divisiblity by 3 (a problem in combinatorics)
Probability
Level
pending
Given the digits 0, 1, 2, 3, 4, 7 and 8, how many 5-digit numbers can be formed which are divisible by 3 using only the given digits. Each digit can be used at most once.
(Eg: 12348 and 84321 are two such possible numbers)
The answer is 744.
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1 solution
3p type number (gives 0 as remainder on division with 3) → 0, 3. 3p+1 type (gives remainder 1)→ 1, 4, 7. 3p+2 type (gives remainder 2) → 2, 8
For a number to be divisible by 3, the sum of its digits must be a multiple of 3. Case -I : Taking three (3p+1) type number and two 3p type number. Digits favourable (0,3,1,4,7) Permutations for this case is 4.4! = 96
Case -II Two digits of (3p+2), two digits of (3p+1) and one digit of 3p type number. Arrangements for this case = C(3,2) x 4! x (4 + 5) = 648 No other case is favourable here Therefore ans = 96 + 648 = 744