Permutation-and-combination-quiz-2

Permutation and Combination Quiz-2

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 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

•  3255
•  2¹²
•  3760
•  3720

Solution
(c) The total number of combinations which can be formed of five different green dyes, taking one or more of them is 2⁵-1=31. 

Similarly, by taking one or more of four different red dyes 2⁴-1=15 combinations can be formed. 

The number of combinations which can be formed of three different red dyes, taking none, one or more of them is 2³=8 

 Hence, the required number of combinations of dyes =31×15×8=3720

Q2.If 1/( ⁴C_n )=1/( ⁵C_n )+1/( ⁶C_n ), then n is equal to

•  3
•  2
•  1
•  0

Solution
1/( ⁴C_n )=1/( ⁵C_n )+1/( ⁶C_n ) 

 ⇒ (n!(4-n)!)/4!=(n!(5-n)!)/5!+(n!(6-n)!)/6! 

 ⇒ ((4-n)!)/4!=((4-n)!(5-n))/(5×4!)+((6-n)(5-n)(4-n)!)/(6×5×4!) 

 ⇒ 1=(5-n)/5+((6-n)(5-n))/(6×5) 

 ⇒ n²-17n+30=0 

 ⇒ (n-15)(n-2)=0 

 ⇒ n=2 [∵ ⁴C_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

•   (52 !)/((17 !)³)
•  (51 !)/((17 !)³ 3 !)
•  (51 !)/((17 !)³)
•  52!/((17 !)³ 3 !)

Solution
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. 

First we divide 52 cards into two groups of 1 card and 51 cards. 

This can be done in (52 !)/(1 ! 51 !) ways 

 Now, every group of 51 cards can be divided into 3 groups of 17 each in (51 !)/((17 !)³ 3 !) 

 Hence, the required number of ways =(52 !)/(1 !51 !)∙(51 !)/((17 !)³ 3 !)=(52 !)/((17 !)³ 3 !)

Q4. If the total number of m elements subsets of the set A={a₁,a₂,a₃,…a_n } is λ times the number of 3 elements subsets containing a₄,then n is

•  (m-1)λ
•  0
•  (m+1)λ
•  mλ

Solution
(d) Total number of m-elements subsetcs of A= ⁿC_m …(i) 

 and number of m-elements subsets of A each containing the element a₄= ^n-1C_m-1 

 According to question, ⁿC_m=λ.^n-1C_m-1 

 ⇒ n/m .^n-1C_m-1=λ.^n-1C_m-1 

 ⇒ λ=n/m or n=mλ

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?

•  Both of these
•  Only(2)
•  Only(1)
•  None of these

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

To find the number of ways in which p particular things are together, we consider p particular thing as a group. 

 ∴ Number of ways in which p particular things are together =(m-p+1)!p! 

 So, number of ways in which p particular things are not together =m!-(m-p+1)!p! 

 (2) Each player shall receive 13 cards. 

 Total number of ways =52!/(13!⁴) 

 Hence, both statements are correct

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

•  61
•  63
• 62
•  64

Solution
(b) Required number = ⁶C₁+ ⁶C₂+ ⁶C₃+ ⁶C₄+ ⁶C₅+ ⁶C₆=2⁶-1=63

Q7. Let l₁ and l₂ be two lines intersecting at P. If A₁,B₁,C₁ are points on l₁ and A₂,B₂,C₂,D₂,E₂ are points on l₂ and if none of these coincides with P, then the number of triangles formed by these eight points, is

•  56
•  55
•  45
•  46

Solution
(c) If triangle is formed including point 'P' the other points must be one from l¹ and other point from l². Number of triangle formed with P=³C₁×⁵C₁=15 ways When P is not included.

 Number of triangle formed =³C₂×⁵C₁+³C₁×⁵C₂=15+15=30 

 Total number of triangles=15+30=45

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

•  32
•  104
•  76
•  64

Solution
(b) The required number of points = ⁸C₂×1+ ⁴C₂×2+( ⁸C₁× ⁴C₁)×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

•  22030
•  21000
•  21580
•  20580

Solution
(d) The number of ways of choosing a committee if there is no restriction is ¹⁰C₄∙ ⁹C₅=^(10!) ⁄ _{(4 !6 !)}∙^(9!) ⁄ _{(4 !5 !)}=26460

 The number of ways of choosing the committee if both Mr.A and Ms. B are included in the committee is ⁹C₃∙ ⁸C₄=5880 

 Therefore, the number of ways of choosing the committee when Mr. A and Ms. B are not together =26480-5880=20580

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

•  r²
•  r!
•  rⁿ
• None of these

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

 The natural number will divided by r!