Q. Find the value of ( 0.9)⁶ , correct upto four place of decimals.                 

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Remember} \mathrm{the} \mathrm{formula}{ \\ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}=\frac{\mathrm{n}!}{\mathrm{r}! \left(\mathrm{n}-\mathrm{r}\right)!}, \mathrm{where} \mathrm{n}\in \mathrm{N}, \mathrm{r}\in \mathrm{W} \mathrm{and} \mathrm{n}\ge \mathrm{r} \\ \mathrm{Also} \mathrm{n}!=1.2.3.4.....\mathrm{n} \\ \mathrm{i}.\mathrm{e}. \\ 5!=5.4.3.2.1=120 \\ \mathrm{Also}{ \\ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}{=}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}-\mathrm{r}} \\ \mathrm{e}.\mathrm{g}.{ \\ }^6{\mathrm{C}}_4{=}^6{\mathrm{C}}_{6-4}{=}^6{\mathrm{C}}_2 \\ \mathrm{Now}{ \\ }^6{\mathrm{C}}_2=\frac{6!}{2! \left(6-2\right)!}=\frac{6!}{2! ·4!} \\ =\frac{6·5·4!}{2! ·4!} \\ =\frac{6·5}{2·1} \\ =\frac{30}{2}=15 \\ \mathrm{Important} \mathrm{trick} \mathrm{is} \mathrm{following}{ \\ }^6{\mathrm{C}}_3 \\ \mathrm{As} 3 \mathrm{is} \mathrm{written} \mathrm{in} \mathrm{subscript}, \mathrm{hence} \mathrm{you} \mathrm{open} 6 \mathrm{only} 3 \mathrm{times}{ \\ }^6{\mathrm{C}}_3=\frac{6·5·4}{3·2·1}=20{ \\ }^8{\mathrm{C}}_2=\frac{8·7}{2·1}=28 \end{gathered}$$

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