The total number of numbers, lying between 100 and 1000 that can be formed with the digits 1,2,3,4,5, if the repetition of digits is not allowed and numbers are divisible by either 3 or 5, is
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Solution
12→3,4,5→3!=6 15→2,3,4→3!=6 24→1,3,5→3!=6 45→1,2,3→3!=6
total =24
divisible by 5
=4×3=12
divisible by →15 135;315;345;435 4 numbers
Required Numbers =24+12−4=32