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JEE • MATHEMATICS • PERMUTATION AND COMBINATION

Permutation and combination practice · Mixed difficulty

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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 661-690 of 742 questions

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661 Easy
661 of 742
Application +4 | -1
A man invites 6 non-vegeterian and 5 vegeterian friends for a dinner party. He arrange 6 non-vegeterian friends on one round table and 5 vegeterian friends along another round table. The number of ways this can be done is:
View answer and explanation
Correct option: C
2880
Option C (2880) is correct because arranging the non-vegetarians and vegetarians separately on round tables gives (6-1)! * (5-1)! = 2880 ways.
662 Medium
662 of 742
Application +4 | -1
Ten different letters of an alphabet are given. Words with five letters are formed from these given letters. Then the number of words which have at least one letter repeated, is
View answer and explanation
Correct option: A
69760
The correct answer is A (69760) because it represents the number of 5-letter words with at least one letter repeated, formed from 10 different letters.
663 Easy
663 of 742
Application +4 | -1
The least positive integral value of nn which satisfies the inequality 10Cn−1>2⋅10Cn{ }^{10} {C}_{n-1}>2 \cdot{ }^{10} {C}_n is
View answer and explanation
Correct option: B
8
Option B (8) is correct because it is the smallest integer value of n that satisfies the given inequality.
664 Medium
664 of 742
Analyzing +4 | -1
The value of (2⋅1P0−3⋅2P1+4⋅3P2−…\left(2 \cdot{ }^{1} P_{0}-3 \cdot{ }^{2} P_{1}+4 \cdot{ }^{3} P_{2}-\ldots\right. up to 51th 51^{\text {th }} term )) +(1!−2!+3!−…+\left(1 !-2 !+3 !-\ldots\right. up to 51th 51^{\text {th }} term )) is equal to :
View answer and explanation
Correct option: C
1+(52)1+(52) !
Option C (1+(52)!) is correct because after simplifying the expression, the result equals 1+(52)!
665 Easy
665 of 742
Application +4 | -1
The number of arrangements of the letters of the word BANANA in which the two N's do not appear adjacently, is
View answer and explanation
Correct option: A
40
Option A (40) is correct because it subtracts the number of arrangements where the two N's are adjacent from the total number of arrangements of the word BANANA.
666 Medium
666 of 742
Application +4 | -1
The total number of ways in which five '++' and three '−-' signs can be arranged in a line such that no two '−-' sign occur together, is
View answer and explanation
Correct option: B
20
Option B (20) is correct because arranging 5 '+' signs creates 6 gaps where 3 '-' signs can be placed such that no two '-' signs are together, resulting in 6C3=20{ }^6 C_3 = 20 ways.
667 Easy
667 of 742
Application +4 | -1
The rank of the word MOTHER when the letters of the word are arranged alphabetically as in a dictionary, is
View answer and explanation
Correct option: C
309
Option C (309) is correct because it is the rank calculated by arranging the letters alphabetically and finding the number of words before MOTHER.
668 Medium
668 of 742
Application +4 | -1
Out of thirty points in a plane, eight of them are collinear. The number of straight lines that can be formed by joining these points, is
View answer and explanation
Correct option: C
408
Option C (408) is correct because the total number of lines formed by 30 points with 8 collinear points is 30C2−8C2+1=408{ }^{30} C_2-{ }^8 C_2+1 = 408.
669 Easy
669 of 742
Understanding +4 | -1
A set contains (2n+1)(2 n+1) elements. The number of subsets of the set which contain at most nn elements is
View answer and explanation
Correct option: D
22n2^{2 n}
Option D (22n2^{2 n}) is correct because the sum of binomial coefficients from 0 to n in a set of (2n+1) elements equals half of the total subsets.
670 Medium
670 of 742
Application +4 | -1
If the letters of the word KRISNA are arranged in all possible ways and these words are written out as in a dictionary, then the rank of the word KRISNA is
View answer and explanation
Correct option: A
324
Option A (324) is correct because it accurately calculates the rank of the word KRISNA in a dictionary arrangement.
671 Easy
671 of 742
Application +4 | -1
Let AA be the set of 4-digit numbers a1a2a3a4a_1 a_2 a_3 a_4, where a1>a2>a3>a4a_1 > a_2 > a_3 > a_4, then n(A)n(A) is equal to
View answer and explanation
Correct option: C
210
Option C (210) is correct because it represents the number of ways to choose 4 distinct digits from 10.
672 Easy
672 of 742
Application +4 | -1
How many numbers lying between 999 and 10000 can be formed with the help of the digits 0, 2, 3, 6, 7, 8 when the digits are not be repeated?
View answer and explanation
Correct option: C
300
Option C (300) is correct because it calculates the number of 4-digit numbers formed with the given digits, excluding those starting with 0.
673 Easy
673 of 742
Application +4 | -1
If each of 10 points on a straight line be joined to each of 10 points on a parallel line then the total number of triangles that can be formed with the given points as vertices, is
View answer and explanation
Correct option: B
900
Option B (900) is correct because it represents the total number of triangles that can be formed with the given points as vertices.
674 Easy
674 of 742
Application +4 | -1
The value of 50C4+∑r=1656−rC3{ }^{50} C_4+\sum\limits_{{r=1}}^6{ }^{56-r} C_3 is
View answer and explanation
Correct option: B
56C4{ }^{56} C_4
Option B (56C4{ }^{56} C_4) is correct because the summation can be simplified using the identity nCr+nCr−1=n+1Cr{ }^n C_r + { }^n C_{r-1} = { }^{n+1} C_r repeatedly.
675 Easy
675 of 742
Application +4 | -1
Number of 4 digit numbers of the form N=abcdN=a b c d which satisfy following three conditions:

