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Permutations and Combinations Test 39

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Permutations and Combinations Test 39
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  • Question 1
    1 / -0
    The total number of ways of arranging the letters AAABBBCCDEFAAABBBCCDEF in a row such that letters CC are separated from one another is
    Solution

  • Question 2
    1 / -0
    If 64a2+36b2=400,ab=464a^2+36b^2=400, ab=4 , then 8a+6b8a+6b is: 
    Solution
    Given  64a2+36b2=400,ab=464a^2+36b^2=400,\quad ab=4
    We know 
    (a+b)2=a2+b2+2ab(a+b)^2=a^2+b^2+2ab
    therefore,
    (8a+6b)2=(8a)2+(6b)2+2(8a)(6a)(8a+6b)^2=(8a)^2+(6b)^2+2(8a)(6a)
                        =64a2+36b2+96ab=64a^2+36b^2+96ab
                        =400+96×4=400+96\times4
                        =400+384=400+384
    (8a+6b)2=784(8a+6b)^2=784
    8a+6b=7848a+6b=\sqrt{784}
    8a+6b=288a+6b=28
  • Question 3
    1 / -0
    There is a defined relationship between the pair of figure on either side of : :. Identify the relationship of the given pair and find be the missing figure.

    Solution

  • Question 4
    1 / -0
    The number of positive integral solutions of the equation x1x2x3x4x5=1050x _ { 1 } x _ { 2 } x _ { 3 } x _ { 4 } x _ { 5 } = 1050 is
    Solution
    We have,
    x1x2x3x4x5=1050=2×3×52×7{x_1}{x_2}{x_3}{x_4}{x_5} = 1050 = 2 \times 3 \times {5^2} \times 7
    Now,
    2,32,3 and 77 can be put in any boxes x1,x2,x3,x4{x_1},{x_2},{x_3},{x_4} and x5{x_5}.
    Also 5,55,5 can be distributed in 55 boxes in
    2+51C51=6C4=15^{2 + 5 - 1}{C_{5 - 1}}{ = ^6}{C_4} = 15Ways.
    So total no. of positive integral solutions =15×5×5×5=15 \times5 \times5 \times5
    =1875=1875
    Option DD is correct answer.
  • Question 5
    1 / -0
    If the area of triangle formed by the points (2a,b),(a+b,2b+a)(2a,b),(a+b,2b+a) and (2b,2a)(2b,2a) be λ\lambda then the area of the triangle whose vertices are (a+b,ab),(3ba,b+3a)(a+b,a-b), (3b-a,b+3a) and (3ab,3ba)(3a-b,3b-a) will be
    Solution
    Area of  =12x1(y2y3)+x2(y3y1)+x3(y1y2) \triangle  \,= \dfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right |
    Given area of triangle whose vertices are (2a,b),(a+b,2b+a),(2b,2a) (2a,b), (a+b,2b+a), (2b,2a) is λ \lambda
    λ =122a(2b+a2a)+(a+b)(2ab)+2b(b2ba) \therefore \lambda  = \dfrac{1}{2}\left | 2a(2b+a-2a)+(a+b)(2a-b)+2b(b-2b-a) \right | 
    λ =124ab+2a24a2+2a2ab+2abb2+2b24b22ab \lambda  = \dfrac{1}{2}\left | 4ab+2a^{2}-4a^{2}+2a^{2}-ab+2ab-b^{2}+2b^{2}-4b^{2}-2ab \right |
    λ =123ab3b2 \lambda  = \dfrac{1}{2}\left | 3ab - 3b^{2} \right | __ (1)
    Now area of triangle whose vertices are (a+b,ab),(3ba,b+3a),(3ab,3ba) (a+b,a-b), (3b-a,b+3a), (3a-b,3b-a) is given by
    Area =12(a+b)(b+3a3b+a)+(3ba)(3baa+b)+(3ab)(abb3a) = \dfrac{1}{2}\left | (a+b)(b+3a-3b+a)+(3b-a)(3b-a-a+b)+(3a-b)(a-b-b-3a) \right |
    =12(a+b)(2b+4a)+(3ba)(4b2a)+(3ab)(2a2b) = \dfrac{1}{2}\left | (a+b) (-2b+4a)+(3b-a)(4b-2a)+(3a-b)(-2a-2b) \right |
