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JEE • MATHEMATICS • PERMUTATION AND COMBINATION

Permutation and combination practice · Mixed difficulty

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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 301-330 of 742 questions

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301 Easy
301 of 742
Application +4 | -1
If there are nn number of seats and m number of people have to be seated, then how many ways are possible to do this (m<n)(m < n) ?
View answer and explanation
Correct option: A
  nPm{\;^n}{P_m}
Option A (  nPm{\;^n}{P_m}) is correct because the number of ways to seat mm people in nn seats is given by the permutation formula   nPm{\;^n}{P_m}.
302 Medium
302 of 742
Application +4 | -1
The number of ways five alphabets can be chosen from the alphabets of the word MATHEMATICS, where the chosen alphabets are not necessarily distinct, is equal to:
View answer and explanation
Correct option: D
179
Option D (179) is correct because it accounts for all possible combinations of selecting five alphabets from the word MATHEMATICS, considering repetitions.
303 Easy
303 of 742
Application +4 | -1
How many numbers between 400 and 1000 can be made with the digits 2, 3, 4, 5, 6 and 0?
View answer and explanation
Correct option: C
60
Option C (60) is correct because it calculates the number of 3-digit numbers between 400 and 1000 using the given digits, considering the restriction on the hundreds place.
304 Easy
304 of 742
Application +4 | -1
Three alike red roses and three alike pink roses are used to form a garland. The number of ways is
View answer and explanation
Correct option: A
3
Option A (3) is correct because there are only three distinct arrangements when considering rotational symmetry of the garland.
305 Easy
305 of 742
Application +4 | -1
How many numbers consisting of 5 digits can be formed in which the digits 3, 4 and 7 are used only once and the digit 5 is used twice?
View answer and explanation
Correct option: B
60
Option B (60) is correct because the number of such numbers is calculated by arranging the 5 digits, considering the repetition of the digit 5.
306 Easy
306 of 742
Application +4 | -1
The number of ways of arranging 8 men and 4 women around a circular table such that no two women can sit together, is
View answer and explanation
Correct option: D
7!  8P47!{\;^8}{P_4}
Option D (7!  8P47!{\;^8}{P_4}) is correct because it first arranges the men around the table and then arranges the women in the gaps between the men.
307 Easy
307 of 742
Application +4 | -1
The number of straight lines that can be drawn out of 10 points of which 7 are collinear
View answer and explanation
Correct option: D
24
Option D (24) is correct because it calculates the total number of lines that can be formed, accounting for the collinearity of 7 points.
308 Easy
308 of 742
Application +4 | -1
A letter lock consists of three rings each marked with 10 different letters. No. of ways to make an unsuccessful attempts to open the lock is
View answer and explanation
Correct option: B
999
Option B (999) is correct because it subtracts the one successful attempt from the total possible combinations.
309 Easy
309 of 742
Application +4 | -1
A five digit number divisible by 3 is to be formed using the numbers 0, 1, 2, 3, 4 and 5, without repetition. The total number of ways this can be done, is
View answer and explanation
Correct option: A
216
Option A (216) is correct because it counts the number of 5-digit numbers divisible by 3 that can be formed using the given digits without repetition.
310 Easy
310 of 742
Application +4 | -1
There are nn number of sets and m number of people have to be seated, then how many ways are possible to do this (m<n)?\left( {m < n} \right)?
View answer and explanation
Correct option: A
  nPm{\;^n}{P_m}
Option A (  nPm{\;^n}{P_m}) is correct because it represents the number of ways to select and seat m people from n sets, considering the order.
311 Easy
311 of 742
Application +4 | -1
The number of ways of dividing 10 girls into two groups of 5 each so that two shortest girls are in the different groups is:
View answer and explanation
Correct option: A
70
Option A (70) is correct because after placing the two shortest girls in different groups, we divide the remaining 8 girls into two groups of 4, resulting in 70 ways.
312 Easy
312 of 742
Application +4 | -1
There are 5 roads leading to a town from a village. The number of different ways in which a villager can go to the town and return back, is
View answer and explanation
Correct option: A
25
Option A (25) is correct because it is the product of the number of ways to go to town and the number of ways to return.
313 Easy
313 of 742
Application +4 | -1
Suppose p∈N,n=PC2p \in {N}, n={ }^{{P}} C_2 and m=nC2m={ }^{{n}} C_2. If 4m:n=4 m: n= 18:118: 1, then pp is equal to
View answer and explanation
Correct option: D
5
Option D (5) is correct because substituting the given expressions for n and m into the ratio and solving for p yields p=5.
314 Easy
314 of 742
Application +4 | -1
In an examination, there are 11 papers. A candidate has to pass in at least 6 papers to pass the examination. The number of ways in which the candidate can pass the examination is:
View answer and explanation
Correct option: B
2102^{10}
Option B (2102^{10}) is correct because the number of ways to pass is the sum of combinations from passing 6 to 11 papers, which simplifies to 2102^{10}.
315 Easy
315 of 742
Application +4 | -1
All letters of the word AGAIN are permuted in all possible ways and the words so formed (with or without meaning) are written as in dictionary, then the 50th word is
View answer and explanation
Correct option: C
NAAIG
Option C (NAAIG) is correct because listing the words in dictionary order reveals that the 50th word is NAAIG.
316 Easy
316 of 742
Application +4 | -1
