# JEE Main Mathematics Permutations and Combinations Online Test

## JEE Main Mathematics Permutations and Combinations Online Test

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JEE Main Mathematics Permutations and Combinations Online Test. JEE Main Online Test for Mathematics Permutations and Combinations. JEE Main Full Online Quiz **for Mathematics Permutations and Combinations**. **JEE Main Free Mock Test Paper 2018.** JEE Main 2018 Free Online Practice Test, Take JEE Online Test for All Subjects. JEE Main Question and Answers for Mathematics Permutations and Combinations. In this test You may find JEE Main all subjects Questions with Answers. Check JEE Main Question and Answers in English. This mock Test is free for All Students. Mock Test Papers are very helpful for Exam Purpose, by using below Mock Test Paper you may Test your Study for Next Upcoming Exams. Now Scroll down below n **Take JEE Main Mathematics Permutations and Combinations…**

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- Question 1 of 30
##### 1. Question

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?

CorrectNumber of ways to fill the IIIrd place = 8

Number of ways to fill the IVth place = 7

Number of ways to fill the Vth place = 6

Hence, by FPC total number of ways = 8 × 7 × 6 = 56 × 6 = 336

IncorrectNumber of ways to fill the IIIrd place = 8

Number of ways to fill the IVth place = 7

Number of ways to fill the Vth place = 6

Hence, by FPC total number of ways = 8 × 7 × 6 = 56 × 6 = 336

- Question 2 of 30
##### 2. Question

If the number of ways of selecting n cards out of unlimited number of cards bearing the number 0, 9, 3, so that they cannot be used to write the number 903 is 93, then n is equal to

CorrectWe cannot write 903

If in the selection of n cards, we get either

(9 or 3), (9 or 0), (0 or 3), (only 0), (only 3) or (only 9)

Can be selected = 2 × 2 × 2 × ……… n times = 2

^{n}Similarly, (9 or 0) or (3 or 0) can be selected 2

^{n}In the above selection (only 0) or (only 3) or (only 9) is repeated twice

∴ Total ways = 2

^{n}+ 2^{n}+ 2^{n}– 3 = 93⇒ 3.2

^{n}= 96⇒ 2

^{n}= 32 = 2^{5}∴ n = 5

IncorrectWe cannot write 903

If in the selection of n cards, we get either

(9 or 3), (9 or 0), (0 or 3), (only 0), (only 3) or (only 9)

Can be selected = 2 × 2 × 2 × ……… n times = 2

^{n}Similarly, (9 or 0) or (3 or 0) can be selected 2

^{n}In the above selection (only 0) or (only 3) or (only 9) is repeated twice

∴ Total ways = 2

^{n}+ 2^{n}+ 2^{n}– 3 = 93⇒ 3.2

^{n}= 96⇒ 2

^{n}= 32 = 2^{5}∴ n = 5

- Question 3 of 30
##### 3. Question

Total number of works formed by using 2 vowels and 3 consonants taken from 4 vowels and 5 consonants is equal to

Correct^{4}C_{2}×^{5}C_{3}× 5! = 7200Incorrect^{4}C_{2}×^{5}C_{3}× 5! = 7200 - Question 4 of 30
##### 4. Question

In a shelf there are 5 physics, 4 mathematics and 3 chemistry books. How many combinations are there if books of same subject are different

Correct2

^{5}× 2^{4}× 2^{3}– 1 = 2^{12}– 1 = 4095Incorrect2

^{5}× 2^{4}× 2^{3}– 1 = 2^{12}– 1 = 4095 - Question 5 of 30
##### 5. Question

Find the number of different words that can be formed from the letters of the word TRIANGLE, so that no vowels are together.

CorrectIn a word TRIANGLE, vowels are (A,E, I) and consonants are (G, L, N, R, T).

First, we fix the 5 consonants in alternate position in 5! ways.

_G_L_N_R_T_

In rest of the six blank positions, three vowels can be arranged in

^{6}P_{3}ways.∴ Total number of different words = 5! ×

^{6}P_{3}= 120 × 6 × 5 × 4 = 14400

IncorrectIn a word TRIANGLE, vowels are (A,E, I) and consonants are (G, L, N, R, T).

First, we fix the 5 consonants in alternate position in 5! ways.

