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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 211-240 of 742 questions

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211 Easy
211 of 742
Application +4 | -1
Let AA be a set of n(≥3)n(\geq 3) distinct elements. The number of triplets (x,y,z)(x, y, z) of the set AA in which atleast two coordinates are equal to
View answer and explanation
Correct option: B
n3−nP3n^3-{ }^n P_3
Option B (n3−nP3n^3-{ }^n P_3) is correct because it calculates the total number of triplets minus the number of triplets with all distinct elements.
212 Easy
212 of 742
Application +4 | -1
Out of 18 points in a plane no three are in the same straight line except five points which are collinear. The number of straight lines that can be formed joining them is
View answer and explanation
Correct option: B
144
Option B (144) is correct because it calculates the total number of lines minus the extra lines counted due to collinearity, plus one for the single line formed by the collinear points.
213 Easy
213 of 742
Application +4 | -1
Let TnT_n be the number of all possible triangles formed by joining vertices of an nn-sided regular polygon. If Tn+1−Tn=10T_{n+1}-T_n=10, the value of nn is
View answer and explanation
Correct option: A
5
Option A (5) is correct because the equation Tn+1−Tn=10{T_{n + 1}} - {T_n} = 10 simplifies to nC2=10{^n}{C_2} = 10, which is satisfied when n = 5.
214 Easy
214 of 742
Application +4 | -1
A seven-digit number made up of all distinct digits 8,7,6,4,2,x8,7,6,4,2, x and yy is divisible by 3. Then, possible number of order pair (x,y)(x, y) is
View answer and explanation
Correct option: B
8
Option B (8) is correct because there are 8 possible ordered pairs (x, y) that make the seven-digit number divisible by 3.
215 Easy
215 of 742
Application +4 | -1
If r,  s,  tr,\;s,\;t are prime numbers and p,qp,q are the positive integers such that LCM of p,  qp,\;q is r2s4t2{r^2}{s^4}{t^2}, then the number of ordered pairs (p,  q)\left( {p,\;q} \right) is
View answer and explanation
Correct option: C
225
Option C (225) is correct because the number of ordered pairs (p, q) is found by considering all possible combinations of the prime factors' exponents in the LCM.
216 Easy
216 of 742
Application +4 | -1
How many numbers greater than 40000 can be formed from the digits 2, 4, 5, 5, 7?
View answer and explanation
Correct option: D
48
Option D (48) is correct because it is calculated by subtracting the number of arrangements with 2 at the first place from the total number of arrangements.
217 Easy
217 of 742
Application +4 | -1
How many words can be made from the letters of the word ‘COMMITTEE’?
View answer and explanation
Correct option: B
9!(2!)3!\frac{{9!}}{{{{\left( {2!} \right)}^3}!}}
Option B (9!(2!)3!\frac{{9!}}{{{{\left( {2!} \right)}^3}!}}) is correct because it accounts for the repeated letters (M, T, E) in the word COMMITTEE.
218 Easy
218 of 742
Application +4 | -1
Every one of the 10 available lamps can be switched on to illuminate certain Hall. The total number of ways in which the hall can be illuminated, is
View answer and explanation
Correct option: B
10231023
Option B, 10231023, is correct because each lamp can be either on or off, giving 2102^{10} possibilities, but we must subtract 1 for the case where all lamps are off.
219 Medium
219 of 742
Application +4 | -1
Sixteen men compete with one another in running swimming and riding. How many prize lists could be made if there were altogether 6 prizes of different values, one for running, 2 for swimming and 3 for riding?
View answer and explanation
Correct option: B
163×152×14{16^3} \times {15^2} \times 14
Option B (163×152×14{16^3} \times {15^2} \times 14) is correct because it represents the product of the number of ways to award each prize.
