## Permutation and Combination Quiz-7

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. Let A be a set containing 10 distinct elements. Then, the total number of distinct functions from A to A is
•  210-1
•  210
•  10!
•  1010
Solution
(d) 1010

Q2.In a football championship, there were played 153 matches. Every team played one match with each other. The number of teams participating in the championship is
•  17
•  18
•  9
•  13
Solution
(b) Let there are n teams. Each team play to every other team in nC23 ways
∴ n C2=153 (given)
⇒ n!/(n-2)!2!=153
⇒ n(n-1)=306
⇒ n2-n-306=0
⇒ (n-18)(n+17)=0
⇒ n=18 (∵n is never negative)
Q3.  In how many ways can 15 members of a council sit along a circular table, when the Secretary is to sit on one side of the chairman and the Deputy Secretary on the other side?
•   2×11!
•  2×15!
•  24
•  12!×2
Solution
(d) Since total number are 15, but three special members constitute one member.
Therefore, required number of arrangements are 12!×2, because, chairman remains between the two specified persons and person can sit in two ways

Q4. If in a chess tournament each contestant plays once against each of the other and in all 45 games are played, then the number of participants is
•  9
•  20
•  15
•  10
Solution
(d) Let there be n participants. Then, we have nC2=45 ⇒(n(n-1))/2=45⇒n2-n-9=0⇒n=10

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
•   nC2× nC2
•  2nC2-2× nC+2
•   2nC2-2× nC2
•  None of these
Solution
(a) We observe that a point is obtained between the lines of two of points on first line are joined by line segments to two points on the second line Hence, required number of points = nC2× nC2

Q6. The number of ways in which one can select three distinct integers between 1 and 30, both inclusive, whose sum is even, is
•   1120
•   2030
• 1575
•   None of these
Solution
(b) ∵ Number are either all even or one even and other two odd
∴ Required number of ways = 15 C3+ 15C3× 15C2 =15!/(3!×12!)+15!/14!×15!/(2!×13!) =(15×14×13)/6+(15×15×14)/2 =455+1575=2030

Q7. The number of four-letter words that can be formed (the words need not be meaningful) using the letters of the word MEDITERRANEAN such that the first letter is E and the last letter is R, is
•  11!/2!2!2!
•  56
•  59
•  11!/3!2!2!
Solution
(c) Word MEDITERRANEAN has 2A, 3E, 1D, 1I, 1M, 2N, 2R, 1T In out of four letters E and R is fixed and rest of the two letters can be chosen in following ways
Case I Both letter are of same kind ie, 3 C2 ways, therefore number of words = 3C2×2!/2!=3
Case II Both letters are of different kinds ie, 8 C2 ways, therefore number of words= 8 C2×2!=56 Hence, total number of words=56+3=59

Q7.A person goes for an examination in which there are four papers with a maximum of m marks from each paper. The number of ways in which one can get 2m marks, is
•  2m+1
•  (m+1)(2m2+4m+3)/3
•  (m+1)(2m2+m+1)/3
•  None of these
Solution
(b) Required number of ways =coefficient of x2min (x0+x1+...+xm )4
=coefficient of x2m in ((1-xm+1/(1-x))4
=coefficient of x2m in (1-4xm+1+6x2m+2+...) (1-x)-4
=2m+3 C2m-4m+2 Cm-1
=((2m+1)(2m+2)(2m+3))/6-(4m(m+1)(m+2))/6
=((m+1)(2m2+4m+3))/3

Q9. The number of integers which lie between 1 and 106 and which have the sum of the digits equal to 12, is
•   8550
•   5382
•   8055
•  6062
Solution
(d) Consider the product (x0+x1+x2…+x9 )(x0+x1+x2…+x6 )…6 factors
The number of ways in which the sum of the digits will be equal to 12 is equal to the coefficient of x12 in the above product.
So, required number of ways = Coeff. Of x12 in ((1-x10)/(1-x))6
= Coeff. Of x12 in (1-x10 )6 (1-x)-6
= Coeff. Of x12 in (1-x)-6 (1- 6C1 x10+⋯)
= Coeff. Of x12 in (1-x)-6- 6 C1∙Coeff.of x2 in (1-x)-6
= 12+6-1 C6-1- 6C1× 2+6-1 C6-1= 17C5-6× 7 C5=6062

Q10. The number of divisors of 9600 including 1 and 9600 are
•   60
•   48
•   58
• 46
Solution
(b) ∵ The factors of 9600 =27×31×52
∴ The number of divisors =(7+1)(1+1)(2+1) =8×2×3=48

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