Q.How many numbers having 5 digits can be formed with the digits 0, 2, 3, 4 and 5 if repetition of digits is not allowed. How many of these are divisible by 5?
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Start your 14-day free trial to unlock the full solution →Using digits without repetition, there are valid 5-digit numbers (leading digit ), of which are divisible by 5 (units digit or ).
[!FORMULA] By the multiplication principle with the restriction that the leading digit cannot be : total 5-digit numbers (choices for the first digit, excluding ) (permutations of the remaining 4 digits in the remaining 4 places) . For divisibility by 5, the units digit must be or ; each case is counted separately (with the first-digit restriction reapplied) and the two cases summed.
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Total 5-digit numbers (no repetition), digits available : the first (leftmost) digit cannot be , so it has choices (). The remaining 4 positions are filled by arranging the remaining 4 digits (which now include ) in ways.
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Compute: .
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Divisible by 5 — Case A: units digit . The units place is fixed at . The first digit can be any of the remaining 4 nonzero digits : choices. The middle 3 digits are filled by arranging the remaining 3 digits in ways.
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Case A count: .
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