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If at the end of certain meeting, everyone had shaken hands with everyone else, it was found that 45 handshakes were exchanged, then the number of members present at the meeting, are
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10
Let ' $n$ ' be the number of members in the meeting
$\therefore \quad$ Total number of handshakes $={ }^{\mathrm{n}} \mathrm{C}_2$
$\therefore \quad{ }^{\mathrm{n}} \mathrm{C}_2=45$
$\begin{array}{ll} & \frac{\mathrm{n} !}{2 !(\mathrm{n}-2) !}=45 \\ & \frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2) !}{2 \times(\mathrm{n}-2) !}=45 \\ & \mathrm{n}(\mathrm{n}-1)=90 \\ \therefore \quad & \mathrm{n}^2-\mathrm{n}-90=0 \\ & \mathrm{n}=10 \text { or } \mathrm{n}=-9 \text { (not possible) } \\ \therefore \quad & \mathrm{n}=10\end{array}$
$\therefore \quad$ Total number of handshakes $={ }^{\mathrm{n}} \mathrm{C}_2$
$\therefore \quad{ }^{\mathrm{n}} \mathrm{C}_2=45$
$\begin{array}{ll} & \frac{\mathrm{n} !}{2 !(\mathrm{n}-2) !}=45 \\ & \frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2) !}{2 \times(\mathrm{n}-2) !}=45 \\ & \mathrm{n}(\mathrm{n}-1)=90 \\ \therefore \quad & \mathrm{n}^2-\mathrm{n}-90=0 \\ & \mathrm{n}=10 \text { or } \mathrm{n}=-9 \text { (not possible) } \\ \therefore \quad & \mathrm{n}=10\end{array}$
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