_G_L_N_R_T_

In rest of the six blank positions, three vowels can be arranged in

^{6}P_{3}ways.∴ Total number of different words = 5! ×

^{6}P_{3}= 120 × 6 × 5 × 4 = 14400

- Question 6 of 30
##### 6. Question

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

CorrectThe number of words starting from A are 5! = 120

The number of words starting from 1 are = 5! = 120

The number of words starting from KA are = 4! = 24

The number of words starting from KI are = 4! = 24

The number of words starting from KN are 4! = 24

The number of words starting from KRA are = 3! = 6

The number of words starting from KRIA are 2! = 2

The number of words starting from KRIN are = 2! = 2

The number of words starting from KRISA are = 1! = 1

The number of words starting from KRISNA are = 1! = 1

Hence, rank of the word KRISNA

= 2(120) + 3 (24) + 6 + 2(24) + 6 + 2(2) + (1) = 324

IncorrectThe number of words starting from A are 5! = 120

The number of words starting from 1 are = 5! = 120

The number of words starting from KA are = 4! = 24

The number of words starting from KI are = 4! = 24

The number of words starting from KN are 4! = 24

The number of words starting from KRA are = 3! = 6

The number of words starting from KRIA are 2! = 2

The number of words starting from KRIN are = 2! = 2

The number of words starting from KRISA are = 1! = 1

The number of words starting from KRISNA are = 1! = 1

Hence, rank of the word KRISNA

= 2(120) + 3 (24) + 6 + 2(24) + 6 + 2(2) + (1) = 324

- Question 7 of 30
##### 7. Question

The number of divisors of 9600 including 1 and 9600 are

Correct9600 = 2

^{7}× 3 × 5^{2}∴ No. of divisors = (7 + 1) × (1 + 1) × (2 + 1) = 8 × 2 × 3 = 48

Incorrect9600 = 2

^{7}× 3 × 5^{2}∴ No. of divisors = (7 + 1) × (1 + 1) × (2 + 1) = 8 × 2 × 3 = 48

- Question 8 of 30
##### 8. Question

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

Correct2

^{10}– 1 = 1023Incorrect2

^{10}– 1 = 1023 - Question 9 of 30
##### 9. Question

How many numbers lying between 10 and 1000 can be formed from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 (repetition of digits is allowed)?

Correct**Case I**when number of two digits.Total number of ways =

^{9}C_{1}×^{9}C_{1}= 9 × 9 = 81**Case II**when number in three digitsTotal number of ways =

^{9}C_{1}×^{9}C_{1}×^{9}C_{1}= 9 × 9 × 9 = 729∴ Total number of ways = 81 + 729 = 810

Incorrect**Case I**when number of two digits.Total number of ways =

^{9}C_{1}×^{9}C_{1}= 9 × 9 = 81**Case II**when number in three digitsTotal number of ways =

^{9}C_{1}×^{9}C_{1}×^{9}C_{1}= 9 × 9 × 9 = 729∴ Total number of ways = 81 + 729 = 810

- Question 10 of 30
##### 10. Question

The number of diagonals that can be drawn in an octagon is

Correct^{8}C_{2}–8 = 28 – 8 = 20Incorrect^{8}C_{2}–8 = 28 – 8 = 20 - Question 11 of 30
##### 11. Question

The number of ways in which a team of eleven players can be selected from 22 players always including 2 of them and excluding 4 of them is

Correct∴ Required number of ways =

^{22–4–2}C_{11–2}=^{16}C_{9}Incorrect∴ Required number of ways =

^{22–4–2}C_{11–2}=^{16}C_{9} - Question 12 of 30
##### 12. Question

Three boys of class X, four boys of class XI and five boys of class XII sit in a row. The total number of ways in which these boys can sit together is equal to

CorrectWe can think of three packets. One consisting of three boys of class X, other consisting of four boys of class XI and last one consisting of five boys of class XII. These packets can be arranged in 3! ways and contents of these packets can be further arranged in 3!4! and 5! ways, respectively. Hence, the total number of ways is 3! × 3! × 4! × 5!

IncorrectWe can think of three packets. One consisting of three boys of class X, other consisting of four boys of class XI and last one consisting of five boys of class XII. These packets can be arranged in 3! ways and contents of these packets can be further arranged in 3!4! and 5! ways, respectively. Hence, the total number of ways is 3! × 3! × 4! × 5!

