2,3,4,5,6,71,2,3,4,5,6,7 such that number formed is multiple of 25 , is 12x+y12 x+y, then find the value of value of x+yx+y.
View answer and explanation
Correct option: A
7
Option A (7) is correct because the number of 4-digit multiples of 25 formed using the digits 1 to 7 is 40, and solving 12x + y = 40 gives x + y = 7.
253 Easy
253 of 485
Application +4 | -1
There are nn concurrent lines and another line parallel to one of them. The number of different triangles that will be formed by the (n+1)(n+1) lines, is
View answer and explanation
Correct option: B
(n−1)(n−2)2\frac{(n-1)(n-2)}{2}
Option B ((n−1)(n−2)2\frac{(n-1)(n-2)}{2}) is correct because it represents the number of ways to choose 2 lines from the n−1n-1 concurrent lines to form a triangle with the additional line.
254 Easy
254 of 485
Application +4 | -1
If the letters of the word "nagpur" are written in all possible ways and then are arranged as in a dictionary, then the rank of the word "nagpur" is:
View answer and explanation
Correct option: B
242
Option B (242) is correct. The rank of the word "nagpur" is determined by arranging the letters alphabetically and counting the words that come before it in the dictionary.
255 Easy
255 of 485
Application +4 | -1
If an=∑r=0n1nCra_n=\sum\limits_{{r=0}}^n \frac{1}{{ }^n C_r}, then ∑r=0nrnCr\sum\limits_{{r=0}}^n \frac{r}{{ }^n C_r} equals:
View answer and explanation
Correct option: A
n2an\frac{n}{2} a_n
Option A (n2an\frac{n}{2} a_n) is correct because by using the property nCr=nCn−r{}^nC_r = {}^nC_{n-r}, we can rewrite the sum and solve for the desired expression.
256 Easy
256 of 485
Analyzing +4 | -1
Number of points having position vector ai^+bj^+ck^a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}, where a,b,c∈{1,2,3,4,5}a, b, c \in\{1,2,3,4,5\} such that 2a+3b+5c2^a+3^b+5^c is divisible by 4 , is
View answer and explanation
Correct option: A
70
The correct answer is A (70) because by analyzing the divisibility rule of 4 for the expression 2a+3b+5c2^a + 3^b + 5^c, we find that there are 70 possible combinations of a,b,a, b, and cc that satisfy the condition.
257 Easy
257 of 485
Application +4 | -1
The number of even divisors of 10800 is
View answer and explanation
Correct option: D
48
Option D (48) is correct because it accurately calculates the number of even divisors by considering at least one factor of 2 in the divisors.
258 Easy
258 of 485
Application +4 | -1
If aa denotes the number of permutations of x+2x+2 things taken all at a time, bb the number of permutations of xx things taken 11 at a time and cc the number of permutations of x−11x-11 things taken all at a time such that a=182bca=182 b c, the value of xx is
View answer and explanation
Correct option: B
12
Option D (18) is correct because it solves for x using the given permutation equation.
259 Easy
259 of 485
Application +4 | -1
A set containing nn elements. A subset PP of AA is chosen. The set AA is reconstructed by replacing the element of PP. A subset QQ of AA is again chosen the number of ways of choosing PP and QQ, so that P∩Q=ϕP \cap Q=\phi, is
View answer and explanation
Correct option: D
3n3^n
Option D (3n3^n) is correct because for each element in the set A, there are three possibilities: it belongs to P, it belongs to Q, or it belongs to neither, given that P and Q are disjoint.
260 Easy
260 of 485
Application +4 | -1
The sides AB,BC,CAAB, BC, CA of a ΔABC\Delta ABC have 3, 4, 5 interior points, respectively on them. Total number of triangles that can be formed using these points as vertices, is equal to
View answer and explanation
Correct option: D
205
Option D (205) is correct because the total number of triangles is calculated by subtracting the number of collinear points from the total number of ways to choose 3 points from the 12 interior points: 12C3−3C3−4C3−5C3=205{}^{12}C_3 - {}^3C_3 - {}^4C_3 - {}^5C_3 = 205.
261 Easy
261 of 485
Analyzing +4 | -1
In word 'ENGINEERING', E's and N's come alternately
View answer and explanation
Correct option: C
12×11P5\frac{1}{2} \times{ }^{11} P_5
Option C (12×11P5\frac{1}{2} \times{ }^{11} P_5) is correct because it arranges the G's, I's, and R, then places the E's and N's alternately.
