There are four boxes, each of them contains exactly the same numbers: 1,2,3,...,n. Four different numbers are drawn from the boxes and multiplicated with each other to get a product. What′s the sum of all products? Σ


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Question Number 88372 by mr W last updated on 10/Apr/20

$T h e r e a r e f o u r b o x e s, e a c h o f t h e m$ $c o n t a i n s e x a c t l y t h e s a m e n u m b e r s :$ $1, 2, 3, . . ., n .$ $F o u r d i f f e r e n t n u m b e r s a r e d r a w n$ $f r o m t h e b o x e s a n d m u l t i p l i c a t e d$ $w i t h e a c h o t h e r t o g e t a p r o d u c t .$ ${W h a t}^{'} s t h e s u m o f a l l p r o d u c t s ?$ $\sum_{a \neq b \neq c \neq d} a b c d = ?$
Commented by mr W last updated on 10/Apr/20

$t h a n k y o u s i r!$ $s o r r y, i {d i d n}^{'} t m a k e c l e a r : t h e b o x e s$ $a r e i d e n t i c a l, s o t h e o d e r o f n u m b e r s$ $p l a y e s n o r u l e . e x a m p l e 1, 3, 8, 10$ $s h o u l d b u i l d o n l y o n e s i n g l e p r o d u c t$ $1 \times 3 \times 8 \times 10 i n t h e s u m .$
Commented by mr W last updated on 10/Apr/20

$m a y b e {i t}^{'} s b e t t e r i f i h a d e x p r e s s e d t h e$ $q u e s t i o n l i k e t h i s :$ $a b o x c o n t a i n s t h e n u m b e r s 1, 2, 3, . . ., n .$ $f o u r n u m b e r s a r e d r a w n f r o m t h e$ $b o x a n d m u l t i p l i c a t e d w i t h e a c h o t h e r$ $t o b u i l d a p r o d u c t . f i n d t h e s u m o f$ $t h e p r o d u c t s o f a l l p o s s i b l e c o m b i n a t i o n s .$
Commented by jagoll last updated on 10/Apr/20

$(1.2.3.4) + (1.3.4.5) + (1.4.5.6) + . . .$ $((n - 3) . (n - 2) . (n - 1) . n) like this ?$
Commented by mr W last updated on 10/Apr/20

