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JEE • MATHEMATICS • PERMUTATION AND COMBINATION

Permutation and combination practice · Medium

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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 211-240 of 250 questions
211 Medium
211 of 250
Application +4 | -1
There are ' nn ' numbered seats around a round table. Total num- ber of ways in which n1(n1<n)n_{1}\left(n_{1} < n\right) persons can sit around the round table is equal to
View answer and explanation
Correct option: B
nPn1{ }^{n} P_{n_{1}}
Option B (nPn1{ }^{n} P_{n_{1}}) is correct because when seats are numbered, the arrangement is a linear permutation.
212 Medium
212 of 250
Application +4 | -1
nn - digit numbers are formed using only three digits 2,5 and 7. The smallest value of nn for which 900 such distinct numbers can be formed, is
View answer and explanation
Correct option: D
7
Option D (7) is correct because 37 is the smallest power of 3 that exceeds 900, ensuring that at least 900 distinct numbers can be formed.
213 Medium
213 of 250
Application +4 | -1
Total number of four digit odd number that can be formed using 0,1,2,3,5,70,1,2,3,5,7 and having distinct digits are
View answer and explanation
Correct option: D
192
Option D (192) is correct because it accounts for the restrictions on the first and last digits to form a four-digit odd number with distinct digits.
214 Medium
214 of 250
Application +4 | -1
There are (n+1)(n+1) white similar balls and (n+1)(n+1) black balls of different size. No. of ways the balls can be arranged in a row so that adjacent balls are of different colours is
View answer and explanation
Correct option: A
2[(n+1)!]22[(n+1) !]^2
Option A (2[(n+1)!]22[(n+1) !]^2) is correct because there are two possible arrangements (black first or white first), and each color's balls can be arranged in (n+1)! ways.
215 Medium
215 of 250
Application +4 | -1
In a club election, the number of contestants is one more than the number of maximum candidates for which a voter can vote. If the total number of ways in which a voter can vote be 62, then the number of candidates is
View answer and explanation
Correct option: C
6
Option C (20) is correct because it represents the number of ways to arrange the boys and girls in decreasing order of height.
216 Medium
216 of 250
Analyzing +4 | -1
If the number of ways of selecting KK coupons out of an unlimited number of coupons bearing the letters A, T, M, so that they cannot be used to spell the word MAT is 93, then KK is equal to
View answer and explanation
Correct option: B
5
Option B (5) is correct because it satisfies the condition that the number of ways of selecting K coupons such that they cannot spell MAT is 93.
217 Medium
217 of 250
Analyzing +4 | -1
The permutations of nn objects taken

