How many numbers can you form using the digits 1, 5, 6 and 9 without repetition of digits such that the value of the digit 6 in all those nu

Question

How many numbers can you form using the digits 1, 5, 6 and 9 without repetition of digits such that the value of the digit 6 in all those number is 100 times the value of the digit 6 in the number 9561?

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bonexptip 6 months 2021-08-15T13:50:59+00:00 1 Answers 16 views 0

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    2021-08-15T13:52:23+00:00

    Answer:

    6 times

    Step-by-step explanation:

    The value of 6 in the final number should be 100 times the value of 6 in 9561

    in 9561 , the value of 6 is 60

    100 times 60 is 6000

    Hence, the value of 6 in the final number should be 6000

    Now, we have the value of one digit out of 4 for our final number

    So, our number will look like: 6 _ _ _

    these dashes need to be filled by the remaining numbers: 1, 5 , 9

    So, we will permute 1 , 5 and 9 in these 3 places since the position of 6 is permanent

    3P3 = 3! = 6

    Therefore, there are 6 possible 4-digit numbers which have the value of 6 as 6000 and no number are repeated from the given set of (1, 5, 6 , 9)

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