Regular polygon | Combinatorics | PRMO-2019 | Problem 15

The above diagram is a diagram of Regular Polygon .we have to draw the diagonals as shown in above.we joined the diagonals such that all the diagonals will be parallel

Can you now finish the problem ..........

If we joined the diagonals (shown in Fig. 1), i.e \((P_3 \to P_10)\),\((P_4\to P_9)\),\((P_5 \to P_8)\) then then we have 3 diagonals.so we have 5\(4 \choose 2\) ways=\(15\) ways.

If we joined the diagonals (shown in Fig.2), i.e \((P_1 \to P_3)\),\((P_10\to P_4)\),\((P_9\to P_5)\),\((P_8\to P_6)\)then we have \(4\)diagonals.so we have 5\(3 \choose 2\) ways=\(30\) ways.

Therefore total numbers of ways that can a pair of parallel diagonals of a regular polygon of \(10\) sides be selected is \(15+30=45\)

The above diagram is a diagram of Regular Polygon .we have to draw the diagonals as shown in above.we joined the diagonals such that all the diagonals will be parallel

Can you now finish the problem ..........

If we joined the diagonals (shown in Fig. 1), i.e \((P_3 \to P_10)\),\((P_4\to P_9)\),\((P_5 \to P_8)\) then then we have 3 diagonals.so we have 5\(4 \choose 2\) ways=\(15\) ways.

If we joined the diagonals (shown in Fig.2), i.e \((P_1 \to P_3)\),\((P_10\to P_4)\),\((P_9\to P_5)\),\((P_8\to P_6)\)then we have \(4\)diagonals.so we have 5\(3 \choose 2\) ways=\(30\) ways.

Therefore total numbers of ways that can a pair of parallel diagonals of a regular polygon of \(10\) sides be selected is \(15+30=45\)