
Everybody in a room shakes hands with everybody else. The total number of handshakes is 66. The total number of persons in the room is
Answer
483.9k+ views
Hint: Here we will use the concept that 2 people are required to complete 1 handshake. If there are \[n\] people in the room, the total number of handshakes in the room will be the number of 2-combinations from a set of \[n\] people. We will find the total number of people by using the formula of combinations.
Formula used: We will use the following formulas to solve the question:
1. The number of \[r\]- combinations from a set of \[n\] elements is given by \[^n{C_r}\] and \[^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}\].
2. The roots of a quadratic equation \[a{x^2} + bx + c = 0\] are given by \[x = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{2},\dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{2}\] .
Complete step-by-step answer:
The total number of handshakes is 66, so we will substitute 66 for \[^n{C_r}\] in the formula \[^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}\]. A handshake requires 2 people, so we will substitute 2 for \[r\] in the formula.
\[\begin{array}{l}\dfrac{{n!}}{{\left( {n - 2} \right)!2!}} = 66\\ \Rightarrow \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)!}}{{\left( {n - 2} \right)!2}} = 66\end{array}\]
We will now cancel the like terms.
\[ \Rightarrow \dfrac{{n\left( {n - 1} \right)}}{2} = 66\]
On cross multiplication, we get
\[ \Rightarrow n\left( {n - 1} \right) = 66 \times 2\]
Now simplifying the above equation, we get
\[\begin{array}{l} \Rightarrow {n^2} - n = 132\\ \Rightarrow {n^2} - n - 132 = 0\end{array}\]
Substituting 1 for \[a\], \[ - 1\] for \[b\] and \[ - 132\] for \[c\] in the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{2}\], we get
\[n = \dfrac{{ - \left( { - 1} \right) \pm \sqrt {{{\left( { - 1} \right)}^2} - 4\left( 1 \right)\left( {132} \right)} }}{{2 \cdot 1}}\]
Simplifying the equation, we get
\[\begin{array}{l}n = \dfrac{{1 \pm \sqrt {1 + 528} }}{2}\\n = \dfrac{{1 \pm \sqrt {529} }}{2}\\n = \dfrac{{1 \pm 23}}{2}\end{array}\]
On simplifying the above equation, we get
\[n = 12, - 11\]
The number of people cannot be negative, so \[n = 12\].
\[\therefore\] Total number of people in the room is 12.
Note: Let us assume that there are \[n\] people in the room. The \[{n^{th}}\] person will shake hands with \[n - 1\] people (all people excluding himself). The \[{\left( {n - 1} \right)^{th}}\] person will shake hands with \[\left( {n - 2} \right)\] people and so on. The second last person will shake hands with only 1 person who is the last person and the last person will not have to shake hands with anyone as he would have already shaken hands with everyone. So the total number of handshakes will be \[\left( {n - 1} \right) + \left( {n - 2} \right) + \left( {n - 3} \right) + ... + 1 + 0\] . Substitute 66 here.
\[\left( {n - 1} \right) + \left( {n - 2} \right) + \left( {n - 3} \right) + ... + 1 + 0 = 66\]
The sum of first \[k\] natural numbers is \[\dfrac{{k\left( {k + 1} \right)}}{2}\] . Substituting \[n - 1\] in place of \[k\] , we get
\[\begin{array}{l} \Rightarrow \dfrac{{\left( {n - 1} \right)\left( {n - 1 + 1} \right)}}{2} = 66\\ \Rightarrow \dfrac{{\left( {n - 1} \right)\left( n \right)}}{2} = 66\end{array}\]
Simplifying the equation, we get
\[\begin{array}{l} \Rightarrow {n^2} - n = 66 \times 2\\ \Rightarrow {n^2} - n - 132 = 0\end{array}\]
Substituting 1 for \[a\], \[ - 1\] for \[b\] and \[ - 132\] for \[c\] in the formula \[x = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{2},\dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{2}\].
\[n = \dfrac{{ - \left( { - 1} \right) + \sqrt {{{\left( { - 1} \right)}^2} - 4\left( 1 \right)\left( {132} \right)} }}{{2 \cdot 1}},\dfrac{{ - \left( { - 1} \right) - \sqrt {{{\left( { - 1} \right)}^2} - 4\left( 1 \right)\left( {132} \right)} }}{{2 \cdot 1}}\]
Simplify the equation.
\[\begin{array}{l}n = \dfrac{{1 + \sqrt {1 + 528} }}{2},\dfrac{{1 - \sqrt {1 + 528} }}{2}\\n = \dfrac{{1 + \sqrt {529} }}{2},\dfrac{{1 - \sqrt {529} }}{2}\end{array}\]
Simplifying the expression, we get
\[\begin{array}{l}n = \dfrac{{1 + 23}}{2},\dfrac{{1 - 23}}{2}\\n = 12, - 11\end{array}\]
The number of people cannot be negative, so \[n = 12\].
\[\therefore\] Total number of people in the room is 12.
