urgent
please explain clearly with proper steps so that i am able to understand....
out of 5 apples 10 mangoes and 15 oranges the number of ways of distributing 15 friuts each to 2 persons is 

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{If} \mathrm{somehow} \mathrm{we} \mathrm{can} \mathrm{figure} \mathrm{out} \mathrm{how} \mathrm{to} \mathrm{distribute} 15 \mathrm{fruits} \mathrm{out} \mathrm{of} 5 \mathrm{apples} 10 \mathrm{mangoes} \mathrm{and} 15 \\ \mathrm{oranges} \mathrm{to} \mathrm{one} \mathrm{person}, \mathrm{our} \mathrm{question} \mathrm{will} \mathrm{be} \mathrm{solved}. \mathrm{Once} \mathrm{we} \mathrm{give} 15 \mathrm{fruits} \\ \mathrm{to} \mathrm{one} \mathrm{person}, \mathrm{other} \mathrm{will} \mathrm{get} \mathrm{remaining} 15 \mathrm{fruits}. \\ \mathrm{So} \mathrm{we} \mathrm{want} \\ \mathrm{A}+\mathrm{M}+\mathrm{O}=15 \\ \mathrm{There} \mathrm{are} 5 \mathrm{Apples}\left(\mathrm{A}\right) \\ \mathrm{There} \mathrm{are} 10 \mathrm{Mangoes}\left(\mathrm{M}\right) \\ \mathrm{There} \mathrm{are} 15 \mathrm{oranges}\left(\mathrm{O}\right) \\ \mathrm{Hence} \mathrm{we} \mathrm{need} \mathrm{to} \mathrm{collect} \\ \mathrm{Coefficient} \mathrm{of} {\mathrm{x}}^{15} \mathrm{in} \left({\mathrm{x}}^0+{\mathrm{x}}^1+{\mathrm{x}}^2+....+{\mathrm{x}}^5\right)\left({\mathrm{x}}^0+{\mathrm{x}}^1+{\mathrm{x}}^2+....+{\mathrm{x}}^{10}\right)\left({\mathrm{x}}^0+{\mathrm{x}}^1+{\mathrm{x}}^2+....+{\mathrm{x}}^{15}\right) \\ =\mathrm{Coefficient} \mathrm{of} {\mathrm{x}}^{15} \mathrm{in}\left(1+\mathrm{x}+{\mathrm{x}}^2+.....+{\mathrm{x}}^5\right)\left(1+\mathrm{x}+{\mathrm{x}}^2+.....+{\mathrm{x}}^{10}\right)\left(1+\mathrm{x}+{\mathrm{x}}^2+.....+{\mathrm{x}}^{15}\right) \\ =\mathrm{Coefficient} \mathrm{of} {\mathrm{x}}^{15} \mathrm{in} \left\{\frac{\left(1-{\mathrm{x}}^6\right)}{\left(1-\mathrm{x}\right)}\right\}\left\{\frac{\left(1-{\mathrm{x}}^{11}\right)}{\left(1-\mathrm{x}\right)}\right\}\left\{\frac{\left(1-{\mathrm{x}}^{16}\right)}{\left(1-\mathrm{x}\right)}\right\} \left[\mathrm{All} \mathrm{brackets} \mathrm{are} \mathrm{in} \mathrm{G}.\mathrm{P}.\right] \\ =\mathrm{Coefficient} \mathrm{of} {\mathrm{x}}^{15} \mathrm{in} {\left(1-\mathrm{x}\right)}^{-3}\left\{\left(1-{\mathrm{x}}^6-{\mathrm{x}}^{11}+{\mathrm{x}}^{17}\right)\left(1-{\mathrm{x}}^{16}\right)\right\} \\ =\mathrm{Coefficient} \mathrm{of} {\mathrm{x}}^{15} \mathrm{in} {\left(1-\mathrm{x}\right)}^{-3}\left\{\left(1-{\mathrm{x}}^6-{\mathrm{x}}^{11}+{\mathrm{x}}^{17}-{\mathrm{x}}^{16}\left(1-{\mathrm{x}}^6-{\mathrm{x}}^{11}+{\mathrm{x}}^{17}\right)\right)\right\} \\ \mathrm{As} \mathrm{we} \mathrm{are} \mathrm{looking} \mathrm{for} \mathrm{coefficient} \mathrm{of} {\mathrm{x}}^{15}, \mathrm{hence} \mathrm{powers} \mathrm{greater} \mathrm{than} 15 \mathrm{are} \mathrm{irrelevant} \mathrm{to} \mathrm{us}. \\ =\mathrm{Coefficient} \mathrm{of} {\mathrm{x}}^{15} \mathrm{in} {\left(1-\mathrm{x}\right)}^{-3}\left\{\left(1-{\mathrm{x}}^6-{\mathrm{x}}^{11}\right)\right\} \\ 0+15=15 \\ 6+9=15 \\ 11+4=15 \\ =1\times \mathrm{Coefficient} \mathrm{of} {\mathrm{x}}^{15} \mathrm{in} {\left(1-\mathrm{x}\right)}^{-3}-1\times \mathrm{Coefficient} \mathrm{of} {\mathrm{x}}^9 \mathrm{in} {\left(1-\mathrm{x}\right)}^{-3}-1\times \mathrm{Coefficient} \mathrm{of} {\mathrm{x}}^4 \mathrm{in} {\left(1-\mathrm{x}\right)}^{-3} \\ \mathrm{Now} \mathrm{coefficient} \mathrm{of} {\mathrm{x}}^{\mathrm{n}} \mathrm{in} \mathrm{expansion} \mathrm{of} {\left(1-\mathrm{x}\right)}^{-\mathrm{r}}{=}^{\mathrm{n}+\mathrm{r}-1}{\mathrm{C}}_{\mathrm{r}-1} \\ =1{\times }^{15+3-1}{\mathrm{C}}_{3-1}-1{\times }^{9+3-1}{\mathrm{C}}_{3-1}-1{\times }^{4+3-1}{\mathrm{C}}_{3-1} \\ {=}^{17}{\mathrm{C}}_2{-}^{11}{\mathrm{C}}_2{-}^6{\mathrm{C}}_2 \\ =\frac{17\times 16}{2\times 1}-\frac{11\times 10}{2\times 1}-\frac{6\times 5}{2\times 1} \\ =136-55-15 \\ =66 \end{gathered}$$

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