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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 181-210 of 485 questions

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181 Easy
181 of 485
Application +4 | -1
The positive integer rr, such that 15C3r=15Cr+3{ }^{15} {C}_{3 r}={ }^{15} {C}_{r+3}, is equal to
View answer and explanation
Correct option: A
3
Option A (3) is correct because solving the equation 15C3r=15Cr+3{}^{15}C_{3r} = {}^{15}C_{r+3} leads to r=3r=3 as the only positive integer solution.
182 Easy
182 of 485
Application +4 | -1
The number of numbers of 4 digits which are not divisible by 5, are
View answer and explanation
Correct option: A
7200
Option A (7200) is correct because it is the total number of 4-digit numbers minus the number of 4-digit numbers divisible by 5.
183 Easy
183 of 485
Application +4 | -1
ABCD is a quadrilateral. 3,4,53,4,5 and 6 points are marked on the sides AB,BC,CD\mathrm{AB}, {BC}, {CD} and DA{DA} respectively. The number of triangles with vertices on different sides is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the number of triangles formed is 342, which is not among the options.
184 Easy
184 of 485
Application +4 | -1
Let QnQ_{{n}} be the number of possible quadrilaterals formed by joining vertices of an nn sided regular polygon. If Qn+1−Qn=20Q_{n+1}-Q_n=20, then value of nn is:
View answer and explanation
Correct option: C
6
Option C (6) is correct because substituting n=6 into the equation Qn+1−Qn=20Q_{n+1}-Q_n=20 after expressing QnQ_n as nC4{}^nC_4 satisfies the equality.
185 Easy
185 of 485
Application +4 | -1
The number of permutations by taking all letters and keeping the vowels of the word COMBINE in the odd places is
View answer and explanation
Correct option: D
576
Option D (576) is correct because the vowels can be arranged in odd places in   4P3{\;^4}{P_3} ways and the remaining letters in 4! ways.
186 Easy
186 of 485
Application +4 | -1
Out of 18 points in a plane, no three are in the same straight line except five points which are collinear. The number of straight lines that can be formed by joining them is
View answer and explanation
Correct option: B
144
Option B (144) is correct because it calculates the total number of lines minus the extra lines counted due to collinearity, plus one for the single line formed by the collinear points.
187 Easy
187 of 485
Application +4 | -1
How many words can be formed from the letters of the word DOGMATIC, if all the vowels remain together?
View answer and explanation
Correct option: B
4320
Option B (4320) is correct because when vowels are together, we treat them as one unit, and then arrange the units and consonants, followed by arranging the vowels themselves.
188 Easy
188 of 485
Application +4 | -1
The number of all possible selections of one or more questions from 10 given questions, each question having an alternative is
View answer and explanation
Correct option: C
310−1{3^{10}} - 1
Option C (310−1{3^{10}} - 1) is correct because each question has 3 options (select the question, select the alternative, or reject the question), and we subtract 1 to exclude the case where all questions are rejected.
189 Easy
189 of 485
Application +4 | -1
All possible four-digit numbers are formed using the digits 0,1,2,3 so that no number has repeated digits. The number of even number among them is
View answer and explanation
Correct option: C
10
Option C (10) is correct because the number of even numbers is calculated by considering the cases where the last digit is 0 or 2, and then arranging the remaining digits accordingly.
190 Easy
190 of 485
Application +4 | -1
In how many ways can 21 English and 19 Hindi books be placed in a row so that no two Hindi books are together?
View answer and explanation
Correct option: A
1540
Option A (1540) is correct because the number of ways to arrange the books is calculated by placing the Hindi books in the spaces between the English books.
191 Easy
191 of 485
Application +4 | -1
The number of numbers lying in (2,3)(2,3), where all the digits after decimal are non-zero and in increasing order, is
View answer and explanation
Correct option: D
1023
Option D (1023) is correct because it represents the total number of subsets (excluding the empty set) that can be formed from the digits 1 to 9, ensuring the digits are in increasing order.
192 Easy
192 of 485
Application +4 | -1
If 33!33 ! is divisible by 2n2^n, then the maximum value of nn is equal to
View answer and explanation
Correct option: B
31
Option B (31) is correct because it represents the highest power of 2 that divides 33!, calculated by summing the quotients of successive divisions by 2.
193 Easy
193 of 485
Application +4 | -1
12 persons are to be arranged to a round table. If two particular persons among them are not to be side by side, the total number of arrangements is
View answer and explanation
Correct option: A
9  (10  !)9\;\left( {10\;!} \right)
Option A (9  (10  !)9\;\left( {10\;!} \right)) is correct because it calculates the total arrangements minus arrangements where the two particular persons are side by side.
194 Easy
194 of 485
Application +4 | -1
The number of rectangles excluding squares from a rectangle of size 9×69 \times 6 is
View answer and explanation
Correct option: B
791
Option B (791) is correct because it is the result of subtracting the number of squares from the total number of rectangles in a 9x6 grid.
195 Easy
195 of 485
Application +4 | -1
If eleven member of a committee sit at a round table so that the President and Secretary always sit together, then the number of arrangements is
View answer and explanation
Correct option: C
9!×29! \times 2
The correct answer is C (9!×29! \times 2) because it is the number of arrangements of 11 committee members around a round table with the President and Secretary always sitting together.
196 Easy
196 of 485
