Let S= {1, 2, 3, 4, ..., 28, 29, 30} How many subgroups of 3 elements are in the S so that their sum is a multiple of 3 The answer is 1 360 subgroups


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Question Number 178213 by Acem last updated on 14/Oct/22

$L e t S = {1, 2, 3, 4, . . ., 28, 29, 30}$ $H o w m a n y s u b g r o u p s o f 3 e l e m e n t s a r e i n t h e S$ $s o t h a t t h e i r s u m i s a m u l t i p l e o f 3$ $T h e a n s w e r i s 1 360 s u b g r o u p s$
Commented by Acem last updated on 14/Oct/22

$H i n t : n m o d 3 = {0, 1, 2} \forall n \in N$
	Answered by nikif99 last updated on 14/Oct/22
	$Let S_0 ={3, 6, 9, ..., 30} n=10 S_1 ={1, 4, 7, ..., 28} n=10 S_2 ={2, 5, 8, ..., 29} n=10 Sum divisible by 3: a) 1 element from each S_0 , S_1 , S_2 : 10^3 b) 3 elements from S_0 : C_3 ^(10) =120 c) 3 elements from S_1 : C_3 ^(10) =120 d) 3 elements from S_2 : C_3 ^(10) =120 Total 1360$
	$L e t S_{0} = {3, 6, 9, . . ., 30} n = 10$ $S_{1} = {1, 4, 7, . . ., 28} n = 10$ $S_{2} = {2, 5, 8, . . ., 29} n = 10$ $S u m d i v i s i b l e b y 3 :$ $a) 1 e l e m e n t f r o m e a c h S_{0}, S_{1}, S_{2} : 10^{3}$ $b) 3 e l e m e n t s f r o m S_{0} : C_{3}^{10} = 120$ $c) 3 e l e m e n t s f r o m S_{1} : C_{3}^{10} = 120$ $d) 3 e l e m e n t s f r o m S_{2} : C_{3}^{10} = 120$ $T o t a l 1360$
	Commented by Acem last updated on 14/Oct/22

	$A b o u q u e t o f f l o w e r s!$
	Commented by nikif99 last updated on 14/Oct/22

	$T h a n k y o u .$
	Commented by Tawa11 last updated on 15/Oct/22

	$Great sir$
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