220 Medium
220 of 742
Application +4 | -1
A person predicts the outcome of 20 cricket matches of his home team. Each match can result either in a win, loss or tie for the home team. Total number of ways in which he can make the predictions, so that exactly 10 predictions are correct, is equal to
View answer and explanation
Correct option: A
20C10⋅210{ }^{20} C_{10} \cdot 2^{10}
Option A (20C10⋅210{ }^{20} C_{10} \cdot 2^{10}) is correct because it represents choosing the 10 correct matches and then the possibilities for the incorrect ones.
221 Easy
221 of 742
Application +4 | -1
A four digit number of distinct digits is formed by using the digits 2,3,4,5,6,7,82,3,4,5,6,7,8. The number of such numbers which are divisible by 25, is
View answer and explanation
Correct option: B
40
Option B (40) is correct because it considers the two possible endings (25 and 75) and calculates the permutations for the remaining digits.
222 Easy
222 of 742
Application +4 | -1
The least positive integer nn for which n−1C5+n−1C6<nC7{ }^{n-1} C_5+{ }^{n-1} C_6 < { }^n C_7 is
View answer and explanation
Correct option: A
14
Option A (14) is correct because it is the smallest positive integer that satisfies the inequality n−1C5+n−1C6<nC7{ }^{n-1} C_5+{ }^{n-1} C_6 < { }^n C_7, which simplifies to n > 13.
223 Easy
223 of 742
Application +4 | -1
The exponent of 3 in 100 !, is
View answer and explanation
Correct option: C
48
Option A (72) is correct because it calculates the number of words formed with at least one vowel.
224 Medium
224 of 742
Application +4 | -1
Number of ways is which 50 identical books can be distributed in 100 students if each student can get atmost one book is
View answer and explanation
Correct option: A
(51)(52)…….(100)50!\frac{(51)(52) \ldots \ldots .(100)}{50 !}
Option A ((51)(52)...(100)/50!) is correct because it represents the number of ways to choose 50 students out of 100 to receive the books.
225 Easy
225 of 742
Understanding +4 | -1
There are 10 true-false questions. The number of ways, in which they can be answered is
View answer and explanation
Correct option: B
2102^{10}
Option B (2102^{10}) is correct because each of the 10 questions can be answered in 2 ways (true or false), leading to 2102^{10} total possibilities.
226 Easy
226 of 742
Understanding +4 | -1
The value of 10C5/11C6{ }^{10} C_5 /{ }^{11} C_6, when numerator and denominator takes their greatest value, is
View answer and explanation
Correct option: A
611\frac{6}{11}
Option A (611\frac{6}{11}) is correct because it is the simplified ratio of the two binomial coefficients.
227 Easy
227 of 742
Application +4 | -1
If all permutations of the letters of the word PENCIL are arranged as in a dictionary, then 413th word is
View answer and explanation
Correct option: D
LICPEN
Option D (LICPEN) is correct because after arranging the permutations of the letters in the word PENCIL in dictionary order, the 413th word is LICPEN.
228 Easy
228 of 742
Application +4 | -1
A seven-digit numbers divisible by 9 is to be formed by using 7 out of numbers {1,2,3,4,5,6,7,8,9}\{1,2,3,4,5, 6,7,8,9\}. The number of ways in which this can be done, is
View answer and explanation
Correct option: D
4⋅7!4 \cdot 7!
Option D (4⋅7!4 \cdot 7!) is correct because there are 4 pairs of digits that can be excluded to make the sum of the remaining 7 digits divisible by 9, and each set of 7 digits can be arranged in 7! ways.
229 Easy
229 of 742
Application +4 | -1
How many four digit numbers can be formed using the digits 1, 2, 3, 4, 5 such that at least one of the digit is repeated?
View answer and explanation
Correct option: D
54−5!{5^4} - 5!
Option D (54−5!{5^4} - 5!) is correct because it subtracts the number of four-digit numbers with no repetition from the total number of four-digit numbers that can be formed.
