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NDA - Mathematic P & C Test 1188
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NDA - Mathematic P & C Test 1188
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  • Question 1/5
    1 / -0.33
    A tea party is arranged for 16 people along two sides of a long table with eight chairs on each side. Four particular men wish to sit on one particular side and two particular men on the other side. The number of ways they can be seated is
    Solutions

    Since four particular men want to sit on a particular side X (say) and two other particular men on the other side Y. So, we are left with 10 guests out of which we can choose 4 for side A and 6 for side B

    Hence, the number of selection for the two sides = ¹⁰C₄ × ⁶C₆

    Now, 8 persons on each side of the table can be arranged among themselves in 8! ways.

    Hence, the total number of arrangements.

    = ¹⁰C₄ × ⁶C₆ × 8! × 8!

    =  × 1 × 8! × 8!

    =  × (8!)²

    = 210 × (8!)²

  • Question 2/5
    1 / -0.33
    There are 9 students appearing for an examination out of which 3 are appearing for physics and remaining 6 are appearing for different subjects. In how many ways can they sit in a row such that no two physics candidates are together.
    Solutions

    From 9 students, 6 are appearing for different subjects so they can arrange in  Ways.

    Now, these 6 students will create 7 gaps between them.

     

    As no two candidates appearing for physics, sit together so these 3 candidates can sit in these gaps in 7P3 ways

    So total number of required ways =  7P3 ways = 720 ×210 = 151200

  • Question 3/5
    1 / -0.33
    What is equal to?
    Solutions

    47C4+51C3+50C3+49C3+48C3+47C3

    We know that nCr+nCr+1 = n+1Cr+1

    47C4+47C3+48C3+49C3+50C3+51C3

    48C4+48C3+49C3+50C3+51C3

    49C4+49C3+50C3+51C3

    50C4+50C3+51C3

    51C3+51C3

    52C4

  • Question 4/5
    1 / -0.33
    If an =  then value of  is _________
    Solutions
    Given
     an =

    Formula Used
    Concept of summation simplification

    Calculation

     =  –

    =  -

    = 0

    Hence Option D is correct.

  • Question 5/5
    1 / -0.33
    What is the maximum number of points of intersection of 5 non-overlapping circles?
    Solutions

    Maximum number of points of intersection of n non-overlapping circles  =

    Here, n = 5

    Hence, Maximum number of points of intersection of 5 non-overlapping circles  =

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