The number 1,2,3,4,...,7 are randomly divided into two non −empty subsets. The probability that the sum of the numbers in the two subsets being equal is (r/s) expressed in the lowest term. Find r+s ?


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Question Number 116397 by john santu last updated on 03/Oct/20

$T h e n u m b e r 1, 2, 3, 4, . . ., 7 a r e r a n d o m l y$ $d i v i d e d i n t o t w o n o n - e m p t y s u b s e t s .$ $T h e p r o b a b i l i t y t h a t t h e s u m o f t h e$ $n u m b e r s i n t h e t w o s u b s e t s b e i n g$ $e q u a l i s \frac{r}{s} e x p r e s s e d i n t h e l o w e s t$ $t e r m . F i n d r + s ?$
	Answered by mr W last updated on 03/Oct/20
	$1+2+3+4+5+6+7=28 each subset should have the sum 14. subsets with 1 number & 6 numbers: n=C_1 ^7 =7 possibilities n_S =0 subsets with 2 numbers & 5 numbers: n=C_2 ^7 =21 possibilities n_S =0 subsets with 3 numbers & 4 numbers: n=C_3 ^7 =35 possibilities 7+3+4, 7+2+5, 7+1+6, 6+5+3=14 n_S =4 p=((0+0+4)/(7+21+35))=(4/(63))=(r/s) ⇒r+s=4+63=67 we can also get the total number of possibilities via {_2 ^7 }=63.$
	$1 + 2 + 3 + 4 + 5 + 6 + 7 = 28$ $e a c h s u b s e t s h o u l d h a v e t h e s u m 14 .$ $s u b s e t s w i t h 1 n u m b e r & 6 n u m b e r s :$ $n = C_{1}^{7} = 7 p o s s i b i l i t i e s$ $n_{S} = 0$ $s u b s e t s w i t h 2 n u m b e r s & 5 n u m b e r s :$ $n = C_{2}^{7} = 21 p o s s i b i l i t i e s$ $n_{S} = 0$ $s u b s e t s w i t h 3 n u m b e r s & 4 n u m b e r s :$ $n = C_{3}^{7} = 35 p o s s i b i l i t i e s$ $7 + 3 + 4, 7 + 2 + 5, 7 + 1 + 6, 6 + 5 + 3 = 14$ $n_{S} = 4$ $p = \frac{0 + 0 + 4}{7 + 21 + 35} = \frac{4}{63} = \frac{r}{s}$ $\Rightarrow r + s = 4 + 63 = 67$ $w e c a n a l s o g e t t h e t o t a l n u m b e r o f$ $p o s s i b i l i t i e s v i a {_{2}^{7}} = 63 .$
	Commented by john santu last updated on 04/Oct/20

	$y e s . . t h a n k y o u$
	Answered by john santu last updated on 04/Oct/20
	$The total number of ways of dividing the seven numbers into two non−empty subsets is ((2^7 −2)/2) = 63. Note that since 1+2+3+...+7=28 ,the sum of the number in each of the two groups is 14. Note also that numbers 5,6,7 cannot be in the same group since 5+6+7=18>14 case(1) only 6 &7 in the same group and 5 the other group. {2,3,4,5},{1,6,7} case(2) only 5 &6 in the same group and 7 in the other group . {1,2,5,6}, {3,4,7} {3,5,6}, { 1,2,4,7 } case(3) only 5 &7 in the same group and 6 in the other group . {2,5,7}, {1,3,4,6} Hence there are 4 such possibilities Thus the required probability is (4/(63)) yielding that p+q = 67$
	$T h e t o t a l n u m b e r o f w a y s o f d i v i d i n g$ $t h e s e v e n n u m b e r s i n t o t w o n o n - e m p t y$ $s u b s e t s i s \frac{2^{7} - 2}{2} = 63 . N o t e t h a t s i n c e$ $1 + 2 + 3 + . . . + 7 = 28, t h e s u m o f t h e$ $n u m b e r i n e a c h o f t h e t w o g r o u p s i s 14 .$ $N o t e a l s o t h a t n u m b e r s 5, 6, 7 c a n n o t$ $b e i n t h e s a m e g r o u p s i n c e 5 + 6 + 7 = 18 > 14$ $c a s e (1) o n l y 6 & 7 i n t h e s a m e g r o u p a n d$ $5 t h e o t h e r g r o u p . {2, 3, 4, 5}, {1, 6, 7}$ $c a s e (2) o n l y 5 & 6 i n t h e s a m e g r o u p a n d$ $7 i n t h e o t h e r g r o u p . {1, 2, 5, 6}, {3, 4, 7}$ ${3, 5, 6}, {1, 2, 4, 7}$ $c a s e (3) o n l y 5 & 7 i n t h e s a m e g r o u p a n d$ $6 i n t h e o t h e r g r o u p . {2, 5, 7}, {1, 3, 4, 6}$ $H e n c e t h e r e a r e 4 s u c h p o s s i b i l i t i e s$ $T h u s t h e r e q u i r e d p r o b a b i l i t y i s$ $\frac{4}{63} y i e l d i n g t h a t p + q = 67$
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