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Permutations and Combinations Test 46

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Permutations and Combinations Test 46
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  • Question 1
    1 / -0
    1212 people came for a carroms tournament. All were divided into pairs of 22 each. After all the matches if it found that half of the selected pairs had to play three matches to decide the winner in a best of three process, how many matches have been held?
     
    Solution

  • Question 2
    1 / -0
    If 19C3r=19Cr+3^{19}C_{3r} = ^{19}C_{r+3} , then r is equal to:
    Solution
    19C3r=19Cr+3^{ 19 }{ C }_{ 3r }=^{ 19 }{ C }_{ r+3 }
    We Know 
    nCr=nCnrr+(nr)=n3r+r+3=194r=16r=4^{ n }{ C }_{ r }=^{ n }{ C }_{ n-r }\\ \Rightarrow r+(n-r)=n\\ \Rightarrow 3r+r+3=19\\ \Rightarrow 4r=16\\ \Rightarrow r=4
    Therefore  The value of  rr is 44
  • Question 3
    1 / -0
    How many alphabets need to be there in a language if one were to make 11 million distinct 33 digit initials using the alphabets of the language?
    Solution
    Digits with initials=3=3
    Total number of initials=1million=1000000=1million=1000000
    Let the alphabets in language be xx
    initials
    =x3=1000000=106x=102={ x }^{ 3 }=1000000={ 10 }^{ 6 }\\ \Rightarrow x={ 10 }^{ 2 }
    Therefore 100100 alphabets are present
  • Question 4
    1 / -0
     After every get-together every person present shakes the hand of every other person. If there were 105 handshakes in all, how many persons were present in the party?
    Solution
    Total number of handshakes =105=105
    Let number of person be =x=x
    Number of handshakes 
    =x(x1)2=105x(x1)=210x=15=\cfrac { x(x-1) }{ 2 } =105\\ \Rightarrow x(x-1)=210\\ x=15
    Hence the correct answer is 1515
  • Question 5
    1 / -0
    80648064 is resolved into all possible product of two factors. Find the number of ways in which this can be done?
    Solution
    8064=27×32×78064=2^{ 7 }\times { 3 }^{ 2 }\times 7
    Resolution of a number into factors (and thier products).
    depends on powers of primes.
    Number of possible products are 
    =(7+1)(2+1)(1)=8×3×1=24=\left( 7+1 \right) \left( 2+1 \right) \left( 1 \right) \\ =8\times 3\times 1\\ =24
  • Question 6
    1 / -0
    How many straight lines can be formed from 1111 points in a plane out of which no three points are collinear?
    Solution
    Number of points =11=11
    No three are collinear
    For a straight lines Two end points are needed
    Number of straight lines needed
    11C211!2!×9!11×10255lines\Rightarrow ^{ 11 }{ C }_{ 2 }\\ \Rightarrow \cfrac { 11! }{ 2!\times 9! } \\ \Rightarrow \cfrac { 11\times 10 }{ 2 } \\ \Rightarrow 55 lines
    Option :C: C
  • Question 7
    1 / -0
    A college offers 77 courses in the morning and 55 in the evening. Find possible number of choices with the student who want to study one course in the morning and one in the evening.
    Solution
    Number of courses in the morning =7=7 
    Selecting one of the 7=7C17=^{ 7 }{ C }_{ 1 } Ways =7=7 ways 
    Number of courses in the evening =5= 5
    Selecting one of the 5=5C15=^{ 5 }{ C }_{ 1 } Ways=5 = 5 ways 
    Total number of ways =7×5  =7\times 5 Ways =35=35 Ways 
  • Question 8
    1 / -0
    When six fair coins are tossed simultaneously, in how many of the outcomes will at most three of the coins turn up as heads?
    Solution

    Total number of coins =6=6

    (1) '0' heads =1=1 chance(6C0) \left( ^{ 6 }{ C }_{ 0 } \right)

    (2) '1' heads =6=6 chance(6C1) \left( ^{ 6 }{ C }_{ 1 } \right)

    (3) '2' heads=15 =15 chance (6C2) \left( ^{ 6 }{ C }_{ 2 } \right)

    (4) '3' heads=20 =20 chance (6C3) \left( ^{ 6 }{ C }_{ 3 } \right)

     Hence total outcome =42=42

  • Question 9
    1 / -0
    The number of ways in which four letters can be selected from the word 'DEGREE' is
    Solution
    There are 66 letters in the word 'DEGREE', namely 3E’s3\text{E's} and D, G, R\text{D, G, R}. Four letters out of six can be selected in the following ways
    (i) 33 like letters and 11 different, viz. EEE + D, G, R\text{EEE + D, G, R}
    (ii) 22 like letters and 22 different, viz. EEE +\text{EEE}\ + any two of D, G, R\text{D, G, R}.
    (iii) All different letters, viz. EDGR\text{EDGR}.
    So, the total number of ways
    =3C1+3C2+1=3+3+1=7=^3C_1 + ^3C_2 + 1 = 3 + 3 + 1 = 7
  • Question 10
    1 / -0
    If 15C3r=15Cr+3^{15}C_{3r} = ^{15}C_{r+3}, then find the value of rr:
    Solution
    15C3r=15Cr+3^{ 15 }{ C }_{ 3r }=^{ 15 }{ C }_{ r+3 }
    We Know 
    15C3r=15Cr+3nCr=nCnrn=r+(nr)=3r+r+3=4r+34r+3=154r=12r=3^{ 15 }{ C }_{ 3r }=^{ 15 }{ C }_{ r+3 }\\ ^{ n }{ C }_{ r }=^{ n }{ C }_{ n-r }\\ n=r+(n-r)\\ =3r+r+3\\ =4r+3\\ \Rightarrow 4r+3=15\\ \Rightarrow 4r=12\\ \Rightarrow r=3
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