262 Easy
262 of 485
Application +4 | -1
If 18!+19!=x10!\frac{1}{8 !}+\frac{1}{9 !}=\frac{x}{10 !}, then xx is equal to
View answer and explanation
Correct option: A
100
Option A (100) is correct because solving the equation by finding a common denominator and simplifying leads to x = 100.
263 Easy
263 of 485
Application +4 | -1
If nPr=720,nCr=120{ }^n P_r=720,{ }^n C_r=120, then (n,r)(n, r) is
View answer and explanation
Correct option: D
(10,3)(10,3)
Option D ((10,3)) is correct because it solves for n and r using the given permutation and combination values.
264 Easy
264 of 485
Application +4 | -1
If Pm{P_m} stands for   mPm{\;^m}{P_m}, then 1+1P1+2P2+3P3+...+n.Pn1 + 1{P_1} + 2{P_2} + 3{P_3} + ... + n.{P_n} is equal to
View answer and explanation
Correct option: D
(n+1)!\left( {n + 1} \right)!
Option D ((n+1)!\left( {n + 1} \right)!) is correct because the given sum simplifies to (n+1)!.
265 Easy
265 of 485
Understanding +4 | -1
2m2 m white counters and 2n2 n red counters are arranged in a straight line with (m+n)(m+n) counters on each side of a central mark. The number of ways of arranging the counters, so that the arrangements are symmetrical with respect to the central mark, is
View answer and explanation
Correct option: A
(m+n)!m!n!\frac{(m+n) !}{m ! n !}
Option A ((m+n)!m!n!\frac{(m+n) !}{m ! n !}) is correct because it represents the number of ways to arrange m white and n red counters on one side of the central mark, determining the entire symmetrical arrangement.
266 Easy
266 of 485
Application +4 | -1
Number of ways of selection of 4 letters of the word "PERFECT" is
View answer and explanation
Correct option: C
2525
Option C (25) is correct because it correctly counts the number of ways to select 4 letters from the word \"PERFECT\", considering the repeated letter E.
267 Easy
267 of 485
Analyzing +4 | -1
The value of 2n[1.3.5…(2n−3)(2n−1)]{2^n}\left[ {1.3.5 \ldots \left( {2n - 3} \right)\left( {2n - 1} \right)} \right] is
View answer and explanation
Correct option: A
(2n)!n!\frac{{\left( {2n} \right)!}}{{n!}}
Option A ((2n)!n!\frac{{(2n)!}}{{n!}}) is correct because the expression can be manipulated to match the factorial form.
268 Easy
268 of 485
Application +4 | -1
There are stalls for 10 animals in a ship. The number of ways the shipload can be made if there are cows, calves and horses to be transported, animals of each kind being not less than 10 , is
View answer and explanation
Correct option: A
59049
Option A (59049) is correct because each stall can be filled in 3 ways, and there are 10 stalls, resulting in 3 to the power of 10.
269 Easy
269 of 485
Analyzing +4 | -1
The number of ways of arranging mm numbers out of 1,2,3,…,n1,2,3, \ldots, n, so that maximum is (n−2)(n-2) and minimum is 2 (repetitions of numbers is allowed) such that maximum and minimum both occur exactly once (n>5,m>3)(n > 5, m > 3), is
View answer and explanation
Correct option: C
m(m−1)(n−5)m−2m(m-1)(n-5)^{m-2}
Option C (m(m−1)(n−5)m−2m(m-1)(n-5)^{m-2}) is correct because it represents the number of ways to arrange mm numbers with the given constraints, considering the positions of the minimum and maximum values and the choices for the remaining numbers.
270 Easy
270 of 485
Understanding +4 | -1
The number of ways in which 10 candidates A1,A2,…A_1, A_2, \ldots, A10A_{10} can be ranked so that A1A_1 is always above A2A_2, is
View answer and explanation
Correct option: B
10 ! 2\frac{10 \text { ! }}{2}
The correct answer is B (10 ! 2\frac{{10 \text { ! }}}{{2}}) because half of the total permutations will have A1A_1 above A2A_2, and the other half will have A2A_2 above A1A_1.
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