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Permutation and Combination Quiz-5

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. How many different words can be formed by jumbling the letters in the word MISSISSIPPI in which no two S are adjacent?
  •  7.6C4.8C4
  •  8.6C4.7C4
  •  6.7.8C4
  •  6.8.7C4
Solution
(a) Given word is MISSISSIPPI Here, I=4 times, S=4 times, P=2 times, M=1 time _M_I_I_I_I_P_P_ Required number of words = ^8 C_4×7!/4!2! = 8C4×(7×6!)/4!2!=7. 8C4.6C4

Q2.Let A={x1,x2,x3,x4,x5,x6 }, B={y1,y2,y3,y4,y5,y6 }.Then the number of one –one mapping from A to B such that f(xi )≠yv. i=1,2,3,4,5,6 is
  •  720
  •  265
  •  360
  •  145
Solution
(b) ∵f(xi)≠yi ie, no object goes to its scheduled place. Then, number of one-one mappings =6!(1-1/1!+1/2!-1/3!+1/4!-1/5!+1/6!) =6!(1/2!-1/3!+1/4!-1/5!+1/6!) =360-120+30-6+1=265

Q3.  The number of diagonals in a polygon of m sides is
  •   m(m-5)/2! 
  •  m(m-1)/2! 
  •   m(m-3)/2!
  •  m(m-2)/2!
Solution
(c) Required number of diagonals =mC2-m =(m(m-1))/2!-m =m(m-3)/2!

Q4. If nCr= nCr-1 and nPr=nPr+1, then the value of n is
  •  3
  •  4
  •  2
  •  5
Solution
(a) Given, nCr = nCr-1 and nPr=nPr+1

⇒n-r+1=r and n-r=1 ⇒n-2r+1=0 and n-r-1=0 On solving, we get n=3,r=2


Q5. A committee of 12 is to be formed from 9 women and 8 men in which at least 5 women have to be included in a committee. Then the number of committees in which the women are in majority and men are in majority are respectively
  •  4784, 1008
  •  2702, 3360
  •  6062, 2702
  •  2702, 1008
Solution
 (d) The number of ways in which at least 5 women can be included in a committee = 9C5× 8C7+ 9C6× 8C6+ 9C7× 8C5+ 9C8× 8C4+ 9C9× 8C3 (i) Women are in majority, then number of ways =9C7× 8C5+ 9C8× 8C4+ 9C9× 8C3 =2016+630+56=2702 (ii) Men are in majority, then number of ways =9C5× 8C7 =126×8=1008

Q6. The number of triangles which can be formed by using the vertices of a regular polygon of (n+3) sides is 220. Then, n is equal to
  •  8
  •  9
  • 10
  •  11
Solution
(b) Number of triangles=n+3C3=220 ⇒ ((n+3)!)/3!n!=220 ⇒ (n+1)(n+2)(n+3)=1320 =12×10×11 =(9+1)(9+2)(9+3) ∴ n=9

Q7. The number of permutations of the letters of the word ‘CONSEQUENCE’ in which all the three E’s are together, is
  •  9!3!
  •  9!/2!2!
  •  9!/2!2!3!
  •  9!/2!3!
Solution
(b) The letters in the word ‘CONSEQUENCE’ are 2C, 3E, 2N, 1O, 1Q, 1S, 1U ∴ Required number of permutations=9!/2!2!

Q8. All the words that can be formed using alphabets A, H, L, U, R are written as in a dictionary (no alphabet is replaced). Then, the rank of the word RAHUL is
  •  70
  •  71
  •  72
  •  74
Solution
(d) In the word RAHUL the letters are (A, H, L, R, U) Number of words starting with A=4!=24 Number of words starting with H=4!=24 Number of words starting with L=4!=24 In the starting with R first one is RAHLU and next one is RAHUL. ∴ Rank of the word RAHUL=3(24)+2=74

Q9. The number of ways in which a pack of 52 cards be divided equally amongst four players in order is
  •  52C13
  •  52C4
  •  (52 !)/(13 !4)
  •  (52 !)/((13 !4) 4 !)
Solution
(c) As the players who are to receive the cards are different So, the required number of ways =(52 !)/(13 !4)

Q10. In how many ways can 5 red and 4 white balls be drawn from a bag containing 10 red and 8 white balls
  •  8C5×10C4
  •  10C5×8C4
  •  18C9
  • None of these
Solution
(b) 10C5×8C4

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