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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 451-480 of 485 questions

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451 Easy
451 of 485
Application +4 | -1
The number of ways in which three girls and nine boys can be seated in two cars, each having numbered seats, 3 in the front and 4 at the back, is
View answer and explanation
Correct option: A
14P12^{14}{P_{12}}
Option A (14P12{}^{14}{P_{12}}) is correct because we are choosing 12 people out of 14 and arranging them in specific numbered seats.
452 Easy
452 of 485
Application +4 | -1
There are 5 historical moments, 6 gardens and 7 shopping malls in the city. In how many ways a tourist can visit the city, if he visits at least one shopping mall?
View answer and explanation
Correct option: A
25.26.(27−1){2^5}{.2^6}.\left( {{2^7} - 1} \right)
Option A (25.26.(27−1){2^5}{.2^6}.\left( {{2^7} - 1} \right)) is correct because each historical moment and garden can either be visited or not, and at least one shopping mall must be visited.
453 Easy
453 of 485
Application +4 | -1
The total number of ways in which 4 boys and 4 girls can form a line, with boys and girls alternating, is
View answer and explanation
Correct option: C
2(4!)22{\left( {4!} \right)^2}
Option C (  4P2{{\;^4}{P_2}}) is correct because the thousandth place must be 3 and the unit place must be 5, leaving 4 digits to arrange in the remaining two places.
454 Easy
454 of 485
Application +4 | -1
Nine hundred distinct n\mathrm{n}-digit number s\mathrm{s} are to be forme using exactly the three digits. 2, 5 and 7 . The smallest value  of n\mathrm{n} for which this is possible is.
View answer and explanation
Correct option: A
7
Option A (7) is correct because it is the smallest number of digits needed to form 900 distinct numbers using only 2, 5, and 7.
455 Easy
455 of 485
Application +4 | -1
A gentleman invites 13 guests to a dinner and places 8 of them at one table and remaining 5 at the other, the tables being round. The number of ways he can arrange the guests is
View answer and explanation
Correct option: D
13!40\frac{13 !}{40}
Option A (60) is correct because the number of ways to arrange beads in a necklace is half the number of ways to arrange them in a circle due to the necklace being able to be flipped.
456 Easy
456 of 485
Application +4 | -1
The sum of the digits in the unit place of all numbers formed with the help of 3, 4, 5, 6 taken al, at a time, is
View answer and explanation
Correct option: B
108
Option B (108) is correct because it calculates the sum of digits in the unit place of all numbers formed with the given digits.
457 Easy
457 of 485
Application +4 | -1
The total number of natural numbers of six digits that can be made with digits 1,2,3,4, if all digits are to appear in the same number at least once, is
View answer and explanation
Correct option: A
1560
Option A (1560) is correct because it sums the number of arrangements where one digit repeats thrice and where two digits repeat twice.
458 Easy
458 of 485
Application +4 | -1
The number of numbers less than 1000 that can be formed out of the digits 0,1,2,3,40,1,2,3,4 and 5 , no digit being repeated, is
View answer and explanation
Correct option: B
131
Option B (131) is correct because it counts the number of 1, 2, and 3-digit numbers that can be formed using the given digits without repetition.
459 Easy
459 of 485
Application +4 | -1
The number of all four digit numbers which are divisible by 4 that can be formed from the digits 1,2,3,4 and 5 is
View answer and explanation
Correct option: A
125
Option A (125) is correct because it considers all possible combinations of the first two digits for each valid last two digits.
460 Easy
460 of 485
Application +4 | -1
The number of ways in which the letters of the word 'SUCCESSFUL' be arranged such that the S's and U's will come together is
View answer and explanation
Correct option: B
7!2!\frac{7 !}{2 !}
Option B (7!2!\frac{7 !}{2 !}) is correct because when the 3 'S's and 2 'U's are treated as one unit each, there are 7 units to arrange, accounting for repetitions of 'C'.
461 Easy
461 of 485
Application +4 | -1
The number of numbers, strictly between 5000 and 10000 can be formed using the digits 1,3,5,7,91,3,5,7,9 without repetition, is
View answer and explanation
Correct option: C
72
Option C (72) is correct because it is the total number of 4-digit numbers between 5000 and 10000 that can be formed using the given digits without repetition.
462 Easy
462 of 485
Remembering +4 | -1
There are 20 students in a class, in which 12 boys and 8 girls. The class teacher select either a boy or a girl for monitor of the class. In how many ways, the teacher can make this selection?
View answer and explanation
Correct option: D
20
Option D (20) is correct because the teacher can select either a boy or a girl, and the total number of ways is the sum of the number of boys and girls.
463 Easy
463 of 485
Application +4 | -1
The number of 9 digit numbers which have all distinct digits, is
View answer and explanation
Correct option: D
10!−9!10 !-9 !
Option D (10!−9!10 !-9 !) is correct because it represents the total number of 9-digit permutations minus those that start with zero.
464 Easy
464 of 485
Application +4 | -1
The last digit of (1!+2!+……+2009!)500(1 !+2 !+\ldots \ldots+2009 !)^{500} is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the last digit is determined by the last digit of the sum of factorials, which simplifies to the last digit of 33 raised to the power of 500.
465 Easy
465 of 485
Analyzing +4 | -1
In how many ways the letters of the word ENGINEERING, can be arranged such that all E's come together but no two N's come together
View answer and explanation
Correct option: C
54×7!\frac{5}{4} \times 7 !
Option C (54×7!\frac{5}{4} \times 7 !) is correct because it arranges the other letters and then places the N's in the gaps, considering the E's are together.
