The number of ways of selecting 8 books from a library which has 9 identical books each on mathematics, physics, chemistry and english is? Answer is 165. Pls solve

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$$\begin{gathered} \mathrm{Your} \mathrm{question} \mathrm{is} \mathrm{quite} \mathrm{good}. \mathrm{Now} \mathrm{we} \mathrm{will} \mathrm{denote} \\ \mathrm{Mathematics}\to \mathrm{M} \\ \mathrm{Physics}\to \mathrm{P} \\ \mathrm{Chemistry}\to \mathrm{C} \\ \mathrm{English}\to \mathrm{E} \\ \mathrm{Now} \mathrm{as} \mathrm{each} \mathrm{books} \mathrm{of} \mathrm{each} \mathrm{subject} \mathrm{is} \mathrm{identical}, \mathrm{hence} \mathrm{number} \mathrm{of} \mathrm{ways} \mathrm{to} \mathrm{select} 1 \mathrm{book} \mathrm{or} 2 \mathrm{books} \mathrm{or} \\ .......\mathrm{or} 8 \mathrm{books} \mathrm{of} \mathrm{a} \mathrm{particular} \mathrm{subject} \mathrm{will} \mathrm{be} 1. \\ \mathrm{As} \mathrm{there} \mathrm{is} \mathrm{no} \mathrm{restriction}, \mathrm{hence} \mathrm{any} \mathrm{combination} \mathrm{is} \mathrm{possible} \mathrm{i}.\mathrm{e}. \\ \mathrm{All} 8 \mathrm{books} \mathrm{can} \mathrm{be} \mathrm{of} \mathrm{E}, \mathrm{or} \mathrm{no} \mathrm{books} \mathrm{can} \mathrm{be} \mathrm{of} \mathrm{E} \mathrm{in} \mathrm{the} \mathrm{selected} \mathrm{books} \mathrm{and} \mathrm{so} \mathrm{on}, \mathrm{hence} \\ 0\le \mathrm{E}\le 8 \\ 0\le \mathrm{P}\le 8 \\ 0\le \mathrm{C}\le 8 \\ 0\le \mathrm{M}\le 8 \\ \mathrm{Now} \mathrm{here} \mathrm{upper} \mathrm{limit} \mathrm{is} 8 \mathrm{because} \mathrm{we} \mathrm{have} \mathrm{to} \mathrm{select} 8 \mathrm{books} \mathrm{not} 9. \\ \mathrm{So} \mathrm{our} \mathrm{objective} \mathrm{is} \mathrm{to} \mathrm{find} \mathrm{number} \mathrm{of} \mathrm{integral} \mathrm{solutions} \mathrm{of}: \\ \mathrm{E}+\mathrm{P}+\mathrm{C}+\mathrm{M}=8 \left[\mathrm{Where} 0\le \mathrm{E}\le 8 \mathrm{and} 0\le \mathrm{P}\le 8 \mathrm{and} 0\le \mathrm{C}\le 8 \mathrm{and} 0\le \mathrm{M}\le 8\right] \\ \mathrm{Now} \mathrm{number} \mathrm{of} \mathrm{integrals} \mathrm{solutions}\left(\mathrm{N}\right) \mathrm{of} \\ {\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+....{\mathrm{x}}_{\mathrm{r}}=\mathrm{n} \left[\mathrm{Where} 0\le {\mathrm{x}}_1\le \mathrm{n} \mathrm{and} 0\le {\mathrm{x}}_2\le \mathrm{n} \mathrm{and} 0\le {\mathrm{x}}_3\le \mathrm{n} \mathrm{and}........ 0\le {\mathrm{x}}_{\mathrm{r}}\le \mathrm{n}\right] \\ \mathrm{is} \mathrm{given} \mathrm{by}: \\ \mathrm{N}{=}^{\mathrm{n}+\mathrm{r}-1}{\mathrm{C}}_{\mathrm{r}-1} \\ \mathrm{Here} \mathrm{n}=8 \mathrm{and} \mathrm{r}=4 \\ \mathrm{Hence} \\ \mathrm{Number} \mathrm{of} \mathrm{ways}{=}^{8+4-1}{\mathrm{C}}_{4-1}{=}^{11}{\mathrm{C}}_3=\frac{11.10.9}{3.2.1}=\frac{330}{2}=\mathbf{165}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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