The remainder when x=1!+2!+3!+4!+…+100x = 1! + 2! + 3! + 4!+\ldots+100 ! is divided by 240, is
View answer and explanation
Correct option: A
153
Option A (153) is correct because for factorials greater than 5!, the remainder when divided by 240 is 0, so we only need to consider the sum of the first five factorials.
317 Easy
317 of 742
Application +4 | -1
Out of 10 red and 9 white balls, 5 red and 4 white balls can be drawn in number of ways ==
View answer and explanation
Correct option: A
10C5×10C4{ }^{10} {C}_5 \times{ }^{10} {C}_4
Option A (10C5×10C4{ }^{10} {C}_5 \times{ }^{10} {C}_4) is correct because it represents the number of ways to choose 5 red balls from 10 and 4 white balls from 10.
318 Easy
318 of 742
Application +4 | -1
Four dice are rolled. The number of possible outcomes in which at least one dice shows 2 is
View answer and explanation
Correct option: B
671
The correct answer is 671 because it is the number of possible outcomes when rolling four dice with at least one dice showing a 2.
319 Medium
319 of 742
Application +4 | -1
There are pp letters a,qa, q letters b,rb, r letters cc. The number of ways of selecting kk letters out of these if p<k<q<rp < k < q < r is
View answer and explanation
Correct option: B
12(p+1)(2k−p)\frac{1}{2}(p+1)(2 k-p)
Option B (12(p+1)(2k−p)\frac{1}{2}(p+1)(2 k-p)) is correct because it represents the number of ways of selecting k letters out of p 'a's, q 'b's, and r 'c's, given p < k < q < r.
320 Easy
320 of 742
Application +4 | -1
AA is a set containing nn elements. A subset PP of AA is chosen. The set AA is reconstructed by replacing the elements of PP. A subset QQ of AA is again chosen. The number of ways of choosing PP and QQ, so that P∩Q=ϕP \cap Q=\phi, is
View answer and explanation
Correct option: D
3n3^n
Option D (3n3^n) is correct because each element in set A has three possibilities: belonging to P only, Q only, or neither, ensuring P and Q have no intersection.
321 Easy
321 of 742
Application +4 | -1
On a railway route there are 15 stations. The number of tickets required in order that it may be possible to book a passenger from every station to every other is
View answer and explanation
Correct option: B
210
Option B (210) is correct because it calculates the number of permutations of choosing 2 stations out of 15, which represents the number of tickets needed from every station to every other station.
322 Easy
322 of 742
Analyzing +4 | -1
If 56Pr+6:54Pr+3=30800:1{ }^{56} P_{r+6}:{ }^{54} P_{r+3}=30800: 1, then rr is equal to
View answer and explanation
Correct option: C
41
The correct answer is C (41) because solving the ratio of permutations leads to the equation 56 * 55 * (51-r) = 30800, which simplifies to r = 41.
323 Easy
323 of 742
Application +4 | -1
The number of arrangements of the word "IDIOTS" such that vowels are at the places which from three consecutive terms of an A.P. is :
View answer and explanation
Correct option: D
108
Option D (108) is correct because it considers all possible arrangements of vowels in A.P. and the arrangements of consonants.
324 Medium
324 of 742
Analyzing +4 | -1
If S=(1)(1!)+(2)(2!)+(3)(3!)+…+n(n!)S=(1)(1 !)+(2)(2 !)+(3)(3 !)+\ldots+n(n !), then
View answer and explanation
Correct option: A
S+1n!∈\frac{S+1}{n !} \in integer
Option A (S+1n!∈\frac{{S+1}}{{n !}} \in integer) is correct because the expression S can be simplified to (n+1)! - 1, making (S+1) equal to (n+1)!, which is divisible by n!.
325 Easy
325 of 742
Application +4 | -1
How many 5-digit telephone numbers can be constructed using the digits 0 to 9, if each number starts with 67 and no digit appears more than once?
View answer and explanation
Correct option: B
336
Option B (336) is correct because after fixing the first two digits, we arrange the remaining 3 digits from the remaining 8 digits without repetition.
326 Medium
326 of 742
Application +4 | -1
A three digit number nn is such that the last two digits of it are equal and differ from the first. The number of such n′n' is
View answer and explanation
Correct option: C
81
Option C (81) is correct because there are 9 choices for the first digit (1-9) and 9 choices for the last two digits (00, 11, ..., 99) excluding the case where the first digit is equal to the last two.
327 Easy
327 of 742
Application +4 | -1
The number of signals that can be sent using 5 flags of different colours, taking one or more at a time, is
View answer and explanation
Correct option: D
325
Option D (325) is correct because it sums the permutations of selecting 1, 2, 3, 4, or 5 flags from the 5 available.
328 Easy
328 of 742
Application +4 | -1
The number of ways can 10 letters be placed in 10 marked envelopes, so that no letter is in the right envelope are
View answer and explanation
Correct option: A
10!(1−11!+12!−13!+...+110!)10!\left( {1 - \frac{1}{{1!}} + \frac{1}{{2!}} - \frac{1}{{3!}} + ... + \frac{1}{{10!}}} \right)
Option A (10!(1−11!+12!−13!+...+110!)10!\left( {1 - \frac{1}{{1!}} + \frac{1}{{2!}} - \frac{1}{{3!}} + ... + \frac{1}{{10!}}} \right)) is correct because it represents the number of derangements of 10 objects.
329 Medium
329 of 742
Application +4 | -1
Total number of positive integral solutions of x1+x3+2x3=15x_1+x_3+2 x_3=15 is equal to
View answer and explanation
Correct option: A
42
Option A (42) is correct because it represents the total number of positive integral solutions for the given equation after considering the possible values of x3.
330 Easy
330 of 742
Application +4 | -1
If the four letter words (need not be meaningful) are to be formed using the letters from the word "MEDITERRANEAN" such that the first letter is R and the fourth letter is EE, then the total number of all such words is :
View answer and explanation
Correct option: B
59
Option B (59) is correct because it accounts for the 3 possibilities with identical letters (EE, AA, NN) and the 56 possibilities with distinct letters, summing to 59.
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