    =122ab+4a22b2+4ab+12b26ab4ab+2a26a26ab+2ab+2b2 = \dfrac{1}{2}\left | -2ab+4a^{2}-2b^{2}+4ab+12b^{2}-6ab-4ab+2a^{2}-6a^{2}-6ab+2ab+2b^{2} \right |
    =1212b212ab = \dfrac{1}{2}\left | 12b^{2}-12ab \right |
    =42(3ab3b2) = \dfrac{4}{2}\left | -(3ab-3b^{2}) \right |
    =4.123ab3b2 = 4.\dfrac{1}{2}\left | 3ab-3b^{2} \right |
    =4λ. = 4 \lambda .
  • Question 6
    1 / -0
    If (1+x+x2)n=r=02narxr(1+x+x^2)^n=\displaystyle\sum^{2n}_{r=0}a_rx^r, then a0a2ra1a2r+1+a2a2r+2....=?a_0a_{2r}-a_1a_{2r+1}+a_2a_{2r+2}-....=?
  • Question 7
    1 / -0
    If repetitions are not allowed, the number of numbers consisting of 44 digits and divisible by 55 and formed out of 0,1,2,3,4,5,60,1,2,3,4,5,6 is 
    Solution
    \rightarrow For divisibility by 55, unit's place should be 00 or 55
    with 00, ----00
             66 choices 6C3×3!\Rightarrow ^6C_3\times 3!
                               =6!3!3!×3!=6.5.4=\dfrac{6!}{3!3!}\times 3!=6.5.4
                                                     =120=120
    with 55, ----55
             66 choices =120=120
                               5C2×2!=10×2=20^5C_2\times 2!=10\times 2=20
                                120+12020=220\Rightarrow 120+120-20=220
                                                      (A)\rightarrow (A)
  • Question 8
    1 / -0
    A Printer number the pages of a book starting with 11 and used 31933193 digit in all. How many pages does the book have:
    Solution
    No. of digits in 1-digit page no. 1×9=9\Rightarrow 1\times 9=9
    No. of digits in 2-digit page no. 2×90=180\Rightarrow 2\times 90= 180
    No. of digits in 3-digit page no. 3×900=2700\Rightarrow 3\times 900= 2700
    No. of digits in 4-digit page no. 300 \Rightarrow  300
    \therefore No. of pages with 4- digit page no. =(300/4)=75= (300/4)=75
    Hence, total no. of pages in the book =(999+75)=1074= (999+75)= 1074
  • Question 9
    1 / -0
    The number of permutations of letters of the word "PARALLAL" atken four at a time must be, 
    Solution
    Permutation of word PARALLAL taken four at a time
    Four letter word might be LLLLLL_
    So total of 1616 words [As 44ways, 44spots]
    For AALLAALL, 66 different arrangements 
    AAAA_ _, can be 4C2{ 4 }_{ { C }_{ 2 } } ways =6=6 ways
    Arranged in 1212 ways is total of 7272 words
    And extra words can be formed in 120120 ways
    16+6+144+120=28616+6+144+120=286 ways
  • Question 10
    1 / -0
    k=110k.k!=\sum _{ k=1 }^{ 10 }{ k.k! } =
    Solution
    Given,

    k=110kk!\sum _{k=1}^{10}kk!

    ak=kk!a_k=kk!

    a1=11!=1a_1=1\cdot \:1!=1

    a2=22!=4a_2=2\cdot \:2!=4

    Similarly,

    a3=18a_3=18

    a4=96a_4=96

    a5=600a_5=600

    a6=4320a_6=4320

    a7=35280a_7=35280

    a8=322560a_8=322560

    a9=3265920a_9=3265920

    a10=36288000a_{10}=36288000

    =1+4+18+96+600+4320+35280+322560+3265920+36288000=1+4+18+96+600+4320+35280+322560+3265920+36288000

    =39916799=39916799

    =11!1=11!-1
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