How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7, 9 which are divisible by 5 and none of the digits is repeated?

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+2 Mathematics Permutation and Combinations

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For all the numbers the last digit must be 5. So, the 1st, and the 2nd places are occupied by 2, 3, 6, 7, and 9 without any repetition. So, this can be done by taking 2 digits at a time from the above 6 digits. The number of ways this can be done is 6P2 = !6/!(6-2) = !6/!4 = (6x5x4x3x2x1)/(4x3x2x1) =...
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For all the numbers the last digit must be 5. So, the 1st, and the 2nd places are occupied by 2, 3, 6, 7, and 9 without any repetition. So, this can be done by taking 2 digits at a time from the above 6 digits. The number of ways this can be done is 6P2 = !6/!(6-2) = !6/!4 = (6x5x4x3x2x1)/(4x3x2x1) = 6x5 = 30. (Answer) read less
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