466 Easy
466 of 485
Application +4 | -1
The number of arrangements that can be formed out of 'LOGARITHM' so that "No two vowels come together is
View answer and explanation
Correct option: A
6!7P36 !^7 P_3
Option A (6!7P36 !^7 P_3) is correct because it represents the number of arrangements where consonants are arranged first, and then vowels are placed in the gaps between them.
467 Easy
467 of 485
Application +4 | -1
If m+nP2=90{ }^{{m}+{n}} P_2=90 and m−nP2=30{ }^{{m}-{n}} P_2=30, then (m,n)(m, n) is equal to
View answer and explanation
Correct option: B
(8,2)(8,2)
Option B ((8,2)) is correct because solving the system of equations derived from the permutation formulas yields m=8 and n=2.
468 Easy
468 of 485
Application +4 | -1
The number of different matrices that can be formed with elements 0,1,20,1,2 or 3,3, each matrix having 4 elements, is
View answer and explanation
Correct option: C
3×443 \times 4^4
Option D (None of these) is correct because the calculated number of ways does not match any of the given options.
469 Easy
469 of 485
Application +4 | -1
How many numbers lying between 10 and 1000 can be formed from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 (repetition of digits is allowed)?
View answer and explanation
Correct option: B
810
Option B (810) is correct because it calculates the total number of 2-digit and 3-digit numbers that can be formed using the given digits with repetition allowed.
470 Easy
470 of 485
Application +4 | -1
The number of arrangements that can be formed out of GANESHPURI so that "The letter 'G' always occurs in first place is
View answer and explanation
Correct option: A
9 !
Option B (9 !) is correct because with 'G' fixed in the first position, the remaining 9 letters can be arranged in 9! ways.
471 Easy
471 of 485
Application +4 | -1
If the letters of the word MOTHER are written in all possible orders and these words are written out as in a dictionary then the rank of the word MOTHER is
View answer and explanation
Correct option: D
309
Option D is correct because the rank of the word MOTHER in a dictionary is found by counting the number of words that come before it, based on alphabetical order.
472 Easy
472 of 485
Analyzing +4 | -1
The number of ways of choosing triplets (x,y,z)(x, y, z) such that z≥max⁡{x,y}z \geq \max \{x, y\} and x,y,z∈{1,2,…,n,n+1}x, y, z \in\{1,2, \ldots, n, n+1\}, is
View answer and explanation
Correct option: A
n+1C3+n+2C3{ }^{n+1} C_3+{ }^{n+2} C_3
Option A (n+1C3+n+2C3{ }^{n+1} C_3+{ }^{n+2} C_3) is correct because it accurately counts the number of triplets satisfying the given condition.
473 Easy
473 of 485
Application +4 | -1
The sum of all four digit numbers that can be formed by using digits 2,4,6,82,4,6,8 (when repetition of digits is not allowed) is
View answer and explanation
Correct option: A
133320
Option A (133320) is correct because it is the sum of all possible four-digit numbers formed by the digits 2, 4, 6, and 8 without repetition.
474 Easy
474 of 485
Application +4 | -1
If   12Pr=1320{\;^{12}}{P_r} = 1320, then rr is equal to
View answer and explanation
Correct option: C
3
Option C (3) is correct because solving the permutation equation   12Pr=1320{\;^{12}}{P_r} = 1320 leads to r=3r = 3.
475 Easy
475 of 485
Application +4 | -1
The number of ways in which the letters of the word BALLOON can be arranged so that two L's do not come together, is
View answer and explanation
Correct option: C
900
Option C (900) is correct because it subtracts the arrangements where the two L's are together from the total possible arrangements.
476 Easy
476 of 485
Application +4 | -1
At an election, a voter may vote for any number of candidates not greater than the number to be chosen. There are 10 candidates and 5 members to be selected. The number of ways of in which a voter can vote is:
View answer and explanation
Correct option: C
637
Option C (637) is correct because the voter can vote for 1 to 5 candidates, and the sum of the combinations 10C1{}^{10} C_1 to 10C5{}^{10} C_5 equals 637.
477 Easy
477 of 485
Application +4 | -1
The number of numbers that can be formed by using digits 1,2,3,4,3,2,1 so that odd digits always occupy odd places
View answer and explanation
Correct option: C
18
The correct answer is 18 because it is the number of numbers that can be formed using the digits 1,2,3,4,3,2,1 such that odd digits always occupy odd places.
478 Easy
478 of 485
Application +4 | -1
If   n+2C8  :  n−2P4=57  :16,{\;^{n + 2}}{C_8}\;:{\;^{n - 2}}{P_4} = 57\;:16, then the value of nn is
View answer and explanation
Correct option: B
19
Option B (19) is correct because it is the solution to the equation derived from the given ratio of combination and permutation after simplification and solving the quadratic equation.
479 Easy
479 of 485
Application +4 | -1
The number of integers greater than 6000 that can be formed using the digits 3,5,6,73,5,6,7 and 8 without repetition, is
View answer and explanation
Correct option: D
192
Option D (192) is correct because it sums the number of possible 4-digit numbers starting with 6, 7, or 8 and the number of possible 5-digit numbers using the given digits.
480 Easy
480 of 485
Application +4 | -1
Ashok, Usha, Rani, Sonu are to give speeches in a class. The teacher can arrange the order of their presentation in
View answer and explanation
Correct option: B
24 ways
Option B (24 ways) is correct because it represents the number of ways to arrange 4 distinct people in a sequence.
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