Answer
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Hint: The formula for the number of handshakes possible with various people say “n” can be solved by a formula \[\dfrac{{n \times \left( {n - 1} \right)}}{2}\] . This is because each of the “n” people can shake hands with \[n - 1\] people and they will not shake their own hand. The handshake between two people should not be counted twice.
Complete step-by-step answer:
Let the number of people who shake hands be “n”. Since, the person will not do a handshake with himself so “n” people can shake hands with \[n - 1\] people.
The handshake between two people should not be counted twice. When a person lets “a” shake hands with another person let “b” and when “b” shakes hands with “a” then that will not be counted as another handshake, it’s the same handshake.
Hence the formula for the number of handshake possible with “n” people is given below:
\[\dfrac{{n \times \left( {n - 1} \right)}}{2}\]
This formula can be used for any number of people. Shaking hands in a group involves pairing of two people in all possible ways.
Hence we have,
\[
\dfrac{{n \times \left( {n - 1} \right)}}{2} \\
\Rightarrow \dfrac{{20 \times \left( {20 - 1} \right)}}{2} \\
\Rightarrow \dfrac{{20 \times 19}}{2} \\
\Rightarrow 10 \times 19 \\
\Rightarrow 190 \;
\]
Therefore, there will be \[190\] handshakes in total.
So, the correct answer is “ \[190\] ”.
Note: This question can also be solved by the formula \[{}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}\] where “n” is the total number of persons, “r” is the number of handshakes. According to this question “n” is \[20\] and “r” is \[2\] . Inputting these values into the formula will give the same answer. This is a normal simple combination formula. By every handshake two persons are involved but handshake is done only once among these two.
Complete step-by-step answer:
Let the number of people who shake hands be “n”. Since, the person will not do a handshake with himself so “n” people can shake hands with \[n - 1\] people.
The handshake between two people should not be counted twice. When a person lets “a” shake hands with another person let “b” and when “b” shakes hands with “a” then that will not be counted as another handshake, it’s the same handshake.
Hence the formula for the number of handshake possible with “n” people is given below:
\[\dfrac{{n \times \left( {n - 1} \right)}}{2}\]
This formula can be used for any number of people. Shaking hands in a group involves pairing of two people in all possible ways.
Hence we have,
\[
\dfrac{{n \times \left( {n - 1} \right)}}{2} \\
\Rightarrow \dfrac{{20 \times \left( {20 - 1} \right)}}{2} \\
\Rightarrow \dfrac{{20 \times 19}}{2} \\
\Rightarrow 10 \times 19 \\
\Rightarrow 190 \;
\]
Therefore, there will be \[190\] handshakes in total.
So, the correct answer is “ \[190\] ”.
Note: This question can also be solved by the formula \[{}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}\] where “n” is the total number of persons, “r” is the number of handshakes. According to this question “n” is \[20\] and “r” is \[2\] . Inputting these values into the formula will give the same answer. This is a normal simple combination formula. By every handshake two persons are involved but handshake is done only once among these two.
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