- Question 13 of 30
##### 13. Question

The total number of seven digit numbers the sum of whose digits is even, is

CorrectSuppose a

_{1}a_{2}a_{3}a_{4}a_{5}a_{6}a_{7}represents a seven digit number. Then a_{1}takes the value 1, 2, 3, …… , 9 and a_{2},a_{3}, ….. , a_{7}all take values 0, 1, 2, 3, …., 9. If we keep a_{1}a_{2}a_{3}…. a_{6}fixed, then the sum a_{1}+a_{2}+ a_{3}+ … …. + a_{6}is either even or odd. Since a_{7}takes 10 values 0, 1, 2, ……, 9 five of the numbers so formed will be even and five odd.Hence, the required number of numbers

= 9∙10 ∙10 ∙10∙10∙10∙5 = 45 × 10

^{5}IncorrectSuppose a

_{1}a_{2}a_{3}a_{4}a_{5}a_{6}a_{7}represents a seven digit number. Then a_{1}takes the value 1, 2, 3, …… , 9 and a_{2},a_{3}, ….. , a_{7}all take values 0, 1, 2, 3, …., 9. If we keep a_{1}a_{2}a_{3}…. a_{6}fixed, then the sum a_{1}+a_{2}+ a_{3}+ … …. + a_{6}is either even or odd. Since a_{7}takes 10 values 0, 1, 2, ……, 9 five of the numbers so formed will be even and five odd.Hence, the required number of numbers

= 9∙10 ∙10 ∙10∙10∙10∙5 = 45 × 10

^{5} - Question 14 of 30
##### 14. Question

The sum of the digits in unit place of all the numbers formed with the help of 3, 4, 5 and 6 taken all at a time is

CorrectThe sum of the digit in unit place of all the numbers formed = 3! (3 + 4 + 5 + 6) = 6 × 18

= 108

IncorrectThe sum of the digit in unit place of all the numbers formed = 3! (3 + 4 + 5 + 6) = 6 × 18

= 108

- Question 15 of 30
##### 15. Question

An n digit number is a positive number with exactly n digits. Nine hundred distinct n digit numbers are to be formed using only the three digits 2, 5 and 7. The smallest value of n for which this is possible is

CorrectEach place can be filled in 3 ways

∴ n places can be filled in 3

^{n}∴ 3

^{n}≥ 900∴ 3

^{6}= 729 and 3^{7}= 2187Hence n = 7

IncorrectEach place can be filled in 3 ways

∴ n places can be filled in 3

^{n}∴ 3

^{n}≥ 900∴ 3

^{6}= 729 and 3^{7}= 2187Hence n = 7

- Question 16 of 30
##### 16. Question

Three boys and three girls are to be seated around a table, in a circle. Among them the boy x does not want any girl neighbor and the girl y does not want any boy neighbor. The number of such arrangement is

Correctx, 1, 2, are boys and y, 3, 4 are girls

No. of arrangement = 2! × 2! = 4

Incorrectx, 1, 2, are boys and y, 3, 4 are girls

No. of arrangement = 2! × 2! = 4

- Question 17 of 30
##### 17. Question

The number of ways in which we can choose a committee from four men and six women so that the committee includes at least two men and exactly twice as many women as men is

CorrectThe number of ways in which we can choose a committee

= Choose two men and four women + choose three men and six women

=

^{4}C_{2}×^{6}C_{4}+^{4}C_{3}×^{6}C_{6}= 6 × 15 + 4 × 1 = 90 + 4 = 94

IncorrectThe number of ways in which we can choose a committee

= Choose two men and four women + choose three men and six women

=

^{4}C_{2}×^{6}C_{4}+^{4}C_{3}×^{6}C_{6}= 6 × 15 + 4 × 1 = 90 + 4 = 94

- Question 18 of 30
##### 18. Question

A question paper is divided into two parts A and B and each part contains 5 questions. The number of ways in which a candidate can answer 6 questions selecting at least two questions from each part is

CorrectThe number of ways that the candidate may select

(i) if 2 questions from A and 4 questions from B =

^{5}C_{2}×^{5}C_{4}= 50(ii) 3 questions from A and 3 questions from B =

^{5}C_{2}×^{5}C_{3}= 100and (iii) 4 questions from A and 2 questions from B =

^{5}C_{4}×^{5}C_{2}= 50Hence, total number of ways = 50 + 100 + 50 = 200

IncorrectThe number of ways that the candidate may select

(i) if 2 questions from A and 4 questions from B =

^{5}C_{2}×^{5}C_{4}= 50(ii) 3 questions from A and 3 questions from B =

^{5}C_{2}×^{5}C_{3}= 100and (iii) 4 questions from A and 2 questions from B =

^{5}C_{4}×^{5}C_{2}= 50Hence, total number of ways = 50 + 100 + 50 = 200

- Question 19 of 30
##### 19. Question

No. of ways of arranging exactly 4 fruits from 5 apples, 4 mangoes and 2 oranges.

