How many 5-digit numbers divisible by 6 can be formed using 0,1,2,3,4 and 5, if repitition of degit is not allowed?

This question can be solved by taking 2 cases: Case I : when 0 is not included then the no. 5,4,3,2,1 can be chosen in 48 ways(last place filled in 2 ways i.e. by digits 2&4) Case II : when zero is included(3 must be excluded to keep the sum of digits 1,2,4,5,0 divisible by 3) and is placed at last then 4! ways = 24 ways & when zero is not placed at last then 36 ways(last place in two ways i.e. 2&4 and first in 3 ways as 0 cannot come at first place and remaining 3 places in 3! ways) SO TOTAL NO. OF WAYS = 48+24+36 = 108 ways PLEASE APPROVE MY ANSWER IF YOU ARE SATISFIED BY IT BY CLICKING YES BELOW THE ANSWER AND BEST OF LUCK FOR YOUR PREPARATION

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