# IIT JEE Main Solved Practice Paper Set 1.1: Permutation and Combination

Find solved practice paper from the chapter Permutation and Combination for IIT JEE Main Exam. This paper consists 10 questions.

Find **Solved Practice Paper for IIT JEE Main** in this article. This practice paper consists of 10 questions from the chapter Permutations and Combinations. We have tried to include all important concepts and formulae in these questions. Also, detailed solution of each question has been provided. All questions are very important from examination point of view.

**Q1. ** The total number of positive integral solution for *x, y, z* such that *xyz* = 24, is

(a) 30 (b) 60

(c) 90 (d) 120

**Correct Option: (a)**

**Sol.**

We have,

*xyz* = 24

*xyz* = 2^{3} × 3^{1}

The number of ways of distributing '*n*' identical balls into '*r*' different boxes is ^{(n + r }^{− }^{1)}C_{(r }_{− }_{1)}

Here we have to group 4 numbers into three groups

Number of integral positive solutions

= ^{(3 + 3 }^{− }^{1)}C_{(3 }_{− }_{1)} × ^{(1+ 3 }^{− }^{1)}C_{(3 }_{− }_{1)}

= ^{5}C_{2} × ^{3}C_{2} = 30

**Q2.** A class has n students, we have to form a team of the students including at least two

students and also excluding at least two students. The number of ways of forming the

team is

(a) 2* ^{n }*− 2

*n*(b) 2

*− 2*

^{n}*n*− 2

(c) 2* ^{n}* − 2

*n*− 4 (d) 2

*− 2*

^{n}*n*− 6

**Correct Option: (b)**

**Sol.**

Required number of ways

**UPSEE 2017 Solved Sample Paper Set-1**

**Q3.** Let *T _{n}* be the number of all possible triangles formed by joining vertices

** **of an *n*-sided regular polygon.* T _{n+}*

_{1}−

*T*= 10, then the value of

_{n}*n*is:

(a) 7

(b) 5

(c) 10

(d) 8

**Sol.**

**Correct Option: (b)**

We have,

**Q4. ** Two numbers are chosen from 1, 3, 5, 7,… 147, 149 and 151 and multiplied together in

all possible ways. The number of ways which will give us the product a multiple of 5,

is

(a) 75 (b) 1815

(c) 95 (d) 1030

**Correct Option: (b)**

**Sol.**

In the given numbers 1, 3,5,7,…, 147, 148, 151the numbers which are multiple of 5 are 5, 15, 25,

35,…, 145. which are an arithmetic sequence

*T _{n} = a + *(

*n*− 1)

*d*

145 = 5 + (*n* – 1) 10

*n* = 15

and if total number of terms in the given sequence is *m, *then

151 = 1 + (*m* – 1) ´ 2

*m* = 76

So, the number of ways in which product is a multiple of 5 = ( both two numbers from 5, 10, 15, 20, 25, 30, 35, … , 150) or ( one number from 5, 10, 15, 20, 25, 30, 35, … , 150 and one from remaining numbers)

**JEE Main Mathematics Solved Sample Paper Set-VII**

**Q5. **The total number of ways in which 5 balls of different color can be

** **distributed among 3 persons so that each person gets at least one ball is:

(a) 75

(b) 150

(c) 210

(d) 243

**Sol.**

**Correct Option: (b)**

Here 5 distinct balls are distributed amongst 3 persons so that each gets at least one ball.

So, we have the possible combinations:

1, 2, 2 or 1, 3, 1

Now, number of ways to distribute 5 balls

**Q6.** If the number of ways of selecting *n* cards out of unlimited number of cards bearing the

numbers 0, 2, 3, so that it cannot be used to write the number 203 is 93, then *n* equal to

(a) 3 (b) 4

(c) 5 (d) 6

**Correct Option: (b)**

**Sol.**

We can not write 203, if in the selection of *n* cards, we get either (2 or 3), (3 or 0), ( 0 or 2), (only 0), (only 2) or (only 3)

**Q7. **The number of seven digit integers, with sum of the digits equal to 10 and

formed by using the digits 1, 2 and 3 only, is

(a) 55

(b) 66

(c) 77

(d) 88

**Sol.**

**Correct Option: (c)**

There are two possible cases:

Case 1: When the digits of the integers are 1, 1, 1, 1, 2, 2, 2

Case 2: When the digits of the integers are 1, 1, 1, 1, 1, 2, 3

Total number of seven digit numbers whose sum is 10 using numbers 1, 2, 3 is 35 + 42 = 77

**Q8.** If *a, b*, c are odd positive integers, then number of integral solutions of *a + b + c* = 13, is

(a) 14 (b) 21

(c) 28 (d) 56

**Correct Option: (b)**

**Sol.**

Let *a* = 2*k* + 1, *b* = 2*l* + 1, *c* = 2*m* + 1

where *k*, *l*, *m are* whole number

*a + b + c* = 13

2*k* + 1 + 2*l* + 1 + 2*z* + 1 = 13

or, *x + y + z* = 5

**JEE Main Physics Solved Sample Paper Set-VII**

**Q9. **Let *n*_{1} < *n*_{2} < *n*_{3} < *n*_{4} < *n*_{5 } be positive integers such that *n*_{1} + *n*_{2} + *n*_{3} + *n*_{4} + *n*_{5} = 20. The numbers of such distinct arrangements (*n*_{1}, *n*_{2}, *n*_{3}, *n*_{4}, *n*_{5}) is

(a) 5 (b) 6

(c) 7 (d) 8

**Correct Option: (c)**

**Sol.**

As,

**Q10. **Let *n* ≥ 2 be an integer. Take n distinct points on a circle and join each pair of points by a line segment. Colour the line segment joining every pair of adjacent points by blue and the rest by red. If the number of red and blue line segment are equal, then the value of *n* is

(a) 5 (b) 6

(c) 7 (d) 8

**Correct Option: (a)**

**Sol.**