CorrectCoeff. of x

^{4}in : (1 + x + x^{2}+ x^{3}+ x^{4}+ x^{5}) (1 + x +x^{2}+ x^{3}+x^{4}) (1 +x + x^{2}): (1–x

^{6}) (1 –x^{5}) (1–x^{3}) (1–x)^{–3}: [1–x

^{3}] [1 – x]^{–3}: [

^{3+}^{4–1}C_{4}–^{3+}^{4–1}C_{1}]: 15 – 3

: 12

IncorrectCoeff. of x

^{4}in : (1 + x + x^{2}+ x^{3}+ x^{4}+ x^{5}) (1 + x +x^{2}+ x^{3}+x^{4}) (1 +x + x^{2}): (1–x

^{6}) (1 –x^{5}) (1–x^{3}) (1–x)^{–3}: [1–x

^{3}] [1 – x]^{–3}: [

^{3+}^{4–1}C_{4}–^{3+}^{4–1}C_{1}]: 15 – 3

: 12

- Question 20 of 30
##### 20. Question

In an steamer, there are stalls for 12 animals and there are horses, cows and calves (not less than 12 each) ready to be shipped in how many ways can the ship load be made?

CorrectFirst stall can be filled in 3 ways, second stall can be filled in 3 ways and so on.

∴ Number of ways of loading steamer

=

^{3}C_{1}×^{3}C_{1}× …..×^{3}C_{1}(12 times)= 3 × 3 × ……× 3 (12 times) = 3

^{12}IncorrectFirst stall can be filled in 3 ways, second stall can be filled in 3 ways and so on.

∴ Number of ways of loading steamer

=

^{3}C_{1}×^{3}C_{1}× …..×^{3}C_{1}(12 times)= 3 × 3 × ……× 3 (12 times) = 3

^{12} - Question 21 of 30
##### 21. Question

The number of different ways the letters of the word VECTOR can be placed in the 8 boxes of the given below such that no row empty is equal to

CorrectTotal ways = (8C2 –2) × ⌊6

= (28 – 2) × ⌊6

= 26 × ⌊6

IncorrectTotal ways = (8C2 –2) × ⌊6

= (28 – 2) × ⌊6

= 26 × ⌊6

- Question 22 of 30
##### 22. Question

The prime ministers of 9 countries meet to discuss the terrorism problem. The number of ways they can seat themselves at a round table so that the Pakistan and India prime ministers sit together is (Pearson)

CorrectTreat India and Pakistan prime ministers as one (I, P) + 7 others we have to arrange 8 person round a table no. of ways = 7! but corresponding to each arrangement India and Pakistan prime ministers total way = 7!.2!

IncorrectTreat India and Pakistan prime ministers as one (I, P) + 7 others we have to arrange 8 person round a table no. of ways = 7! but corresponding to each arrangement India and Pakistan prime ministers total way = 7!.2!

- Question 23 of 30
##### 23. Question

The number of mappings (functions) from the set A = {1, 2, 3} into the set B = {1, 2, 3, 4, 5, 6, 7} such that

*f(i)*, whenever__<__f(j)*i < j*, isCorrectIf the function is one one, then select any three from the set B in

^{7}C_{3}ways i.e., 35 ways.If the function is many one, then there are two possibilities, All three corresponds to same element number of such functions =

^{7}C_{1}= 7 ways. Two corresponds to same element. Select any two from the set B. The larger one corresponds to the larger and the smaller one corresponds to the smaller the third may corresponds to any two. Number of such functions =^{7}C_{2}× 2 = 42So, the required number of mapping = 354 + 7 + 42 = 84

IncorrectIf the function is one one, then select any three from the set B in

^{7}C_{3}ways i.e., 35 ways.If the function is many one, then there are two possibilities, All three corresponds to same element number of such functions =

^{7}C_{1}= 7 ways. Two corresponds to same element. Select any two from the set B. The larger one corresponds to the larger and the smaller one corresponds to the smaller the third may corresponds to any two. Number of such functions =^{7}C_{2}× 2 = 42So, the required number of mapping = 354 + 7 + 42 = 84

- Question 24 of 30
##### 24. Question

The letter of the word RANDOM are written in all possible orders and these words are written out as in a dictionary. Then the rank of the work RANDOM is

CorrectWords beginning with A, D, M, N and O are 5! each. Words beginning with RAD are 3! and also for RAM are 3!. Then comes RAND. First we shall have RANDMO and the RANDOM.

∴ rank is 5(5!) + 2(3!) + 2 = 614

IncorrectWords beginning with A, D, M, N and O are 5! each. Words beginning with RAD are 3! and also for RAM are 3!. Then comes RAND. First we shall have RANDMO and the RANDOM.

