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

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

$$Therearefourboxes,eachofthem$$$$containsexactlythesamenumbers:$$$$1,2,3,\dots ,n.$$$$Fourdifferentnumbersaredrawn$$$$fromtheboxesandmultiplicated$$$$witheachothertogetaproduct.$$$${What}^{\prime }sthesumofallproducts?$$$$\sum _{a\ne b\ne c\ne d}abcd=?$$

Commented by mr W last updated on 10/Apr/20

$$thankyousir!$$$$sorry,i{didn}^{\prime }tmakeclear:theboxes$$$$areidentical,sotheoderofnumbers$$$$playesnorule.example1,3,8,10$$$$shouldbuildonlyonesingleproduct$$$$1\times 3\times 8\times 10inthesum.$$

Commented by mr W last updated on 10/Apr/20

$$maybe{it}^{\prime }sbetterifihadexpressedthe$$$$questionlikethis:$$$$aboxcontainsthenumbers1,2,3,\dots ,n.$$$$fournumbersaredrawnfromthe$$$$boxandmultiplicatedwitheachother$$$$tobuildaproduct.findthesumof$$$$theproductsofallpossiblecombinations.$$

Commented by jagoll last updated on 10/Apr/20

$$(1.2.3.4)+(1.3.4.5)+(1.4.5.6)+\dots$$$$((\mathrm{n}-3).(\mathrm{n}-2).(\mathrm{n}-1).\mathrm{n})\mathrm{like}\mathrm{this}?$$

Commented by mr W last updated on 10/Apr/20

$$yes.$$

Commented by jagoll last updated on 10/Apr/20

$$\mathrm{i}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{have}\mathrm{idea}\mathrm{sir}.\mathrm{how}\mathrm{to}$$$$\mathrm{solve}\mathrm{it}?$$