$y e s .$
Commented by jagoll last updated on 10/Apr/20

$i {don}^{'} t have idea sir . how to$ $solve it ?$
Commented by Tony Lin last updated on 10/Apr/20
$first consider Σ_(a≠b) ab Σ_(a≠b) ab =1×2+1×3+∙∙∙+1×n +2×1+2×3+∙∙∙+2×n ∙∙∙ +n×1+n×2+∙∙∙+n(n−1)r =1[((n(n+1))/2)]−1^2 +2[((n(n+1))/2)]−2^2 +∙∙∙ +n[((n(n+1))/2)]−n^2 =[((n(n+1))/2)]^2 −((n(n+1)(2n+1))/6) =((n(n−1)(n+1)(3n+2))/(12)) Σ_(a≠b≠c) abc =1×2×3+1×2×4+∙∙∙+1×2×n +1×3×2+1×3×4+∙∙∙+1×n(n−1) ∙∙∙ +1×n×2+1×n×3+∙∙∙+1×n(n−1) ∙∙∙and Σ2×∙∙∙ Σ3×∙∙∙ to Σn×∙∙∙ =((n(n−1)(n+1)(3n+2))/(12))−2[((n(n+1))/2)−1^2 ] +((n(n−1)(n+1)(3n+2))/(12))−2[((2n(n+1))/2)−2^2 ] ∙∙∙ +((n(n−1)(n+1)(3n+2))/(12))−2[((n^2 (n+1))/2)−n^2 ] =((n^2 (n−1)(n+1)(3n+2))/(12))−((2n(n−1)(n+1)(3n+2))/(12)) =((n(n−1)(n−2)(n+1)(3n+2))/(12)) Σ_(a≠b≠c≠d) abcd =1×2×3×4+1×2×3×5+∙∙∙+1×2×3×n +1×2×4×3+1×2×4×5+∙∙∙+1×2×4×n ∙∙∙ +1×2×n×3+1×2×n×4+∙∙∙+1×2×n(n−1) ∙∙∙and Σ1×3×∙∙∙to Σ1×n×∙∙∙ and Σ2×∙∙∙ to Σn×∙∙∙ =((n(n−1)(n−2)(n+1)(3n+2))/(12)) −3{((n(n−1)(n+1)(3n+2))/(12))−2[((n(n+1))/2)−1^2 ]} +((n(n−1)(n−2)(n+1)(3n+2))/(12)) −3{((n(n−1)(n+1)(3n+2))/(12))−2[((2n(n+1))/2)−2^2 ] +∙∙∙ =((n^2 (n−1)(n−2)(n+1)(3n+2))/(12)) −3[((n^2 (n−1)(n+1)(3n+2))/(12))−((2n(n−1)(n+1)(3n+2))/(12))] =((n^2 (n−1)(n−2)(n+1)(3n+2))/(12)) − ((3n(n−1)(n−2)(n+1)(3n+2))/(12)) =((n(n−1)(n−2)(n−3)(n+1)(3n+2))/(12)) from the rule Σ_(a_1 ≠a_2 ≠a_3 ∙∙∙≠a_k ) Π_(j=1) ^k a_j ,1<k<n =((P_k ^n (n+1)(3n+2))/(12)) , P_k ^n =C_k ^n ×k!$
$f i r s t c o n s i d e r \sum_{a \neq b} a b$ $\sum_{a \neq b} a b$ $= 1 \times 2 + 1 \times 3 + \cdot \cdot \cdot + 1 \times n$ $+ 2 \times 1 + 2 \times 3 + \cdot \cdot \cdot + 2 \times n$ $\cdot \cdot \cdot$ $+ n \times 1 + n \times 2 + \cdot \cdot \cdot + n (n - 1) r$ $= 1 [\frac{n (n + 1)}{2}] - 1^{2} + 2 [\frac{n (n + 1)}{2}] - 2^{2} + \cdot \cdot \cdot$ $+ n [\frac{n (n + 1)}{2}] - n^{2}$ $= {[\frac{n (n + 1)}{2}]}^{2} - \frac{n (n + 1) (2 n + 1)}{6}$ $= \frac{n (n - 1) (n + 1) (3 n + 2)}{12}$ $\sum_{a \neq b \neq c} a b c$ $= 1 \times 2 \times 3 + 1 \times 2 \times 4 + \cdot \cdot \cdot + 1 \times 2 \times n$ $+ 1 \times 3 \times 2 + 1 \times 3 \times 4 + \cdot \cdot \cdot + 1 \times n (n - 1)$ $\cdot \cdot \cdot$ $+ 1 \times n \times 2 + 1 \times n \times 3 + \cdot \cdot \cdot + 1 \times n (n - 1)$ $\cdot \cdot \cdot a n d Σ 2 \times \cdot \cdot \cdot Σ 3 \times \cdot \cdot \cdot t o Σ n \times \cdot \cdot \cdot$ $= \frac{n (n - 1) (n + 1) (3 n + 2)}{12} - 2 [\frac{n (n + 1)}{2} - 1^{2}]$ $+ \frac{n (n - 1) (n + 1) (3 n + 2)}{12} - 2 [\frac{2 n (n + 1)}{2} - 2^{2}]$ $\cdot \cdot \cdot$ $+ \frac{n (n - 1) (n + 1) (3 n + 2)}{12} - 2 [\frac{n^{2} (n + 1)}{2} - n^{2}]$ $= \frac{n^{2} (n - 1) (n + 1) (3 n + 2)}{12} - \frac{2 n (n - 1) (n + 1) (3 n + 2)}{12}$ $= \frac{n (n - 1) (n - 2) (n + 1) (3 n + 2)}{12}$ $\sum_{a \neq b \neq c \neq d} a b c d$ $= 1 \times 2 \times 3 \times 4 + 1 \times 2 \times 3 \times 5 + \cdot \cdot \cdot + 1 \times 2 \times 3 \times n$ $+ 1 \times 2 \times 4 \times 3 + 1 \times 2 \times 4 \times 5 + \cdot \cdot \cdot + 1 \times 2 \times 4 \times n$ $\cdot \cdot \cdot$ $+ 1 \times 2 \times n \times 3 + 1 \times 2 \times n \times 4 + \cdot \cdot \cdot + 1 \times 2 \times n (n - 1)$ $\cdot \cdot \cdot a n d Σ 1 \times 3 \times \cdot \cdot \cdot t o Σ 1 \times n \times \cdot \cdot \cdot$ $a n d Σ 2 \times \cdot \cdot \cdot t o Σ n \times \cdot \cdot \cdot$ $= \frac{n (n - 1) (n - 2) (n + 1) (3 n + 2)}{12}$ $- 3 {\frac{n (n - 1) (n + 1) (3 n + 2)}{12} - 2 [\frac{n (n + 1)}{2} - 1^{2}]}$ $+ \frac{n (n - 1) (n - 2) (n + 1) (3 n + 2)}{12}$ $- 3 {\frac{n (n - 1) (n + 1) (3 n + 2)}{12} - 2 [\frac{2 n (n + 1)}{2} - 2^{2}]$ $+ \cdot \cdot \cdot$ $= \frac{n^{2} (n - 1) (n - 2) (n + 1) (3 n + 2)}{12}$ $- 3 [\frac{n^{2} (n - 1) (n + 1) (3 n + 2)}{12} - \frac{2 n (n - 1) (n + 1) (3 n + 2)}{12}]$ $= \frac{n^{2} (n - 1) (n - 2) (n + 1) (3 n + 2)}{12}$ $- \frac{3 n (n - 1) (n - 2) (n + 1) (3 n + 2)}{12}$ $= \frac{n (n - 1) (n - 2) (n - 3) (n + 1) (3 n + 2)}{12}$ $f r o m t h e r u l e$ $\sum_{a_{1} \neq a_{2} \neq a_{3} \cdot \cdot \cdot \neq a_{k}} \underset{j = 1}{\prod^{k}} a_{j}, 1 < k < n$ $= \frac{P_{k}^{n} (n + 1) (3 n + 2)}{12}, P_{k}^{n} = C_{k}^{n} \times k!$
Commented by Tony Lin last updated on 10/Apr/20

