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Permutations and Combinations Test 43

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Permutations and Combinations Test 43
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  • Question 1
    1 / -0
    Find : 13,23,43,83,163,?13,23,43,83,163 , ?
  • Question 2
    1 / -0
    The number of seven letter words that can be formed by using the letters of the word  SUCCESSSUCCESS  that the two  CC are together but no two  SS  are together is
    Solution
    using the letters of the word  SUCCESSSUCCESS  that the two  CC are together but no two  SS  are together 
    let two C's be 1 unit 
    \therefore no. of ways =6!4!=30=\frac{6!}{4!}=30
    We have to put 1 letter between one S
    So, No of ways=2×3!=12=2\times 3!=12
    No. of ways=30-12=18
  • Question 3
    1 / -0
    A committee of 44 persons is to be formed from 22 ladies, 22 old men and 44 young men such that it includes at least 11 lady. at least 11 old man and at most 22 young men. Then the total number of ways in which this committee can be formed is :
    Solution

  • Question 4
    1 / -0
    The value of 47C4+j=15 (52j) C3^{47}C_{4}+\displaystyle \sum _{ j=1 }^{ 5 }\ ^{ \left( 52-j \right)  } { C }_{ 3 } is
    Solution
    We  have,
    47C4+51C3+50C3+49C3+48C3+47C3=nCr+nCr1=n+1Cr=48C4+48C3+49C3+50C3+51C3=49C4+49C3+50C3+51C3\begin{array}{l} ^{ 47 }{ C_{ 4 } }{ +^{ 51 } }{ C_{ 3 } }{ +^{ 50 } }{ C_{ 3 } }{ +^{ 49 } }{ C_{ 3 } }{ +^{ 48 } }{ C_{ 3 } }{ +^{ 47 } }{ C_{ 3 } } \\ { =^{ n } }{ C_{ r } }{ +^{ n } }{ C_{ r-1 } }{ =^{ n+1 } }{ C_{ r } } \\ { =^{ 48 } }{ C_{ 4 } }{ +^{ 48 } }{ C_{ 3 } }{ +^{ 49 } }{ C_{ 3 } }{ +^{ 50 } }{ C_{ 3 } }{ +^{ 51 } }{ C_{ 3 } } \\ { =^{ 49 } }{ C_{ 4 } }{ +^{ 49 } }{ C_{ 3 } }{ +^{ 50 } }{ C_{ 3 } }{ +^{ 51 } }{ C_{ 3 } } \end{array}
    Similarly,
    =52C4=^{52}{C_4}.
  • Question 5
    1 / -0
    496:204::329:?496 : 204 : : 329 : ?
  • Question 6
    1 / -0
    Find the missing number.

    Solution

     Step 1: Observing the difference between the adjacent numbers \textbf{ Step 1: Observing the difference between the adjacent numbers}

                     If we notice the adjacent numbers, we see 4×21=7 \text{If we notice the adjacent numbers, we see } 4\times 2 -1 =7

                     Similarly, 7×21=13 \text{Similarly, } 7 \times 2-1=13

                     Similarly, 13×21=25 \text{Similarly, } 13\times2 -1=25

                     Similarly, 25×21=49 \text{Similarly, } 25\times2-1=49

    Step 2: Calculating the missing number \textbf{Step 2: Calculating the missing number}

                    Thus, looking at the above observations, we can conclude the missing number will be: \text{Thus, looking at the above observations, we can conclude the missing number will be:}

                    49×21=97 \Rightarrow 49\times 2-1=97

    Thus, the missing number is D 97\textbf{Thus, the missing number is D 97}

  • Question 7
    1 / -0
    Find xx, if 14!1x=15!\dfrac {1}{4!}-\dfrac {1}{x}=\dfrac {1}{5!}.
    Solution

  • Question 8
    1 / -0
    Find the sum of first 1515 terms of the sequence whose nth{n}^{th} term is 3+4n3+4n.
    Solution
    nthn^{th} term = 3+4 n
    1st1^{st} term = 3+4(1)=7
    2nd2^{nd} term = 3+4(2)=11
    3rd3^{rd} term 3+4(3)=15
    It is an AP with a=7, d=4
    S15=152[2(7)+14(4)]S_{15}=\frac{15}{2}[2(7)+14(4)]
    =15[7+28]= 15[7+28]
    =525= 525

  • Question 9
    1 / -0
    solve

    Solution
    Pattens is as follows:-

    (209)2=121, (20 - 9)^{2} = 121 ,                  (2411)2=169(24 - 11)^{2} = 169

    So (2813)2=152=225 (28 - 13)^{2} = 15^{2} = 225

    Hence missing place = 225 
  • Question 10
    1 / -0
    insert the missing number:  5,8,12,17,23,___,38 
    Solution
    5,8,12,17,23,......385,8,12,17,23,......38
    5+3=8;8+4=12,12+5=17,17+6=235+3=8;8+4=12,12+5=17,17+6=23
    23+7=30,30+8=3023+7=30,30+8=30
    30\therefore 30
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