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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 271-300 of 485 questions

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271 Easy
271 of 485
Application +4 | -1
If   18C15+2(  18C16)+  17C16+1=  nC3{\;^{18}}{C_{15}} + 2\left( {{\;^{18}}{C_{16}}} \right) + {\;^{17}}{C_{16}} + 1 = {\;^n}{C_3}, then nn is equal to
View answer and explanation
Correct option: B
20
Option B (20) is correct because simplifying the given expression using combination identities leads to 20C3^{20}C_3, thus n = 20.
272 Easy
272 of 485
Application +4 | -1
In a group of 8 girls, two girls are sisters. The number of ways in which the girls can sit in a row so that two sisters are not sitting together is
View answer and explanation
Correct option: D
30240
Option D (30240) is correct because it calculates the total number of ways 8 girls can sit minus the number of ways the two sisters sit together, resulting in 8! - 2 * 7! = 30240.
273 Easy
273 of 485
Application +4 | -1
Given that n\mathrm{n} is odd, the number of ways in which three numbers in A.P. can be selected from 1,2,3,….n1,2,3, \ldots . \mathrm{n} is
View answer and explanation
Correct option: D
(n−1)24\frac{(\mathrm{n}-1)^2}{4}
Option D ((n−1)24\frac{{(n-1)^2}}{{4}}) is correct because it accounts for selecting two even or two odd numbers from the set to form an arithmetic progression.
274 Easy
274 of 485
Application +4 | -1
The number of ways in which 10 examination papers can be arranged so that the best and the worst papers never come together, is
View answer and explanation
Correct option: B
8.9 !
Option B (8.9 !) is correct because it calculates the total arrangements minus the arrangements where the best and worst papers are together.
275 Easy
275 of 485
Application +4 | -1
The number of mappings (functions) from the set A={1,  2,  3}A = \left\{ {1,\;2,\;3} \right\} into the set B={1,  2,  3,  4,  5,  6,  7}B = \left\{ {1,\;2,\;3,\;4,\;5,\;6,\;7} \right\} such that f(i)≤f(j)f\left( i \right) \le f\left( j \right) whenever i<ji < j, is
View answer and explanation
Correct option: A
84
Option A (84) is correct because it represents the number of non-decreasing functions from A to B, calculated by stars and bars method or combinations with repetition.
276 Easy
276 of 485
Application +4 | -1
S1,  S2,…,S10{S_1},\;{S_2}, \ldots ,{S_{10}} are the speakers in a conference. If S1{S_1} addresses only after S2{S_2}, then the number of ways the speakers address is
View answer and explanation
Correct option: D
10!2!\frac{{10!}}{{2!}}
Option D (10!2!\frac{{10!}}{{2!}}) is correct because S1{S_1} addresses only after S2{S_2}, so the total arrangements are divided by 2.
277 Easy
277 of 485
Application +4 | -1
4 buses runs between Bhopal and Gwalior. If a man goes from Gwalior to Bhopal by a bus and comes back to Gwalior by another bus, then the total possible ways are
View answer and explanation
Correct option: A
12
Option A (12) is correct because the man has 4 choices for going and then 3 choices for returning on a different bus.
278 Easy
278 of 485
Application +4 | -1
A delegation of five students is to be formed from a group of 10 students. If three particular students want to remain together whereas two particular students do not want to remain together, then the number of selection is
View answer and explanation
Correct option: B
20
Option B (20) is correct because by considering the cases where the three particular students are always together and the two particular students are never together, the total number of selections is 20.
279 Easy
279 of 485
Application +4 | -1
The number of permutations of nn different objects taken r(≥3)r(\geq 3) at a time which include 3 particular objects, is
View answer and explanation
Correct option: D
rP3,n−3Pr−3{ }^r P_3,{ }^{n-3} P_{r-3}
The correct answer is D (rP3,n−3Pr−3{ }^r P_3,{ }^{n-3} P_{r-3}) because it represents the number of ways to arrange 3 particular objects in r places and then arrange the remaining objects.
280 Easy
280 of 485
Application +4 | -1
Rakshit is allowed to select 1 or more books out of (2n+1)(2 n+1) distinct books. If the number of ways in which he may not select all of them is 126, then value of n{n} is ___.
View answer and explanation
Correct option: A
3
Option A (3) is correct because the value of n is found by solving the equation derived from the given condition about selecting books.
281 Easy
281 of 485
Application +4 | -1
The number of all the possible selections which a student can make for answering one or more questions out of eight given questions in a paper, when each question has an alternative is
View answer and explanation
Correct option: B
6560
Option B (6560) is correct because each question has 3 options (answer the first part, answer the second part, or omit the question), and we subtract 1 to exclude the case where all questions are omitted.
282 Easy
282 of 485
Application +4 | -1
Find the no. of Permutations of all the letters of the word MATHEMATICS which starts with consonants only
View answer and explanation
Correct option: A
7×10!8\frac{7 \times 10 !}{8}
Option A (7×10!8\frac{7 \times 10 !}{8}) is correct because it calculates the number of permutations of MATHEMATICS starting with a consonant.
283 Easy
283 of 485
Application +4 | -1
A man has got seven friends. The number of ways in which he can invite one or more of his friends to dinner, is
View answer and explanation
Correct option: C
127
Option B (1024) is correct because each of the 5 men can vote for any of the 4 candidates, leading to 454^5 possible ways.
284 Easy
284 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 given by
View answer and explanation
Correct option: D
(8,2)
Option D ((8,2)) is correct because it is derived by solving the permutation equations for m and n.
