Skip to main content
Back to PERMUTATION AND COMBINATION

Your Progress

0 / 742 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • PERMUTATION AND COMBINATION

Permutation and combination practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

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.

Take a chapter quiz →
Showing 451-480 of 742 questions

Preparing practice and saved progress…

451 Easy
451 of 742
Application +4 | -1
The number of ways in which a couple can sit around a table with 6 guests, if the couple take consecutive seats, is
View answer and explanation
Correct option: A
1440
Option A (1440) is correct because the couple can be considered as one unit, and they can be arranged among themselves in 2! ways. The total arrangements are 6! * 2! = 1440.
452 Medium
452 of 742
Application +4 | -1
The value of (7C0+7C1)+(7C1+7C2)+…+(7C6+7C7)\left({ }^7 C_0+{ }^7 C_1\right)+\left({ }^7 C_1+{ }^7 C_2\right)+\ldots+\left({ }^7 C_6+{ }^7 C_7\right) is
View answer and explanation
Correct option: A
28−22^8-2
Option A (28−22^8-2) is correct because the given expression can be simplified using binomial coefficient identities to equal 28−22^8-2.
453 Easy
453 of 742
Application +4 | -1
There are three copies each of 4 different books. The number of ways in which they be arranged on a shelf is
View answer and explanation
Correct option: A
12!(3!)4\frac{12 !}{(3 !)^4}
Option A (12!(3!)4\frac{{12 !}}{{(3 !)^4}}) is correct because it accounts for the repetition of each of the 4 different books.
454 Medium
454 of 742
Application +4 | -1
Ten persons are arranged in a row. The number of ways of selecting four persons so that no two persons sitting next to each other are selected is
View answer and explanation
Correct option: C
35
Option C (35) is correct because the number of ways to select four non-adjacent persons from ten is equivalent to arranging 4 ones and 6 zeros with no two ones adjacent, which can be done in 7C4=35{ }^7 C_4 = 35 ways.
455 Easy
455 of 742
Application +4 | -1
How many words can be formed from the letters of the word ARTICLE, if vowels always comes at the odd places?
View answer and explanation
Correct option: B
576
Option B (576) is correct because it calculates the number of ways to arrange the letters of 'ARTICLE' such that vowels occupy odd places.
456 Easy
456 of 742
Application +4 | -1
The total number of 4-digit numbers that are greater than 3000, that can be formed using the digits 1,2,3,4,5,61, 2, 3, 4, 5, 6 (no digit is being repeated in any number) is equal to
View answer and explanation
Correct option: B
240
Option B (240) is correct because it is the number of 4-digit numbers greater than 3000 that can be formed without repetition.
457 Medium
457 of 742
Analyzing +4 | -1
The number of ways can 14 identical toys be distributed among three boys, so that each one gets atleast one toy and no two boys get equal number of toys, is
View answer and explanation
Correct option: C
60
Option C (60) is correct because it accurately counts the number of ways to distribute the toys such that each boy gets at least one and no two boys get the same number of toys.
458 Medium
458 of 742
Application +4 | -1
If all permutations of the letters of the word AGAIN are arranged as in dictionary, then fifteen word is
View answer and explanation
Correct option: C
NAAIG
Option C (NAAIG) is correct because when arranging the permutations in dictionary order, NAAIG is the 50th word.
459 Easy
459 of 742
Application +4 | -1
Number of number greater than 1000 but not greater than 4000 which can be formed with the digits 0, 1, 2, 3, 4, are
View answer and explanation
Correct option: B
375
Option B (375) is correct because it accounts for all 4-digit numbers between 1000 and 4000 formed using the given digits without repetition.
460 Easy
460 of 742
Application +4 | -1
Find nn if n−1P3:nP4=1:9{ }^{n-1} P_3:{ }^n P_4=1: 9
View answer and explanation
Correct option: A
9
Option A (9) is correct because the ratio n−1P3:nP4=1:9{ }^{n-1} P_3:{ }^n P_4=1: 9 simplifies to 1n=19\frac{1}{n}=\frac{1}{9}, which gives n=9n=9.
461 Medium
461 of 742
Application +4 | -1
The 3 digit numbers which are divisible either 3 or
4 but not divisible by 48
View answer and explanation
Correct option: D
432
Option D (89) is correct because it represents the rank of the word TOUGH when the letters of GHOTU are arranged alphabetically.
462 Medium
462 of 742
Application +4 | -1
The letters of the word OUGHT are written in all possible ways and these words are arranged as in a dictionary, in a series. Then the serial number of the word TOUGH is
View answer and explanation
Correct option: C
89
Option C (89) is correct because it represents the correct lexicographical order of the word TOUGH among all permutations of OUGHT.
463 Easy
463 of 742
Application +4 | -1
The number of ways in which the letters of 'ARRANGE' be arranged so that "Two R's come together and Two A's come together is
View answer and explanation
Correct option: A
120
Option A (120) is correct because it treats the two R's and two A's as single units and arranges the remaining letters.
464 Medium
464 of 742
Application +4 | -1
The number of times of the digits 3 will be written when list- ing the integers from 1 to 1000 is
View answer and explanation
Correct option: B
300
Option B (300) is correct because it accurately counts the occurrences of the digit 3 in the integers from 1 to 1000.
465 Easy
465 of 742
Application +4 | -1
The total number of permutations of n(>1)n( > 1) different things taken not more than r at a time, when each thing may be repeated any number of times is
View answer and explanation
Correct option: C
n(nr−1)n−1\frac{{n\left( {{n^r} - 1} \right)}}{{n - 1}}
Option C (\frac{{n\left( {{n^r} - 1} \right)}}{{n - 1}}) is correct because the total number of permutations is the sum of a geometric series.
