## Permutation and Combination Quiz-8

Dear Readers,

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 5 letters and 5 different envelopes. The number of ways in which all the letters can be put in wrong envelope, is
•  119
•  44
•  59
•  40
Solution

(b) Required numbers=5![1-1/1!+1/2!-1/3!+1/4!-1/5!] =44
Q2.Let Tn denote the number of triangles which can be formed using the vertices of a regular polygon of n sides. If T(n+1)-Tn=21, then n equals
•  5
•  7
•  6
•  None of these
Solution
(b) Tn=nC3 and Tn+1-Tn=21
n+1C3-nC3=21
nC2+nC3nC3=21
⇒ nC2=21
⇒ (n(n-1))/2=21
⇒ n2-n-42=0
⇒ (n-7)(n+6)=0
∴ n=7 [∵ ≠-6]

Q3. The number of diagonals of a polygon of 20 sides is
•  210
•  190
•  180
•  170
Solution
(d) Number of diagonals in a polygon of n sides = n C2-n
Here, n=20
∴ required number of diagonals = 20 C2-20 =(20×19)/(2×1)-20=170

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
•  1540
•  1450
•  1504
•  1405
Solution
(a) First we fix the alternate position of 21 English book, in which 22 vacant places for Hindi books, hence total number of ways are 22 C19=1540

Q5.In a group of boys, two boys are brothers and in this group, 6 more boys are there. In how many ways, they can sit if the brothers are not to sit alongwith each other :
•  4820C3
•  1410
•  2830
•  None of these
Solution
(d) Required number of ways = Total number of ways in which 8 boys can sit - Number of ways in which two brothers sit together =8 !-7 !×2 !=7 !×6=30240

Q6.In a chess tournament where the participants were to play one game with one another, two players fell ill having played 6 games each, without playing among themselves. If the total number of games is 117, then the number of participants at the beginning was
•  15
•  16
• 17
•  None of these
Solution
(a) Let the number of participants at the beginning was n
∴ (n(n-1))/2=117-12
⇒ n(n-1)=2×105
⇒ n2-n-210=0
⇒ (n-15)(n+14)=0
⇒ n=15 [∵n≠-14]

Q7.If a denotes the number of permutations of x+2 things taken all at a time, b the number of permutations of x things taken 11 at a time and c the number of permutations of x-11 things taken all at a time such that a=182 bc, then the value of x is
•  15
•  12
•  10
•  None of these
Solution

x! ⇒(x+2)(x+1)=182
⇒x2+3x-180=0
⇒(x-12)(x+15)=0
⇒x=12,-15
∴ Neglect the negative value of x. ⇒ x=12

Q8.The number of ways in which a committee can be formed of 5 members from 6 men and 4 women if the committee has at least one woman, is
•  186
•  246
•  252
•  244
Solution

Q9.The number of ways in which seven persons can be arranged at a round table, if two particular persons may not sit together is
•  480
•  120
•  80
•  None of these
Solution

Q10. The total numbers of ways of dividing 15 things into groups of 8,4 and 3 respectively is
•  (15 !)/(8 !4 !(3 !)2 )
•  (15 !)/(8 !4 !3 !)
•  (15 !)/(8 !4 !)
• None of these
Solution
(b) No solution available.

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