(i) atleast rr objects at a time

(ii) atmost rr objects at a time
(if repetition of the objects is allowed) are respectively
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the correct formulas for permutations with repetition allowed are different from the given options.
218 Medium
218 of 250
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.
219 Medium
219 of 250
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)!
220 Medium
220 of 250
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.
221 Medium
221 of 250
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.
222 Medium
222 of 250
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.
223 Medium
223 of 250
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!}.
224 Medium
224 of 250
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.
225 Medium
225 of 250
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.
226 Medium
226 of 250
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.
227 Medium
227 of 250
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.
228 Medium
228 of 250
Application +4 | -1
ABCD is a convex quadrilateral. 3,4,53,4,5 and 6 points are marked on the sides AB,BC,CD\mathrm{AB}, \mathrm{BC}, \mathrm{CD} and DA\mathrm{DA} respectively. The number of triangles with vertices on different sides is
View answer and explanation
Correct option: A
342
Option A (342) is correct because it sums the number of triangles formed by choosing one vertex from each of the three sides in all possible combinations.
229 Medium
229 of 250
Application +4 | -1
Nine hundred distinct N –digit numbers are to be formed by using 6, 8 and 9 only. The smallest
value of N for which this is possible is
View answer and explanation
Correct option: B
7
The correct answer is D (1+icot⁡(π/2n)1+i \cot (\pi / 2 n)) because it is the simplified form of the geometric series using complex number properties.
230 Medium
230 of 250
Application +4 | -1
Nine toys are to be packed in 9 boxes. If 5 of them are too big for 3 boxes, then the number of ways in which they can be packed is
View answer and explanation
Correct option: A
6P5⋅4{ }^6 P_5 \cdot 4 !
Option A (6P5⋅4!{ }^6 P_5 \cdot 4 !) is correct because it represents the number of ways to pack the 5 big toys into the 6 suitable boxes, and then pack the remaining 4 toys into the remaining 4 boxes.
231 Medium
231 of 250
Application +4 | -1
The number of ways in which the letters of the word VOWEL can be arranged so that the letters O, E occupy only even places, is
View answer and explanation
Correct option: A
12
Option A (12) is correct because it calculates the number of ways to arrange the letters of the word VOWEL such that O and E occupy only even places.
232 Medium
232 of 250
Analyzing +4 | -1
4nC2n:2nCn{ }^{4 n} C_{2 n}:{ }^{2 n} C_n is equal to
View answer and explanation
Correct option: A
{1⋅3⋅5…(4n−1)}{1⋅3⋅5…(2n−1)}2\frac{\{1 \cdot 3 \cdot 5 \ldots(4 n-1)\}}{\{1 \cdot 3 \cdot 5 \ldots(2 n-1)\}^2}
Option A is correct because the ratio of combinations can be simplified to the given expression by expanding the factorials and cancelling terms.
233 Medium
233 of 250
Application +4 | -1
Two teachers are taking 6 students to a zoo. The teachers decide to split up. Each student must choose one of the teachers, with the condition that each teacher must take at least one student. Number of possible ways of doing this is:
View answer and explanation
Correct option: B
62
Option B (62) is correct because it represents the total number of ways the students can be divided between the two teachers, ensuring each teacher has at least one student.
234 Medium
234 of 250
Understanding +4 | -1
Let an=a_n= number of nn-digit positive integers formed by the digits 0, 1 or both such that no consecutive digits in them is 0.
Let bn=b_n= number of such nn digit integers ending with digit 1, and cn=c_n= the numner of such nn digit integers ending with digit 0, then for n≥3n \geq 3,
View answer and explanation
Correct option: D
an=bn+cna_n=b_n+c_n
Option D (an=bn+cna_n=b_n+c_n) is correct because the total number of n-digit integers is the sum of those ending in 1 and those ending in 0.
235 Medium
235 of 250
Application +4 | -1
The number of ways in which the digits 1,2,3,4,31,2,3,4,3, 2,1 can be arranged so that the odd digits always occupy the odd places, is
View answer and explanation
Correct option: C
18
Option C (18) is correct because the odd digits (1, 3, 3, 1) can be arranged in odd places in 4!2!2!\frac{{4!}}{{2!2!}} ways, and the even digits (2, 4, 2) can be arranged in even places in 3!2!\frac{{3!}}{{2!}} ways, so the total number of ways is 4!2!2!×3!2!=18\frac{{4!}}{{2!2!}} \times \frac{{3!}}{{2!}} = 18.
236 Medium
236 of 250
Application +4 | -1
The number of arrangements of letters of the word 'BANANA' in which the two N's do not appear together is
View answer and explanation
Correct option: A
40
Option A (40) is correct because it's the total arrangements minus arrangements where the two N's are together.
237 Medium
237 of 250
Application +4 | -1
If two numbers are selected from numbers 1 to 25, then the number of ways that their difference does not exceed 10 is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the calculated number of ways does not match any of the given options.
238 Medium
238 of 250
Application +4 | -1
Six boys and six girls sit along a line alternately in xx ways; and along a circle again alternatively in yy ways, then
View answer and explanation
Correct option: D
x=12yx=12 y
Option D (x=12y) is correct because the ratio of linear to circular arrangements with alternating boys and girls is 12.
239 Medium
239 of 250
Analyzing +4 | -1
In a certain test, there are nn questions. In this test, 2n−i2^{n- i} students gave wrong answers to atleast ii questions, where i=1,2,3,...,n.i =1, 2, 3,..., n. If the total number of wrong answers given is 2047, then nn is equal to
View answer and explanation
Correct option: B
11
Option B (11) is correct because the total number of wrong answers can be expressed as a geometric series, which simplifies to 2^n - 1 = 2047, leading to n = 11.
240 Medium
240 of 250
Application +4 | -1
The number of permutations of all the letters of the word 'EXERCISES' is
View answer and explanation
Correct option: B
30240
The correct answer is B (30240) because it is the number of permutations of the letters in 'EXERCISES' considering repeated letters.
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