230 Easy
230 of 742
Application +4 | -1
A dictionary is printed consisting of 7 letters words only that can be made with a letters of the word CRICKET. If the words are printed at the alphabetical order as in an ordinary dictionary, then the number of words before the word CRICKET is
View answer and explanation
Correct option: A
530
Option A (530) is correct because it calculates the number of words before \"CRICKET\" by considering alphabetical order and accounting for repeated letters.
231 Easy
231 of 742
Application +4 | -1
The number of six-digit numbers that can be formed from the digits 1, 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
The correct answer is D (720) because it correctly calculates the number of six-digit numbers with even terminal digits from the given digits without repetition.
232 Medium
232 of 742
Application +4 | -1
The number of words of four letters containing equal number of vowels and consonants, repetition allowed, is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the calculated number of words does not match any of the given options.
233 Medium
233 of 742
Application +4 | -1
The number of words which can be formed the letters of the word MAXIMUM, if two consonants cannot occur together, is
View answer and explanation
Correct option: A
4!
Option A (4!) is correct because the vowels must occupy alternate positions, and the consonants can then be arranged in the remaining spaces.
234 Easy
234 of 742
Application +4 | -1
The number of ways in which pp positive and nn negative signs may be placed in a row so that no two negative signs shall be together is
View answer and explanation
Correct option: A
p+1Cn{ }^{p+1} {C}_n
Option A (p+1Cn{}^{p+1} {C}_n) is correct because it represents the number of ways to place negative signs such that no two are together.
235 Easy
235 of 742
Analyzing +4 | -1
The number of 6 digits numbers that can be formed using the digits 0, 1, 2, 5, 7 and 9 which are divisible by 11 and no digit is repeated, is
View answer and explanation
Correct option: A
60
Option A (60) is correct because it is the total number of 6-digit numbers that can be formed using the given digits, divisible by 11, with no repetition.
236 Easy
236 of 742
Understanding +4 | -1
Let SS be the set of all functions from the set AA to the set AA. If n(A)=kn(A)=k, then n(S)n(S) is
View answer and explanation
Correct option: B
kkk^{k}
Option B (kkk^{k}) is correct because each of the k elements in A can be mapped to any of the k elements in A.
237 Easy
237 of 742
Application +4 | -1
If 10Cr−1>2(10Cr){ }^{10} C_{{r}-1} > 2\left({ }^{10} C_{\mathrm{r}}\right) then rr is equal to
View answer and explanation
Correct option: C
8, 9, 10
Option C (8, 9, 10) is correct because these values of r satisfy the inequality 10Cr−1>2(10Cr){}^{10} C_{{r}-1} > 2\left({}^{10} C_{\mathrm{r}}\right) after simplification.
238 Easy
238 of 742
Application +4 | -1
If x∈Nx \in {N} and (2x)!3!(2x−3)!:x!2!(x−2)=44:3\frac{(2 x) !}{3 !(2 x-3) !}: \frac{x !}{2 !(x-2)}=44: 3, then xx is equal to
View answer and explanation
Correct option: A
6
Option A (6) is correct because solving the given ratio of permutations leads to the equation 2x-1=11, which gives x=6.
239 Easy
239 of 742
Application +4 | -1
A basket ball team consists of 12 pairs of twin brothers. On the first day of training, all 24 players stand in a circle in such a way that all pairs of twin brothers are neighbours. Number of ways this can be done is :
View answer and explanation
Correct option: B
212(11!)2^{12}(11 !)
Option B (212(11!)2^{12}(11 !)) is correct because there are 11! ways to arrange the 12 pairs in a circle, and each pair can be arranged in 2 ways.
240 Medium
240 of 742
Application +4 | -1
A variable name in certain computer language must be either an alphabet or an alphabet followed by a decimal digit. The total number of different variable names that can exist in that language is equal to
View answer and explanation
Correct option: C
286
Option C (286) is correct because it correctly calculates the total number of variable names by considering alphabets and alphabets followed by digits.
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