∴ rank is 5(5!) + 2(3!) + 2 = 614

- Question 25 of 30
##### 25. Question

Six points in a plane be joined in all possible ways by indefinite straight lines and if no two of them be coincident or parallel and no three pass through the same point (with the exception of the original 6 points). The number of distinct points or intersection is equal to

CorrectNumber of lines from 6 point =

^{6}C_{2}= 15Points of intersection obtained from these lines =

^{15}C_{2}= 105Now, we find the number of times, the original 6 points come.

Consider one point say A

_{1}. Joining A_{1}to remaining 5 points, we get 5 lines and any two lines from these 5 lines gives A_{1}as the point of intersection.∴ A

_{1}is common in^{5}C_{2}= 10 times out of 105 points of intersections.Similar is the case with other five points.

∴ 6 original points come 6 × 10 = 60 times in points of intersection.

Hence, the number of distinct points of intersection

= 105 – 60 + 6 = 51

IncorrectNumber of lines from 6 point =

^{6}C_{2}= 15Points of intersection obtained from these lines =

^{15}C_{2}= 105Now, we find the number of times, the original 6 points come.

Consider one point say A

_{1}. Joining A_{1}to remaining 5 points, we get 5 lines and any two lines from these 5 lines gives A_{1}as the point of intersection.∴ A

_{1}is common in^{5}C_{2}= 10 times out of 105 points of intersections.Similar is the case with other five points.

∴ 6 original points come 6 × 10 = 60 times in points of intersection.

Hence, the number of distinct points of intersection

= 105 – 60 + 6 = 51

- Question 26 of 30
##### 26. Question

Six persons A, B, C, D, E and F are to be seated at a circular table. The number of ways this can be done if A must have either B or C on his right and B must have either C or D on his right is

CorrectCase-I If B is right on A

Sub case –I C is right on B

then no. of ways = (4 – 2)! = 6

Sub case-II if D is right on B

then no. of ways = (4 – 1)! = 6

Case-II if C is right on A

⇒ D must be right on B = (4 – 1)! = 3! = 6

Hence total no. of ways is 6 + 6 + 6 = 18

IncorrectCase-I If B is right on A

Sub case –I C is right on B

then no. of ways = (4 – 2)! = 6

Sub case-II if D is right on B

then no. of ways = (4 – 1)! = 6

Case-II if C is right on A

⇒ D must be right on B = (4 – 1)! = 3! = 6

Hence total no. of ways is 6 + 6 + 6 = 18

- Question 27 of 30
##### 27. Question

If 5 parallel straight lines are intersected by 4 parallel straight lines, then the number of parallelograms, thus formed, is

CorrectNumber of parallelograms =

^{5}C_{2}×^{4}C_{2}= 60IncorrectNumber of parallelograms =

^{5}C_{2}×^{4}C_{2}= 60 - Question 28 of 30
##### 28. Question

The straight lines I

_{1}, I_{2}, I_{3}are parallel and lie in the same plane. A total numbers of*m*points are taken on I_{1}. n points on I_{2}, k points on I_{3}. The maximum number of triangles formed with vertices at these points isCorrectTotal number of points are m + n + k, the triangles formed by these points =

^{m + n +}^{k}C_{3}Joining of three points on the same line gives no triangle, the number of such triangles is

^{m}C_{3}+^{n}C_{3}+^{k}C_{3}∴ Required number of triangles =

^{m + n + }^{k}C_{3}–^{m}C_{3}–^{n}C_{3}–^{k}C_{3}IncorrectTotal number of points are m + n + k, the triangles formed by these points =

^{m + n +}^{k}C_{3}Joining of three points on the same line gives no triangle, the number of such triangles is

^{m}C_{3}+^{n}C_{3}+^{k}C_{3}∴ Required number of triangles =

^{m + n + }^{k}C_{3}–^{m}C_{3}–^{n}C_{3}–^{k}C_{3} - Question 29 of 30
##### 29. Question

In an examination there are three multiple choice questions and each question has 4 choices. Number of ways in which a student can fail to get all answers correct, is

CorrectEach question can be answered in 4 ways and all question can be answered correctly in only one way.

So, required number of ways = 63

IncorrectEach question can be answered in 4 ways and all question can be answered correctly in only one way.

So, required number of ways = 63

- Question 30 of 30
##### 30. Question

Two straight lines intersect at a point R. Points P

_{1}, P_{2}……….P_{n}are taken on one line and points Q_{1}, Q_{2}…..Q_{n}on the other. If the point R is not be used, the number of triangles that can be drawn using these points as vertices, isCorrectIncorrect