Commented by Tony Lin last updated on 10/Apr/20

$$firstconsider\sum _{a\ne b}ab$$$$\sum _{a\ne b}ab$$$$=1\times 2+1\times 3+\cdot \cdot \cdot +1\times n$$$$+2\times 1+2\times 3+\cdot \cdot \cdot +2\times n$$$$\cdot \cdot \cdot$$$$+n\times 1+n\times 2+\cdot \cdot \cdot +n(n-1)r$$$$=1\left[\frac{n(n+1)}{2}\right]-1^2+2\left[\frac{n(n+1)}{2}\right]-2^2+\cdot \cdot \cdot$$$$+n\left[\frac{n(n+1)}{2}\right]-n^2$$$$={\left[\frac{n(n+1)}{2}\right]}^2-\frac{n(n+1)(2n+1)}{6}$$$$=\frac{n(n-1)(n+1)(3n+2)}{12}$$$$\sum _{a\ne b\ne c}abc$$$$=1\times 2\times 3+1\times 2\times 4+\cdot \cdot \cdot +1\times 2\times n$$$$+1\times 3\times 2+1\times 3\times 4+\cdot \cdot \cdot +1\times n(n-1)$$$$\cdot \cdot \cdot$$$$+1\times n\times 2+1\times n\times 3+\cdot \cdot \cdot +1\times n(n-1)$$$$\cdot \cdot \cdot and\mathrm{\Sigma }2\times \cdot \cdot \cdot \mathrm{\Sigma }3\times \cdot \cdot \cdot to\mathrm{\Sigma }n\times \cdot \cdot \cdot$$$$=\frac{n(n-1)(n+1)(3n+2)}{12}-2[\frac{n(n+1)}{2}-1^2]$$$$+\frac{n(n-1)(n+1)(3n+2)}{12}-2[\frac{2n(n+1)}{2}-2^2]$$$$\cdot \cdot \cdot$$$$+\frac{n(n-1)(n+1)(3n+2)}{12}-2[\frac{n^2(n+1)}{2}-n^2]$$$$=\frac{n^2(n-1)(n+1)(3n+2)}{12}-\frac{2n(n-1)(n+1)(3n+2)}{12}$$$$=\frac{n(n-1)(n-2)(n+1)(3n+2)}{12}$$$$\sum _{a\ne b\ne c\ne d}abcd$$$$=1\times 2\times 3\times 4+1\times 2\times 3\times 5+\cdot \cdot \cdot +1\times 2\times 3\times n$$$$+1\times 2\times 4\times 3+1\times 2\times 4\times 5+\cdot \cdot \cdot +1\times 2\times 4\times n$$$$\cdot \cdot \cdot$$$$+1\times 2\times n\times 3+1\times 2\times n\times 4+\cdot \cdot \cdot +1\times 2\times n(n-1)$$$$\cdot \cdot \cdot and\mathrm{\Sigma }1\times 3\times \cdot \cdot \cdot to\mathrm{\Sigma }1\times n\times \cdot \cdot \cdot$$$$and\mathrm{\Sigma }2\times \cdot \cdot \cdot to\mathrm{\Sigma }n\times \cdot \cdot \cdot$$$$=\frac{n(n-1)(n-2)(n+1)(3n+2)}{12}$$$$-3\{\frac{n(n-1)(n+1)(3n+2)}{12}-2[\frac{n(n+1)}{2}-1^2]\}$$$$+\frac{n(n-1)(n-2)(n+1)(3n+2)}{12}$$$$-3\{\frac{n(n-1)(n+1)(3n+2)}{12}-2[\frac{2n(n+1)}{2}-2^2]$$$$+\cdot \cdot \cdot$$$$=\frac{n^2(n-1)(n-2)(n+1)(3n+2)}{12}$$$$-3[\frac{n^2(n-1)(n+1)(3n+2)}{12}-\frac{2n(n-1)(n+1)(3n+2)}{12}]$$$$=\frac{n^2(n-1)(n-2)(n+1)(3n+2)}{12}$$$$-\frac{3n(n-1)(n-2)(n+1)(3n+2)}{12}$$$$=\frac{n(n-1)(n-2)(n-3)(n+1)(3n+2)}{12}$$$$fromtherule$$$$\sum _{a_1\ne a_2\ne a_3\cdot \cdot \cdot \ne a_k}\underset{j=1}{\prod ^k}{a}_j,1<k<n$$$$=\frac{P_k^n(n+1)(3n+2)}{12},P_k^n=C_k^n\times k!$$

Commented by Tony Lin last updated on 10/Apr/20

$$Iguesstheanswermightbe$$$$\frac{C_k^n(n+1)(3n+2)}{12}$$$$whichhasnopermutation$$$$onlycombination$$

Commented by mr W last updated on 10/Apr/20

$$i{don}^{\prime }thabethesolutionfor\sum _{a\ne b\ne c\ne d}abcd.$$$$but\sum _{a\ne b\ne c}abc=\frac{(n-2)(n-1)n^2{(n+1)}^2}{48}.$$$$canyoupleasecheckwithyours?$$

Commented by Tony Lin last updated on 10/Apr/20

$$mysupposeiswrongbutIfound$$$$\mathrm{\Sigma }a=\underset{k=1}{\sum ^n}k=\frac{n(n+1)}{2}$$$$\sum _{a\ne b}ab=\underset{k=1}{\sum ^{n-1}}\frac{k(k+1)}{2}=\frac{n(n-1){(n+1)}^2(3n+2)}{24}$$$$\sum _{a\ne b\ne c}abc=\underset{k=1}{\sum ^{n-1}}\frac{k(k-1){(k+1)}^2(3k+2)}{24}$$$$=\frac{n^2{(n+1)}^2(n-1)(n-2)}{48}$$$$\sum _{a\ne b\ne c\ne d}abcd=\underset{k=1}{\sum ^{n-1}}\frac{k^2{(k+1)}^2(k-1)(k-2)}{48}$$$$=\frac{(n-3)(n-2)(n-1)n(n+1)(15n^3+15n^2-10n-8)}{5760}$$

Commented by mr W last updated on 10/Apr/20

$$veryinteresting,thanksalotsir!$$

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