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27 Mar 2026, 07:56
Kudos
Bookmarks
Adding all = 15 = multiple of 3.
Arranging those 5 numbers = 5! ways = 120
You can also form another sum using a 0 and 1,2,4,5 adding to 12. This somehow happens because you stretch the sum of 3 using two terms and fill the remaining with a 0.
Arranging them = 5! - 4! (Terms formed with 0 at start subtracted) = 96
Total = 120+96 = 216
Answer: Option E
Arranging those 5 numbers = 5! ways = 120
You can also form another sum using a 0 and 1,2,4,5 adding to 12. This somehow happens because you stretch the sum of 3 using two terms and fill the remaining with a 0.
Arranging them = 5! - 4! (Terms formed with 0 at start subtracted) = 96
Total = 120+96 = 216
Answer: Option E
Quote:
25 Apr 2026, 10:40
Kudos
Bookmarks
out of 0,1,2,3,4,5 six digits, 5 digit multiple of 3 is to be formed.
case 1: when using only 1,2,3,4,5 as the sum is divisible by 3 can be arranged among themselves in !5=120 ways
case2: if we add 0 to the above group then we must remove 3 so as to be multiple of 3.
0 can only be used at units, tens, hundreds and thousands place--4 ways
and rest can be arranged at the leftover 4 places in !4 ways=24
total=24*4=96
hence total no. of requisite such numbers=120+96=216
case 1: when using only 1,2,3,4,5 as the sum is divisible by 3 can be arranged among themselves in !5=120 ways
case2: if we add 0 to the above group then we must remove 3 so as to be multiple of 3.
0 can only be used at units, tens, hundreds and thousands place--4 ways
and rest can be arranged at the leftover 4 places in !4 ways=24
total=24*4=96
hence total no. of requisite such numbers=120+96=216
