MHT CET · Maths · Permutation Combination
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
- A 10
- B 15
- C 20
- D 21
Answer & Solution
Correct Answer
(A) 10
Step-by-step Solution
Detailed explanation
Let ' \(n\) ' be the number of members in the meeting
\(\therefore\) Total number of handshakes \(={ }^{\mathrm{n}} \mathrm{C}_2\)
\(\therefore{ }^{\mathrm{n}} \mathrm{C}_2=45\)
\(\frac{\mathrm{n} !}{2 !(\mathrm{n}-2) !}=45 \)
\( \frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2) !}{2 \times(\mathrm{n}-2) !}=45 \)
\( \mathrm{n}(\mathrm{n}-1)=90 \)
\( \therefore \mathrm{n}^2-\mathrm{n}-90=0 \)
\( \mathrm{n}=10 \text { or } \mathrm{n}=-9 \text { (not possible) } \)
\( \therefore \mathrm{n}=10\)
\(\therefore\) Total number of handshakes \(={ }^{\mathrm{n}} \mathrm{C}_2\)
\(\therefore{ }^{\mathrm{n}} \mathrm{C}_2=45\)
\(\frac{\mathrm{n} !}{2 !(\mathrm{n}-2) !}=45 \)
\( \frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2) !}{2 \times(\mathrm{n}-2) !}=45 \)
\( \mathrm{n}(\mathrm{n}-1)=90 \)
\( \therefore \mathrm{n}^2-\mathrm{n}-90=0 \)
\( \mathrm{n}=10 \text { or } \mathrm{n}=-9 \text { (not possible) } \)
\( \therefore \mathrm{n}=10\)
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