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Question Number 134389 by EDWIN88 last updated on 03/Mar/21

Eight dice are tossed. If the dice are identical in  appearance , how many different−looking   (distinguishable) occurrences are there?

$$\mathrm{Eight}\:\mathrm{dice}\:\mathrm{are}\:\mathrm{tossed}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{dice}\:\mathrm{are}\:\mathrm{identical}\:\mathrm{in} \\ $$$$\mathrm{appearance}\:,\:\mathrm{how}\:\mathrm{many}\:\mathrm{different}−\mathrm{looking}\: \\ $$$$\left(\mathrm{distinguishable}\right)\:\mathrm{occurrences}\:\mathrm{are}\:\mathrm{there}? \\ $$

Answered by bramlexs22 last updated on 03/Mar/21

Theorem   If r is nonnegative integer , the number   of n−tupples (r_1 ,r_2 ,...,r_n ) , is r_i  an   integer satisfying the condition         r_i  ≥ 0 (i=1,2,...,n)      r_1 +r_2 +r_3 +...+r_n  = r      is  (((n+r−1)),((      n−1)) )  In this case → { ((n=6)),((r=8)) :}   so  (((6+8−1)),((    6−1)) ) =  (((13)),((  5)) ) = ((13.12.11.10.9)/(5.4.3.2.1))=1287

$$\mathrm{Theorem}\: \\ $$$$\mathrm{If}\:\mathrm{r}\:\mathrm{is}\:\mathrm{nonnegative}\:\mathrm{integer}\:,\:\mathrm{the}\:\mathrm{number}\: \\ $$$$\mathrm{of}\:\mathrm{n}−\mathrm{tupples}\:\left(\mathrm{r}_{\mathrm{1}} ,\mathrm{r}_{\mathrm{2}} ,...,\mathrm{r}_{\mathrm{n}} \right)\:,\:\mathrm{is}\:\mathrm{r}_{{i}} \:\mathrm{an}\: \\ $$$$\mathrm{integer}\:\mathrm{satisfying}\:\mathrm{the}\:\mathrm{condition}\: \\ $$$$\:\:\:\:\:\:\mathrm{r}_{{i}} \:\geqslant\:\mathrm{0}\:\left({i}=\mathrm{1},\mathrm{2},...,\mathrm{n}\right) \\ $$$$\:\:\:\:\mathrm{r}_{\mathrm{1}} +\mathrm{r}_{\mathrm{2}} +\mathrm{r}_{\mathrm{3}} +...+\mathrm{r}_{\mathrm{n}} \:=\:\mathrm{r}\: \\ $$$$\:\:\:\mathrm{is}\:\begin{pmatrix}{\mathrm{n}+\mathrm{r}−\mathrm{1}}\\{\:\:\:\:\:\:\mathrm{n}−\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{In}\:\mathrm{this}\:\mathrm{case}\:\rightarrow\begin{cases}{\mathrm{n}=\mathrm{6}}\\{\mathrm{r}=\mathrm{8}}\end{cases} \\ $$$$\:\mathrm{so}\:\begin{pmatrix}{\mathrm{6}+\mathrm{8}−\mathrm{1}}\\{\:\:\:\:\mathrm{6}−\mathrm{1}}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{13}}\\{\:\:\mathrm{5}}\end{pmatrix}\:=\:\frac{\mathrm{13}.\mathrm{12}.\mathrm{11}.\mathrm{10}.\mathrm{9}}{\mathrm{5}.\mathrm{4}.\mathrm{3}.\mathrm{2}.\mathrm{1}}=\mathrm{1287} \\ $$$$ \\ $$

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