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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 211-240 of 485 questions

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211 Easy
211 of 485
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.
212 Easy
212 of 485
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.
213 Easy
213 of 485
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.
214 Easy
214 of 485
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}.
215 Easy
215 of 485
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.
216 Easy
216 of 485
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.
217 Easy
217 of 485
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.
218 Easy
218 of 485
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.
219 Easy
219 of 485
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.
220 Easy
220 of 485
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.
221 Easy
221 of 485
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.
222 Easy
222 of 485
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 :
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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.
223 Easy
223 of 485
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.
224 Easy
224 of 485
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.
225 Easy
225 of 485
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.
226 Easy
226 of 485
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.
227 Easy
227 of 485
Application +4 | -1
An urn contains 3 red pens, 4 green pens and 6 yellow pens. The number of ways of drawing 4 pens from the urn, if atleast one red pen is to be included in the draw, is (all the pens are different from each other)
View answer and explanation
Correct option: B
505
Option B (505) is correct because it calculates the total ways to choose 4 pens minus the ways to choose 4 pens without any red pens, which is 13C4−10C4=505{}^{13}C_4 - {}^{10}C_4 = 505.
228 Easy
228 of 485
Application +4 | -1
The number of functions ff from the set A={0,1,2}A=\{0,1,2\} into the set B={0,1,2,3,4,5,6,7}B=\{0,1,2,3,4,5,6,7\} such that f(i)≤f(j)f(i) \leq f(j) for i<ji < j and i,j∈Ai, j \in A, is
View answer and explanation
Correct option: C
10C3{ }^{10} C_3
Option C (10C3{ }^{10} C_3) is correct because it accurately represents the number of non-decreasing functions from A to B using combinations with repetition.
229 Easy
229 of 485
Application +4 | -1
How many words can be made from the letters of the word DELHI, if LL comes in the middle of every word?
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Correct option: B
24
Option B (24) is correct because the number of words is calculated by arranging the remaining letters around the fixed letter L.
230 Easy
230 of 485
Application +4 | -1
Let T4\mathrm{T}_4 denotes the number of triangles which can be formed by using the vertices of a regular polygon of nn sides, If Tn+1−Tn=21T_{n+1}-T_n=21, then nn equals.
View answer and explanation
Correct option: A
7
Option A (7) is correct because solving the equation Tn+1−Tn=21T_{n+1} - T_n = 21 where Tn=nC3T_n = nC3 leads to n = 7.
231 Easy
231 of 485
Application +4 | -1
The number of diagonals that can be drawn by joining the vertices of an octagon is
View answer and explanation
Correct option: C
20
Option A (20) is correct because it calculates the number of ways to connect vertices minus the number of sides of the octagon.
232 Easy
232 of 485
Application +4 | -1
Find the number of chains that can be prepared using 5 different colored beads
View answer and explanation
Correct option: D
12×4\frac{1}{2} \times 4 !
Option D (12×4\frac{1}{2} \times 4 !) is correct because the number of ways to arrange n distinct objects in a circle is (n-1)!, and for a chain, we divide by 2 to account for flipping it over.
233 Easy
233 of 485
Application +4 | -1
There are three piles of identical yellow, black and green balls and each pile contains at least 20 balls. The number of ways of selecting 20 balls if the number of black balls to be selected is twice the number of yellow balls, is
View answer and explanation
Correct option: B
7
Option B (7) is correct because it represents the number of possible values for the number of yellow balls that satisfy the given conditions.
234 Easy
234 of 485
Application +4 | -1
nn lines are drawn in a plane such that no two of them are parallel and no three of them are concurrent. The number of different points at which these lines will cut, is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the number of intersection points is given by nC2{}^nC_2, which is not equal to any of the given options.
235 Easy
235 of 485
Application +4 | -1
5 different objects are to be distributed among 3 persons such that no two persons get the same number of objects. Number of ways this can be done, is
View answer and explanation
Correct option: B
90
Option B (90) is correct because the total number of ways to distribute the objects is the sum of the ways for the two possible distributions (0,1,4) and (0,2,3).
236 Easy
236 of 485
Application +4 | -1
The number of ways by which letter of word ASSASSINATION can be arranged such that all vowels come together is:
View answer and explanation
Correct option: C
8!6!4!(2!)23!\frac{8 ! 6 !}{4 !(2 !)^2 3 !}
Option D (8!6!4!3!2!\frac{{8 ! 6 !}}{{4 ! 3 ! 2 !}}) is correct because it represents the number of arrangements of the letters in the word ASSASSINATION with all vowels together, considering repetitions.
237 Easy
237 of 485
Application +4 | -1
A telegraph has 5 arms and each arm is capable of 4 distinct positions, including the position of rest. The total number of signals that can be made is
View answer and explanation
Correct option: B
1023
Option B (1023) is correct because each arm has 4 possible states, and we subtract 1 to exclude the case where all arms are at rest.
238 Easy
238 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: C
58th58 \mathrm{th}
Option C (58th) is correct because it correctly calculates the number of words that come before SMALL in the dictionary order and adds 1.
239 Easy
239 of 485
Application +4 | -1
The total number of words that can be formed using all letters of the word 'RITESH' that neither begins with I nor ends with R\mathrm{R}, is
View answer and explanation
Correct option: A
504
Option A (504) is correct because it is the total number of arrangements minus those starting with I or ending with R, plus those starting with I and ending with R.
240 Easy
240 of 485
Application +4 | -1
The number of ways in which 10 boys and 10 girls can be seated in a row so that boys and girls alternate is
View answer and explanation
Correct option: A
3628800
Option D (None of these) is correct because the correct answer is 2 * (10!)2 which is not among the options.
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