(i) 4000≤N<60004000 \leq N < 6000

(ii) NN is a multiple of 5

(iii) 3≤b<c≤63 \leq b < c \leq 6

is equal to:
View answer and explanation
Correct option: C
24
Option C (24) is correct because the number of possible 4-digit numbers is found by considering the constraints on each digit.
676 Medium
676 of 742
Analyzing +4 | -1
The number of 6-digit numbers that can be made with the digits 1,2,31,2,3 and 44 and having exactly two pairs of digits is
View answer and explanation
Correct option: C
1080
Option C (1080) is correct because the number of ways to select two digits to appear twice is 4C2{}^4C_2, and the number of arrangements is 6!2!×2!\frac{6!}{2! \times 2!}.
677 Easy
677 of 742
Application +4 | -1
There are nn different books and pp copies of each. The number of ways in which a selection can be made from them is
View answer and explanation
Correct option: C
(p+1)n−1{\left( {p + 1} \right)^n} - 1
Option C ((p+1)n−1{\left( {p + 1} \right)^n} - 1) is correct because each book can be selected in p+1 ways (0 to p copies), and we subtract 1 for the case where no book is selected.
678 Easy
678 of 742
Application +4 | -1
If x∈Nx \in {N} and (x+2)!(2x−1)!⋅(2x+1)!(x+3)!=727\frac{(x+2) !}{(2 x-1) !} \cdot \frac{(2 x+1) !}{(x+3) !}=\frac{72}{7}, then xx is equal to
View answer and explanation
Correct option: B
4
Option B (4) is correct because simplifying the given factorial equation leads to a quadratic equation, which has x=4 as a valid solution.
679 Easy
679 of 742
Application +4 | -1
The number of natural numbers smaller than 104,{10^4}, in the decimal notation of which all the digit are different is
View answer and explanation
Correct option: A
5274
The correct answer is A (5274) because it is the number of natural numbers smaller than 104{10^4} with all distinct digits.
680 Medium
680 of 742
Analyzing +4 | -1
The integers from 1 to 1000 are written in order around a circle. Starting at 1, every fifteenth number is marked (i.e. 1, 16, 31 etc.) This process is continued until a number is reached which has already been marked, then unmarked numbers are
View answer and explanation
Correct option: D
800
Option D (800) is correct because after marking every fifteenth number until a repetition, 800 numbers remain unmarked.
681 Easy
681 of 742
Application +4 | -1
There are pp copies each of nn different books. The number of different ways in which a non-empty selection can be made from them, is
View answer and explanation
Correct option: A
(p+1)n−1(p+1)^n-1
Option A ((p+1)n−1(p+1)^n-1) is correct because it represents the number of non-empty selections from n different books with p copies each.
682 Easy
682 of 742
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.
683 Easy
683 of 742
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.
684 Easy
684 of 742
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.
685 Medium
685 of 742
Analyzing +4 | -1
The number of three digit numbers of the form xyzx y z such that x<y,   z⩽yx < y,\,\,\,z \leqslant y and x≠0x \ne 0, is
View answer and explanation
Correct option: C
276
Option C (276) is correct because it represents the total count of three-digit numbers satisfying the given conditions x<yx < y and z⩽yz \leqslant y with x≠0x \ne 0, calculated using summation.
686 Medium
686 of 742
Analyzing +4 | -1
Let n1=x1x2x3n_1=x_1 x_2 x_3 and n2=y1y2y3n_2=y_1 y_2 y_3 be two 3-digit numbers, then the pairs of n1n_1 and n2n_2 can be formed, so that n1n_1 can be subtracted from n2n_2 without borrowing, is
View answer and explanation
Correct option: C
45⋅(55)245 \cdot(55)^2
Option C (45⋅(55)245 \cdot(55)^2) is the correct answer because it represents the total number of ways to form the pairs n1n_1 and n2n_2 such that n1n_1 can be subtracted from n2n_2 without borrowing.
687 Easy
687 of 742
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&nbsp; 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.
688 Easy
688 of 742
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.
689 Easy
689 of 742
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.
690 Medium
690 of 742
Analyzing +4 | -1
nn is selected from the set {1,2,3,…,100}\{1,2,3, \ldots, 100\} and the number 2n+3n+5n2^n+3^n+5^n is formed. Total number of ways of selecting nn, so that the formed number is divisible by 4, is equal to
View answer and explanation
Correct option: B
49
Option B (49) is correct because only even values of n make 2n+3n+5n2^n + 3^n + 5^n divisible by 4, and there are 49 even numbers between 2 and 100 inclusive.
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