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Permutations and Combinations Test - 1

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Permutations and Combinations Test - 1
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  • Question 1
    1 / -0

    The number of all numbers that can be formed by using some or all of the digits 1, 3, 5, 7, 9 (without repetitions) is

  • Question 2
    1 / -0

    The number of even numbers that can be formed by using all the digits 1, 2, 3, 4, and 5 (without repetitions) is

    Solution
    t th th h t o
    1 2 3 4 2

    Since we need an even number , ones place can be occupied by only two numbers 2 and 4 in two ways.

    Since repetition is not allowed tens place is occupied by remaining 4 numbers in 4 ways and the hundred's place by 3 ways and thousands place in 2 ways.

    Hence total number of even numbers can be formed in 1x2x3x4x2 = 48.

     

  • Question 3
    1 / -0

    Find Rank of word ‘wife ‘among the words that can be formed with its letters and arranged as in dictionary is

    Solution

    Explanation:

    1.Arrange all the alphabets in alphabetical order ( E, F , I  , W )

    2. 4 alphabets can form 4! words = 24 words.

     

  • Question 4
    1 / -0

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

    Solution

    Explanation:

    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 5
    1 / -0

    A coin is tossed n times, the number of all the possible outcomes is

    Solution

    Explanation:

    There can either be a heads or tails, therefore for every toss, the possible outcomes are 2. hence for n number of toss the possibilities are 2n.

     

  • Question 6
    1 / -0

    The number of all three digit even numbers such that if 5 is one of the digits then next digit is 7 is

    Solution

    Explanation:

    You have two different kinds of such three-digit even numbers.First is  5 at the hundred'splace and  second 5 is not at the hundred'splace

    •  In first case no is of the form 57x, where x is the unit's digit ,which can  be 0,2,4,6,8 which is just 5 possibilities.Hence the no of possibilities in this case is 1x1x5=5

    • In second case the hundred's digit can be 1,2,3,,4,,6,7,8,or,9 which is 8 ways and the ten's digit  can be  any of the 9 numbers and unit digit can be any of the 5 even numbers .Therfore the no: of ways  will be 8×9×5=360

    So total we  have 360 + 5 = 365 possibilities.

     

  • Question 7
    1 / -0

    Numbers greater than 1000 but not greater than 5000 are to be formed with the digits 0, 1, 2, 3, 5, allowing repetitions, the number of possible numbers is

    Solution

    Explanation:

    th h t o
    3 5 5 5

    One's place can be occupied by any of the  5 numbers, tens place by any of the 5 numbers and hundreds place in 5 ways since repetition is allowed. But thousands place can be occupied by 2,3,1, only since the required number should be greater than 1000 and less than 5000.Hence  total number of arrangement=3x5x5x5=375

     

  • 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

    Explanation:

    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.

    t th th h t o
    4 1 2 3 4

    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.

    th h t o
    4 2 3 4

    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

    The number of different ways in which a man can invite one or more of his 6 friends to dinner is?

    Solution

    Explanation:

    He can invite any one  friend in 6C1 ways= 6 ways:

    He can invite any two friends in 6C2 ways = 15 ways

    He can invite any three friends in 6C3 ways =20 ways

    He can invite any 4 friends in 6C4ways = 15 ways

    He can invite any 5 friends in 6C5 ways = 6 ways

    He can invite  all the 6 friends in 6C6 ways= 1 way.

    Since any one of these could happen total possibilities are, 6+15+20+15+6+1 = 63.

     

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