2
$\begingroup$

Question:

In how many ways can three different integers be selected from the numbers $1$ to $12$,so that their sum can be exactly divided by $3$?

Solution:

if order is not important (as i am not sure about this) total ways= $4\cdot3\cdot2 + (4\cdot4\cdot4)\cdot 3!= 24 + 192 = 216$

if order is important, total ways = $\binom{4}{3} + \binom{4}{1}\cdot\binom{4}{1}\cdot\binom{4}{1} = 4 + 64 = 68$

Is my solution correct?

$\endgroup$
  • $\begingroup$ You should perhaps explain where your calculations are coming from. I would suspect that you are missing a factor of $3$ in the $4 * 3 * 2$ term. $\endgroup$ – Shagnik Aug 18 '16 at 10:22
  • $\begingroup$ I considered three groups: (group a: 3, 6, 9, 12) (group b: 1, 4, 7,10) (group c: 2, 5, 8, 11) now, i can select either 3 from group a, or 1 each from 3 group. so, i can select 3 numbers from group a = 4c3 = 4 ways 1 from each 3 group= 4c1* 4c1* 4c1= 64 Total= 64+4 = 68 ways $\endgroup$ – Mahmudul Hasan Aug 18 '16 at 10:35
  • 1
    $\begingroup$ thank you @Shagnik i understand now. i only considered 3 number that are o module 3, but missed out 3 numbers that are 1 modulo 3 and 3 numbers that are 2 modulo 3 $\endgroup$ – Mahmudul Hasan Aug 18 '16 at 10:52
5
$\begingroup$

Note that in order that the sum of $3$ number will be divided by $3$, you have that their sum modulo $3$ will be $0$, therefore you could have the next possibilities:

  • $3$ numbers that are $0$ modulo $3$
  • $3$ numbers that are $1$ modulo $3$
  • $3$ numbers that are $2$ modulo $3$
  • $3$ numbers such that one is $0$ modulo $3$, one is $1$ modulo $3$ and one is $2$ modulo $3$

it's not hard to calculate each one of those cases, and since they are disjoint the sum of them to get the wanted result.

$\endgroup$
  • 1
    $\begingroup$ thank you, now i get it. i didn't consider 3 numbers that are 1 modulo 3 and 3 numbers that are 2 modulo 3 $\endgroup$ – Mahmudul Hasan Aug 18 '16 at 10:39
2
$\begingroup$

In the range $[1,12]$ we have:

  • Exactly $4$ values of $n$ such that $n\equiv0\pmod3$
  • Exactly $4$ values of $n$ such that $n\equiv1\pmod3$
  • Exactly $4$ values of $n$ such that $n\equiv2\pmod3$

Therefore we can split it into the following disjoint cases:

  • $a,b,c\equiv0,0,0\pmod3\implies\binom43=4$ combinations
  • $a,b,c\equiv1,1,1\pmod3\implies\binom43=4$ combinations
  • $a,b,c\equiv2,2,2\pmod3\implies\binom43=4$ combinations
  • $a,b,c\equiv0,1,2\pmod3\implies4^3=64$ combinations

Therefore we have $4+4+4+64=76$ combinations altogether.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.