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Permutations and Combinations Test - 2

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Permutations and Combinations Test - 2
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Weekly Quiz Competition
  • Question 1
    1 / -0

    5 men and 6 women have to be seated in a straight row, so that no two women are together. Find the number of ways in which this can be done

  • Question 2
    1 / -0

    If all permutations of the letters of the word AGAIN are arranged as in dictionary, then fiftieth word will be

  • Question 3
    1 / -0

    If nC4, nC5 and nC6 are in A.P, then n is equal to

  • Question 4
    1 / -0

    A 5-digit number divisible by 3 is to be formed using the digits 0, 1, 2, 3, 4 and 5 without repetition. The total number of ways in which this can be done is

  • Question 5
    1 / -0

    The number of integral solutions of the equation x1 + x2 + x3 + x4 = 16, xi > 0 is:

  • Question 6
    1 / -0

    A basket contains 12 red, 10 black and 5 pink balls. 5 balls are drawn at random. The number of ways of drawing at least 2 black balls is

  • Question 7
    1 / -0

    The figures 4, 5, 6, 7, 8 are written in every possible order. The number of numbers greater than 56000 is

    Solution

    Total possibilities  of 5 digit numbers which can be formed using the given digits are 5!= 120 ways.

    But since the number should be greater than 56000 we cannot have the numbers starting with 4 or 54

    The combinations in which the number 4 comes at the start  is 4!  = 24 ways.

    The combinations in which the number 54 comes at the start  is 2! = 6

    Hence the numbers greater than 56,000 = 120 - ( 24+ 6) = 120 - 30 = 90 ways.

     

  • Question 8
    1 / -0

    The number of significant numbers which can be formed by using any number of the digits 0, 1, 2, 3, 4 but using each not more than once in each number is

    Solution

    Case 1: for 5 digit numbers Ten Tens thousands place can be occupied in 4 ways since 0 cannot be suitable. For ones place remaining 4 numbers ( since repetition is not allowed ) can be occupied in 4 ways : and the tens place in 3 ways and hundreds place in 2 ways and the thousands place in 1 way. Hence the total number of ways is 4x1x4x2x3= 96.

    Case 2: for 4 digit numbers thousands place can be occupied in 4 ways since 0 cannot be suitable. For ones place remaining 4 numbers ( since repetition is not allowed ) can be occupied in 4 ways : and the tens place in 3 ways and hundreds place in 2 ways . Hence the total number of ways is 4x4x2x3= 96.

    Case 3: for 3digit numbers hundreds place can be occupied in 4 ways since 0 cannot be suitable. For ones place remaining 4 numbers ( since repetition is not allowed ) can be occupied in 4 ways : and the tens place in 3 ways  Hence the total number of ways is 4x4x3= 48

    Case 4: for 2 digit numbers Tens place can be occupied in 4 ways since 0 cannot be suitable. For ones place remaining 4 numbers ( since repetition is not allowed ) can be occupied. Hence the total number of ways is 4x4=16.

    Case 5: For single digit in 4 ways .

    Hence 96 + 96 + 48+ 16+4 = 260

     

  • Question 9
    1 / -0

    In a function, 10 persons including A, B and C have to perform. If A has to perform before B and B has to perform before C, how many different ways are there to arrange the order of their performances?

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