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Question Number 21760 by Tinkutara last updated on 03/Oct/17

$$\mathrm{There}\mathrm{are}m\mathrm{points}\mathrm{on}\mathrm{the}\mathrm{line}AB\mathrm{and}$$$$n\mathrm{points}\mathrm{on}\mathrm{the}\mathrm{line}AC,\mathrm{excluding}\mathrm{the}$$$$\mathrm{point}A.\mathrm{Triangles}\mathrm{are}\mathrm{formed}\mathrm{joining}$$$$\mathrm{these}\mathrm{points}$$$$\left(\mathrm{i}\right)\mathrm{When}\mathrm{point}A\mathrm{is}\mathrm{not}\mathrm{included},$$$$\left(\mathrm{ii}\right)\mathrm{When}\mathrm{point}A\mathrm{is}\mathrm{included}.$$$$\mathrm{The}\mathrm{ratio}\mathrm{of}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{such}\mathrm{triangles}$$$$\mathrm{is}$$

Answered by mrW1 last updated on 03/Oct/17

$$\left(\mathrm{i}\right)$$$${\mathrm{mC}}_2^{\mathrm{n}}+{\mathrm{nC}}_2^{\mathrm{m}}=\frac{\mathrm{mn}(\mathrm{m}+\mathrm{n}-2)}{2}$$$$\left(\mathrm{ii}\right)$$$${\mathrm{mC}}_2^{\mathrm{n}}+{\mathrm{nC}}_2^{\mathrm{m}}+\mathrm{mn}=\frac{\mathrm{mn}(\mathrm{m}+\mathrm{n}-2)}{2}+\mathrm{mn}=\frac{\mathrm{mn}(\mathrm{m}+\mathrm{n})}{2}$$$$$$$$\mathrm{ratio}=\frac{\mathrm{m}+\mathrm{n}-2}{\mathrm{m}+\mathrm{n}}$$

Commented by Tinkutara last updated on 03/Oct/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

Commented by mrW1 last updated on 03/Oct/17

$$\mathrm{there}\mathrm{are}\mathrm{mn}\mathrm{triangles}\mathrm{with}\mathrm{point}\mathrm{A}.$$$$$$$$\mathrm{you}\mathrm{can}\mathrm{also}\mathrm{calculate}\mathrm{the}\mathrm{number}$$$$\mathrm{of}\mathrm{triangles}\mathrm{in}\left(\mathrm{ii}\right)\mathrm{like}\mathrm{this}:$$$${\mathrm{mC}}_2^{\mathrm{n}+1}+{\mathrm{nC}}_2^{\mathrm{m}}=\frac{\mathrm{m}(\mathrm{n}+1)\mathrm{n}}{2}+\frac{\mathrm{nm}(\mathrm{m}-1)}{2}=\frac{\mathrm{mn}(\mathrm{m}+\mathrm{n})}{2}$$

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