As per analysis for previous years, it has been observed that students preparing for JEE MAINS find Mathematics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Mathematics due to their ENGINEERING background.

Furthermore, sections such as Mathematics are dominantly based on theories, laws, numerical in comparison to a section of Engineering which is more of fact-based, Physics, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced.

**Q1.**There are five different green dyes, four different blue dyes and three different red dyes. The total number of combinations of dyes that can be chosen taking at least one green and one blue dye is

(c) The total number of combinations which can be formed of five different green dyes, taking one or more of them is 2

^{5}-1=31.

^{4}-1=15 combinations can be formed.

^{3}=8

**Q2.**If 1/(

^{4}C

_{n})=1/(

^{5}C

_{n})+1/(

^{6}C

_{n}), then n is equal to

1/(

^{4}C

_{n})=1/(

^{5}C

_{n})+1/(

^{6}C

_{n})

^{2}-17n+30=0

^{4}C

_{n}is not meaningful for n=15]

**Q3.**The number of ways in which 52 cards can be divided into 4 sets, three of them having 17 cards each and the fourth one having just one card

Here, we have to divide 52 cards into 4 sets, three of them having 17 cards each and the fourth one having just one card.

^{3}3 !)

^{3}3 !)=(52 !)/((17 !)

^{3}3 !)

**Q4.**If the total number of m elements subsets of the set A={a

_{1},a

_{2},a

_{3},…a

_{n}} is Î» times the number of 3 elements subsets containing a

_{4},then n is

(d) Total number of m-elements subsetcs of A=

^{n}C

_{m}…(i)

_{4}=

^{n-1}C

_{m-1}

^{n}C

_{m}=Î».

^{n-1}C

_{m-1}

^{n-1}C

_{m-1}=Î».

^{n-1}C

_{m-1}

**Q5.**Consider the following statement: 1.The number of ways of arranging m different things taken all at a time in which p≤m perticular things are never together is m!-(m-p+1)!p! 2. A pack of 52 cards can be divided equally among four players in order in 52!/(13!

^{4}) ways Which of these is/are correct?

(c) (1) Total number of ways of arranging m things =m!

^{4})

**Q6.**Ramesh has 6 friends. In how many ways can he invite one or more of them at a dinner?

(b) Required number =

^{6}C

_{1}+

^{6}C

_{2}+

^{6}C

_{3}+

^{6}C

_{4}+

^{6}C

_{5}+

^{6}C

_{6}=2

^{6}-1=63

**Q7.**Let l

_{1}and l

_{2}be two lines intersecting at P. If A

_{1},B

_{1},C

_{1}are points on l

_{1}and A

_{2},B

_{2},C

_{2},D

_{2},E

_{2}are points on l

_{2}and if none of these coincides with P, then the number of triangles formed by these eight points, is

(c) If triangle is formed including point 'P' the other points must be one from l

^{1}and other point from l

^{2}. Number of triangle formed with P=

^{3}C

_{1}×

^{5}C

_{1}=15 ways When P is not included.

^{3}C

_{2}×

^{5}C

_{1}+

^{3}C

_{1}×

^{5}C

_{2}=15+15=30

**Q8.**The greatest possible number of points of intersection of 8 straight lines and 4 circle is

(b) The required number of points =

^{8}C

_{2}×1+

^{4}C

_{2}×2+(

^{8}C

_{1}×

^{4}C

_{1})×2 =28+12+32×2=104

**Q9.**The number of ways choosing a committee of 4 woman and 5 men from 10 women and 9 men, if Mr. A refuses to serve on the committee when Ms. B is a member of the committee, is

(d) The number of ways of choosing a committee if there is no restriction is

^{10}C

_{4}∙

^{9}C

_{5}=

^{(10!)}⁄

_{(4 !6 !)}∙

^{(9!)}⁄

_{(4 !5 !)}=26460

^{9}C

_{3}∙

^{8}C

_{4}=5880

**Q10.**The products of any r consecutive natural numbers is always divisible by

(b) The product of r consecutive natural numbers =1.2.3.4…..r=r!