$I g u e s s t h e a n s w e r m i g h t b e$ $\frac{C_{k}^{n} (n + 1) (3 n + 2)}{12}$ $w h i c h h a s n o p e r m u t a t i o n$ $o n l y c o m b i n a t i o n$
Commented by mr W last updated on 10/Apr/20

$i {d o n}^{'} t h a b e t h e s o l u t i o n f o r \sum_{a \neq b \neq c \neq d} a b c d .$ $b u t \sum_{a \neq b \neq c} a b c = \frac{(n - 2) (n - 1) n^{2} {(n + 1)}^{2}}{48} .$ $c a n y o u p l e a s e c h e c k w i t h y o u r s ?$
Commented by Tony Lin last updated on 10/Apr/20

$m y s u p p o s e i s w r o n g b u t I f o u n d$ $Σ a = \underset{k = 1}{\sum^{n}} k = \frac{n (n + 1)}{2}$ $\sum_{a \neq b} a b = \underset{k = 1}{\sum^{n - 1}} \frac{k (k + 1)}{2} = \frac{n (n - 1) {(n + 1)}^{2} (3 n + 2)}{24}$ $\sum_{a \neq b \neq c} a b c = \underset{k = 1}{\sum^{n - 1}} \frac{k (k - 1) {(k + 1)}^{2} (3 k + 2)}{24}$ $= \frac{n^{2} {(n + 1)}^{2} (n - 1) (n - 2)}{48}$ $\sum_{a \neq b \neq c \neq d} a b c d = \underset{k = 1}{\sum^{n - 1}} \frac{k^{2} {(k + 1)}^{2} (k - 1) (k - 2)}{48}$ $= \frac{(n - 3) (n - 2) (n - 1) n (n + 1) (15 n^{3} + 15 n^{2} - 10 n - 8)}{5760}$
Commented by mr W last updated on 10/Apr/20

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