285 Easy
285 of 485
Application +4 | -1
The number of 5 digits telephone number having at least one of their digits repeated, is
View answer and explanation
Correct option: D
69760
Option D (69760) is correct because it is calculated by subtracting the number of 5-digit telephone numbers with no repeated digits from the total number of 5-digit telephone numbers.
286 Easy
286 of 485
Application +4 | -1
9 balls are to be placed in 9 boxes and 5 of the balls cannot fit into 3 small boxes. The number of ways of arranging one ball in each of the boxes is
View answer and explanation
Correct option: C
17280
Option C (17280) is correct because it accounts for the restriction of 5 balls not fitting into 3 boxes.
287 Easy
287 of 485
Application +4 | -1
The number of different permutations of the word ‘BANANA’ is
View answer and explanation
Correct option: D
60
Option D (60) is correct because it accounts for the repetitions of the letters A and N in the word BANANA.
288 Easy
288 of 485
Application +4 | -1
The sum of all numbers which can be formed by using the digits 1,3,5,71,3,5,7 all at a time and which have no digit repeated is
View answer and explanation
Correct option: C
16×1111×316 \times 1111 \times 3 !
Option C (16×1111×316 \times 1111 \times 3 !) is correct because it calculates the sum of all numbers formed by the digits 1, 3, 5, 7 without repetition.
289 Easy
289 of 485
Application +4 | -1
The number of six digit numbers which have sum of their digits as an odd integer, is
View answer and explanation
Correct option: B
450000
Option B (450000) is correct because it accounts for the possibilities of odd and even sums of the first five digits and then chooses the last digit accordingly.
290 Easy
290 of 485
Application +4 | -1
If 12Pr−11P6+6.11P5{ }^{12} {P}_{{r}}-{ }^{11} {P}_6+6 .{ }^{11} {P}_5, then r{r} is equal to:
View answer and explanation
Correct option: A
6
Option A (6) is correct because 12Pr=11P6+6.11P5{ }^{12} {P}_{{r}}={ }^{11} {P}_6+6 .{ }^{11} {P}_5 simplifies to 12Pr=12P6{ }^{12}P_r = { }^{12}P_6, thus r=6r=6.
291 Easy
291 of 485
Application +4 | -1
The number of six digit numbers that can be formed from the digits 1,2,3,4,5,61,2,3,4,5,6 and 7 , so that digits do not repeat and the terminal digits are even is
View answer and explanation
Correct option: D
720
Option D (720) is correct because there are 3 choices for the first even digit, 2 for the last, and then 5P4 ways to arrange the remaining digits.
292 Easy
292 of 485
Application +4 | -1
The number of points (x,y,z)(x, y, z) in space, whose each coordinate is a negative integer such that x+y+z+12=0x+y+z+12=0, is
View answer and explanation
Correct option: B
55
Option B (55) is correct because it represents the number of points satisfying the given conditions, which translates to finding positive integral solutions of a related equation.
293 Easy
293 of 485
Application +4 | -1
Letter of the word INDIANOIL are arranged in all possible ways. The number of permutations in which A, I, O occur only at odd places, is
View answer and explanation
Correct option: C
240
Option C (240) is correct because it correctly calculates the number of arrangements where A, I, and O are only in odd places, considering the repetitions of I and N.
294 Easy
294 of 485
Application +4 | -1
The value of 50C4+∑r=16      56−rC3  ^{50}{C_4} + \sum \limits_{r = 1}^6 \;\;{\;^{56 - r}}{C_3}\; is
View answer and explanation
Correct option: A
  56C4{\;^{56}}{C_4}
Option A (  56C4{\;^{56}}{C_4}) is correct because the given summation can be simplified using the identity   nCr+  nCr−1=  n+1Cr{\;^n}{C_r} + {\;^n}{C_{r - 1}} = {\;^{n + 1}}{C_r} repeatedly.
295 Easy
295 of 485
Application +4 | -1
The sum of divisors of 2520 is
View answer and explanation
Correct option: A
9360
Option A (9360) is correct because it is the sum of all divisors, calculated by multiplying the sums of powers of each prime factor.
296 Easy
296 of 485
Application +4 | -1
The number of ways are arranging 19 persons, if two particular delegates never sit together
View answer and explanation
Correct option: D
19 !- 18 ! 2 !
Option D (19 !- 18 ! 2 !) is correct because it subtracts the number of arrangements where the two delegates are together (18! * 2!) from the total number of arrangements (19!).
297 Easy
297 of 485
Application +4 | -1
If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and arranged as in a dictionary, then the position of the word SMALL is
View answer and explanation
Correct option: D
58th
The correct answer is D (58th) because the rank of the word SMALL is determined by arranging all possible words in dictionary order.
298 Easy
298 of 485
Application +4 | -1
The number of zeros at the end of 100 ! Is
View answer and explanation
Correct option: C
24
Option C (24) is correct because it represents the highest power of 5 that divides 100!, which determines the number of trailing zeros.
299 Easy
299 of 485
Application +4 | -1
The total number of selections from 4 boys and 3 girls if each selection has to contain at least one boy is
View answer and explanation
Correct option: B
120
Option A (120) is correct because it represents the total number of selections with at least one boy.
300 Easy
300 of 485
Application +4 | -1
The total number of 8 digits numbers which have all different digits is
View answer and explanation
Correct option: A
3265920
Option A (3265920) is correct because we calculate the number of 8-digit numbers with distinct digits, considering that the first digit cannot be 0.
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