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Find the number of 4 digit figures that can be formed using 4 digits which lie between 3500 and 6000 from the numbers 2; 3; 4; 5; 6; 7; 8; 9 without repetition.

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  • 9 years ago
    Favorite Answer

    2*7*6*5 + 5*6*5 = 570

  • Anonymous
    9 years ago

    From 2 to 9 there are total "8" numbers.

    First find all numbers between 3500 and 4000

    Make four boxes to allot numbers,

    In first box you can put 3 in "1" way

    In second box you can put 5,6,7,8,9 in "5" ways..................6 no. remains

    In 3rd and 4th box you can put remaining 6 numbers in 6P2 ways

    Total no. of ways = 1 x 5 x 6P2 = 150

    Now, find all numbers between 4000 and 6000

    In first box you can put 4 and 5 in "2" way...................7 no. remains

    In 3 boxes you can put remaining 7 numbers in 7P3 ways

    Total no. of ways = 2 x 7P3 = 420

    Total numbers b/w 3500 and 6000

    = 150 + 420

    = 570

  • Anonymous
    9 years ago

    1*5*6P2 + 1*7P3 + 1*7P3

    = 570 .... answer

    @ kashish : first digit can be 3 or 4 or 5.

    if the first digit is 3 then the second digit has to be greater than or equal to 5 cuz minimum nuber is 3500 so second digit can be 5,6,7,8,9

    so, 1*5*6P2 ways

    if first digit is 4 or 5 then other digits can be arranged is 7P3 ways.

    so 1*7P3 + 1*7P3.

    so total 1*5*6P2 + 1*7P3 + 1*7P3 = 570 ... answer

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