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Permutation Test - 10
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Permutation Test - 10
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  • Question 1/7
    1 / -0

    Exponent of 12 in 50! is

    Solutions

    12 = 22 × 3

    Exponent of 12 in 50! = min{E2(50!), E3(50!)} 

    = E3 (50!) = 22 

     

  • Question 2/7
    1 / -0

    The number of zero's at the end 60! of is

    Solutions

    Number of zero's at the end of 60! = exponent 10 in 

    60! = min{E2 (60!), E5 (60!)} 

    = E5 (60!) = 14.

     

  • Question 3/7
    1 / -0

    Three dice are rolled. The number of possible outcomes in which at least one die shows 6 is :

    Solutions

    Total number of outcomes = 6 × 6 × 6 = 216

    Number of outcomes in which none of the dice shows 6 = 5 × 5 × 5 = 125

    ∴ Number of outcomes in which at least one die shows 6 = 216 - 125 = 91

     

  • Question 4/7
    1 / -0

    The maximum number of points of intersection of 8 circles, is

    Solutions

    Two circles intersect in 2 points.

    ∴ Maximum number of points of intersection

    = 2 × number of selection of two circles from 8 circles

    = 2 × 8C2 = 2 × 28 

    = 56.

     

  • Question 5/7
    1 / -0

    Let 1 ≤ m < n ≤ p. The number of subsets of the set A = {1, 2, 3,…p} having m,n as the least and the greatest elements respectively, is

    Solutions

    Total number of subject = the number of selections of at least two elements including m, n and natural numbers lying between m and n= total number of selections from n-m-1 different things 2n - m - 1

     

  • Question 6/7
    1 / -0

    The number of odd proper divisors of 3p . 6m. 21n is

    Solutions

    3p. 6m. 21n  =  2m . 3p + m + n . 7n

    ∴ The required number of proper divisors 

    = Number of selection of any number of 3's and 7's

    [∴ For odd divisors 2 must not be selected]

    = (p + m + n + 1)(n + 1) - 1.

     

  • Question 7/7
    1 / -0

    The number of even proper divisors of 1008 is

    Solutions

    1008 = 24 × 32 × 7

    ∴ The required number of even proper divisors

    = total number of selections of at least one 2 and any number of 3's or 7's

    = 4 × (2 + 1) × (1 + 1) - 1 = 23.

     

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