466 Medium
466 of 742
Application +4 | -1
The number of ways in which the letters of word FRACTION be arranged so that no two vowels are together, is
View answer and explanation
Correct option: B
14400
Option B (14400) is correct because it calculates the number of arrangements of the letters in FRACTION such that no two vowels are together.
467 Easy
467 of 742
Analyzing +4 | -1
The number of numbers of 9 distinct digits such that all the digits in the first four places are less than the digit in the middle and all the digits in the last four places are greater than that in the middle is
View answer and explanation
Correct option: B
576
Option B (576) is correct because the middle digit must be 5, and the remaining digits can be arranged in 4! * 4! ways.
468 Medium
468 of 742
Application +4 | -1
The total number of functions, f:{1,2,3,4}→{1,2,3,4,5,6}f:\{1,2,3,4\} \rightarrow\{1,2,3,4,5,6\} such that f(1)+f(2)=f(3)f(1)+f(2)=f(3), is equal to
View answer and explanation
Correct option: B
90
Option B (90) is correct because it is the total number of functions satisfying the given condition, calculated by considering all possible cases.
469 Medium
469 of 742
Application +4 | -1
Number greater than 1000 but less than 4000 is formed using the digits 0,1,2,3,40,1,2,3,4 (repetition allowed). Their number is
View answer and explanation
Correct option: C
374
The correct answer is 374 because the number of integers between 1000 and 4000 that can be formed using the digits 0, 1, 2, 3, 4 with repetition allowed is 374.
470 Easy
470 of 742
Application +4 | -1
The number of ways in which 30 marks can be alloted to 8 questions if each question carries at least 2 marks, is
View answer and explanation
Correct option: C
116280
Option D (None of these) is correct because the calculated value 116280 matches the option provided in the prompt.
471 Easy
471 of 742
Application +4 | -1
For the post of 5 teachers there are 23 applicants, 2 posts are reserved for S. C. candidates and there are 7 S.C. candidates among the applicants. The number of ways in which the selection can be made is
View answer and explanation
Correct option: C
11760
Option C (11760) is correct because it calculates the number of ways to select 2 SC candidates and 3 non-SC candidates for the teacher positions.
472 Easy
472 of 742
Application +4 | -1
The number of ways in which a mixed doubles game in tennis can be arranged from 5 married couples, if no husband and wife play in the same game, is
View answer and explanation
Correct option: C
60
Option C (60) is correct because the number of ways to arrange a mixed doubles game is calculated considering the restrictions.
473 Medium
473 of 742
Application +4 | -1
The number of ways in which a committee of 5 can be chosen from 10 candidates so as to exclude the youngest, if it includes the oldest, is
View answer and explanation
Correct option: A
196
Option D (None of the above) is incorrect. Option A (196) is correct because it calculates the number of ways to choose a committee of 5 from 10 candidates, including the oldest and excluding the youngest.
474 Easy
474 of 742
Application +4 | -1
On a New Year day every student of a class sends a card to every other student. The postman delivers 600 cards. The number of students in the class is
View answer and explanation
Correct option: C
25
Option C (25) is correct because it solves the permutation equation nP2=600{ }^n P_2=600 to find the number of students, n, where each student sends a card to every other student.
475 Easy
475 of 742
Application +4 | -1
The number of ways in which 6 rings can be worn on four fingers of one hand, is
View answer and explanation
Correct option: A
46{4^6}
Option A (46{4^6}) is correct because each of the 6 rings can be worn on any of the 4 fingers.
476 Easy
476 of 742
Application +4 | -1
The value of 1.1!+2.2!+3.3!+...+n.n!1.1! + 2.2! + 3.3! + ... + n.n! is
View answer and explanation
Correct option: C
(n+1)!−1\left( {n + 1} \right)! - 1
Option C ((n+1)!−1(n + 1)! - 1) is correct because the given series can be rewritten and simplified using the properties of factorials.
477 Medium
477 of 742
Application +4 | -1
The total number of 5-digit numbers, formed by using the digits 1, 2, 3, 5, 6, 7 without repetition, which are multiple of 6, is
View answer and explanation
Correct option: D
72
Option D (72) is correct because there are 48 numbers excluding 3 and 24 numbers excluding 6 that are divisible by 6.
478 Easy
478 of 742
Application +4 | -1
The number of ways of arranging 20 boys so that 3 particular boys are separated is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the correct formula for arranging with separation is (17!) * (18! / 15!).
479 Easy
479 of 742
Application +4 | -1
The number of 4 letter words that can be formed from letters of the word 'PART', when:

Statement I: Repetition is not allowed is 24.

Statement II: Repetition is allowed is 256.

Which of the above statement(s) is/are true?
View answer and explanation
Correct option: C
Both I and II
Option C (Both I and II) is correct because with no repetition the number of 4 letter words is 24 and with repetition the number of 4 letter words is 256.
480 Easy
480 of 742
Application +4 | -1
The number of ways we can put 5 different balls in 5 different boxes such that at most three boxes is empty, is equal to
View answer and explanation
Correct option: A
55−55^5-5
The correct answer is A (55-5) because it represents the total number of ways to put 5 balls in 5 boxes minus the ways where all balls are in one box.
Revision notes and key takeaways

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

Showing 451-480 of 742 questions

Your practice

0 / 742 answered · 742 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.