Application +4 | -1
The total numbers of greater than 100 and divisible by 5, that can be formed from the digits 3,4,5,6 if no digit is repeated is
View answer and explanation
Correct option: D
12
The correct answer is D (12) because it is the number of numbers greater than 100 and divisible by 5 that can be formed from the digits 3, 4, 5, 6 without repetition.
197 Easy
197 of 485
Application +4 | -1
From 4 Officers and 8 Jawans, a committee of 6 is to be chosen to include exactly one officer. The number of such committee is
View answer and explanation
Correct option: C
224
Option C (224) is correct because it represents the number of ways to choose exactly one officer from 4 and 5 jawans from 8.
198 Easy
198 of 485
Application +4 | -1
The number of ways in which ten candidates A1,A2,A3,A4,⋯ ,A10A_1, A_2, A_3, A_4, \cdots, A_{10} can be arranged in a row "If A1A_1 is always above A2A_2 is
View answer and explanation
Correct option: B
10!2!\frac{10 !}{2 !}
Option B (10!2!\frac{{10 !}}{{2 !}}) is correct because it considers all possible arrangements of the 10 candidates and then divides by 2! to account for the fact that A1A_1 must always be above A2A_2.
199 Easy
199 of 485
Analyzing +4 | -1
There are three teams each of a chairman, a supervisor and a worker three different companies. There are nine bonus of different denominations to be paid to these nine persons in all. In how many ways can this be done with due respect to superiority is given in every team?
View answer and explanation
Correct option: D
None of these
Option A (16) is correct because it is the result of subtracting the number of triangles with 1 or 2 sides common with the octagon from the total number of triangles.
200 Easy
200 of 485
Application +4 | -1
The number of possible outcomes in a throw of nn ordinary dice in which atleast one of the dice shows an odd number, is
View answer and explanation
Correct option: C
6n−3n6^n-3^n
Option D (None of these) is correct because the calculated number of words does not match any of the given options.
201 Easy
201 of 485
Application +4 | -1
In a class, there are 15 boys and 10 girls. The teacher wants to select a boy and a girl to represent the class in a function, then the total number of selection by teacher is
View answer and explanation
Correct option: D
150
Option D (150) is correct because it is the product of the number of ways to select a boy and the number of ways to select a girl.
202 Easy
202 of 485
Application +4 | -1
If there are nn number of seats and m number of people have to be seated, then how many ways are possible to do this (m<n)(m < n) ?
View answer and explanation
Correct option: A
  nPm{\;^n}{P_m}
Option A (  nPm{\;^n}{P_m}) is correct because the number of ways to seat mm people in nn seats is given by the permutation formula   nPm{\;^n}{P_m}.
203 Easy
203 of 485
Application +4 | -1
How many numbers between 400 and 1000 can be made with the digits 2, 3, 4, 5, 6 and 0?
View answer and explanation
Correct option: C
60
Option C (60) is correct because it calculates the number of 3-digit numbers between 400 and 1000 using the given digits, considering the restriction on the hundreds place.
204 Easy
204 of 485
Application +4 | -1
Three alike red roses and three alike pink roses are used to form a garland. The number of ways is
View answer and explanation
Correct option: A
3
Option A (3) is correct because there are only three distinct arrangements when considering rotational symmetry of the garland.
205 Easy
205 of 485
Application +4 | -1
How many numbers consisting of 5 digits can be formed in which the digits 3, 4 and 7 are used only once and the digit 5 is used twice?
View answer and explanation
Correct option: B
60
Option B (60) is correct because the number of such numbers is calculated by arranging the 5 digits, considering the repetition of the digit 5.
206 Easy
206 of 485
Application +4 | -1
The number of ways of arranging 8 men and 4 women around a circular table such that no two women can sit together, is
View answer and explanation
Correct option: D
7!  8P47!{\;^8}{P_4}
Option D (7!  8P47!{\;^8}{P_4}) is correct because it first arranges the men around the table and then arranges the women in the gaps between the men.
207 Easy
207 of 485
Application +4 | -1
The number of straight lines that can be drawn out of 10 points of which 7 are collinear
View answer and explanation
Correct option: D
24
Option D (24) is correct because it calculates the total number of lines that can be formed, accounting for the collinearity of 7 points.
208 Easy
208 of 485
Application +4 | -1
A letter lock consists of three rings each marked with 10 different letters. No. of ways to make an unsuccessful attempts to open the lock is
View answer and explanation
Correct option: B
999
Option B (999) is correct because it subtracts the one successful attempt from the total possible combinations.
209 Easy
209 of 485
Application +4 | -1
A five digit number divisible by 3 is to be formed using the numbers 0, 1, 2, 3, 4 and 5, without repetition. The total number of ways this can be done, is
View answer and explanation
Correct option: A
216
Option A (216) is correct because it counts the number of 5-digit numbers divisible by 3 that can be formed using the given digits without repetition.
210 Easy
210 of 485
Application +4 | -1
There are nn number of sets and m number of people have to be seated, then how many ways are possible to do this (m<n)?\left( {m < n} \right)?
View answer and explanation
Correct option: A
  nPm{\;^n}{P_m}
Option A (  nPm{\;^n}{P_m}) is correct because it represents the number of ways to select and seat m people from n sets, considering the order.
Revision notes and key takeaways

Review PERMUTATION AND COMBINATION in your textbook or class notes, then use this set to check your understanding.

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