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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.The number of 4-digit even numbers that can be formed using 0,1,2,3,4,5,6 without repetition is
•  420
•  120
•  300
•  20
Solution

(a) An even number has an even digit at unit place ∴ Required number of even numbers = Number of even numbers having 0 at unit’s place + Number of even numbers having a non-zero digit at unit’s place = 6C3×3 !×1+ 3C1 (6C3×3 !- 5C2×2 !) =120+3×(120-20)=420
Q2.A father with 8 children takes 3 at a time to the zoological garden, as often as he can without taking the same 3 children together more than once. The number of times he will go to the garden, is
•  112
•  56
•  336
•  None of these
Solution
(b) The number of times he will go to the garden is same as the number of selecting 3 children from 8 children ∴ The required number of times= 8C3=56

•  2n-1
•  2n-1
•  2n+1
•  2n
Solution
(a) nPr= nCr r ! ⇒nPr/r!= nCr ⇒∑r=1nnPr/r!=∑r=1nnCrr!=nC1+nC2+...+nCn=2n-1

Q4. If the letters of the word ‘SACHIN’ are arranged in all possible ways and these words are written out as in dictionary, then the word ‘SACHIN’ appears at serial number
•  601
•  600
•  603
•  602
Solution
(a) In the word SACHIN order of alphabets is A, C, H, I, N, S. The number of words starting with A, C, H, I, N are each equal to 5! ∴ Total number of wards5×5!=600 The first word starting with S is SACHIN So, word SACHIN appears at serial number 601

Q5.Three straight lines L1,L2,L3 are parallel and lie in the same plane. A total of m points are taken on L1,n points on L2,k points on L3. The maximum number of triangles formed with vertices at these points are
•  m+n+kC3
•  m+n+kC3-mC3-nC3
•  m+n+kC3+mC3+nC3
•  None of these
Solution
(d) Total number of points on a three lines are m+n+k ∴ maximum number of triangles = m+n+kC3-mC3-nC3-kC3 (subtract those triangles in which point on the same line)

Q6.Eight chair are numbered 1 to 8. Two women and three men wish to occupy one chair each. First the women choose the chairs from amongst the chair marked 1 to 4; and then the men select the chairs from amongst the remaining. The number of possible arrangement is
•  6C3x4C2
•  4C2+4P3
• 4P2x6P3
•  None of these
Solution
(c) Since, first the 2 women select the chairs amongst 1 to 4 in 4P2 ways. Now, from the remaining 6 chairs three men could be arranged in 6P3 ways. ∴ Total number of arrangements= 4P2x6P3

Q7.A polygon has 44 diagonals, then the number of its sides are
•  8
•  7
•  11
•  None of these
Solution
(c) Let number of sides of polygon=n ⇒ nC2-n=44 [given] ⇒ n!/2!(n-2)!-n=44 ⇒ n(n-1)-2n=88 ⇒ n2-3n-88=0 ⇒ (n-11)(n+8)=0 ⇒ n=11,-8 Since, sides cannot be negative ∴ n=11

Q8.The total number of flags with three horizontal strips, in order that can be formed using 2 identical red, 2 identical green and 2 identical white strips, is equal to
•  2(4!)
•  4!
•  3(4!)
•  None of these
Solution
(b) All strips are of different colours, then the number of flags =3!=6 When two strips are of same colour, then the number of flags = 3C13!2=18 ∴ Total number of flags=6+18=24=4!

Q9.The value of 47C4+ 5r=152-rC3 is equal to
•  52C4
•  47C6
•  52C5
•  None of these
Solution
(a) 47C4+∑5r=1 52-rC3= 47C4+51C3+50C3+49C3+48C3+47C3 = 51C3+50C3+49C3+48C3+(47C3+47C4) = 52C4

Q10. If there are n number of seats and m number of people have to be seated, then how many ways are possible to do this(m < n)?
•  nCm
•  nCn×(m-1)!
•  n-1Pm-1
• nPm
Solution
(d) ∵ Total number of seats=n and number of people=m Ist person can be seated in n ways IInd person can be seated in (n-1) ways ⋯⋯⋯⋯⋯ ⋯⋯⋯⋯⋯ ⋯⋯⋯⋯ mth person can be seated in (n-m+1)ways ∴ Total number of ways =n(n-1)(n-2)…(n-m+1)=nPm Alternate In out of n seats m people can be seated in nPm ways ## Want to know more

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