Formula used: We will use the following formulas to solve the question:
1. The number of \[r\]- combinations from a set of \[n\] elements is given by \[^n{C_r}\] and \[^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}\].
2. The roots of a quadratic equation \[a{x^2} + bx + c = 0\] are given by \[x = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{2},\dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{2}\] .
Complete step-by-step answer:
The total number of handshakes is 66, so we will substitute 66 for \[^n{C_r}\] in the formula \[^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}\]. A handshake requires 2 people, so we will substitute 2 for \[r\] in the formula.
\[\begin{array}{l}\dfrac{{n!}}{{\left( {n - 2} \right)!2!}} = 66\\ \Rightarrow \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)!}}{{\left( {n - 2} \right)!2}} = 66\end{array}\]
We will now cancel the like terms.
\[ \Rightarrow \dfrac{{n\left( {n - 1} \right)}}{2} = 66\]
On cross multiplication, we get
\[ \Rightarrow n\left( {n - 1} \right) = 66 \times 2\]
Now simplifying the above equation, we get
\[\begin{array}{l} \Rightarrow {n^2} - n = 132\\ \Rightarrow {n^2} - n - 132 = 0\end{array}\]
Substituting 1 for \[a\], \[ - 1\] for \[b\] and \[ - 132\] for \[c\] in the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{2}\], we get
\[n = \dfrac{{ - \left( { - 1} \right) \pm \sqrt {{{\left( { - 1} \right)}^2} - 4\left( 1 \right)\left( {132} \right)} }}{{2 \cdot 1}}\]
Simplifying the equation, we get
\[\begin{array}{l}n = \dfrac{{1 \pm \sqrt {1 + 528} }}{2}\\n = \dfrac{{1 \pm \sqrt {529} }}{2}\\n = \dfrac{{1 \pm 23}}{2}\end{array}\]
On simplifying the above equation, we get
\[n = 12, - 11\]
The number of people cannot be negative, so \[n = 12\].
\[\therefore\] Total number of people in the room is 12.
Note: Let us assume that there are \[n\] people in the room. The \[{n^{th}}\] person will shake hands with \[n - 1\] people (all people excluding himself). The \[{\left( {n - 1} \right)^{th}}\] person will shake hands with \[\left( {n - 2} \right)\] people and so on. The second last person will shake hands with only 1 person who is the last person and the last person will not have to shake hands with anyone as he would have already shaken hands with everyone. So the total number of handshakes will be \[\left( {n - 1} \right) + \left( {n - 2} \right) + \left( {n - 3} \right) + ... + 1 + 0\] . Substitute 66 here.
\[\left( {n - 1} \right) + \left( {n - 2} \right) + \left( {n - 3} \right) + ... + 1 + 0 = 66\]
The sum of first \[k\] natural numbers is \[\dfrac{{k\left( {k + 1} \right)}}{2}\] . Substituting \[n - 1\] in place of \[k\] , we get
\[\begin{array}{l} \Rightarrow \dfrac{{\left( {n - 1} \right)\left( {n - 1 + 1} \right)}}{2} = 66\\ \Rightarrow \dfrac{{\left( {n - 1} \right)\left( n \right)}}{2} = 66\end{array}\]
Simplifying the equation, we get
\[\begin{array}{l} \Rightarrow {n^2} - n = 66 \times 2\\ \Rightarrow {n^2} - n - 132 = 0\end{array}\]
Substituting 1 for \[a\], \[ - 1\] for \[b\] and \[ - 132\] for \[c\] in the formula \[x = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{2},\dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{2}\].
\[n = \dfrac{{ - \left( { - 1} \right) + \sqrt {{{\left( { - 1} \right)}^2} - 4\left( 1 \right)\left( {132} \right)} }}{{2 \cdot 1}},\dfrac{{ - \left( { - 1} \right) - \sqrt {{{\left( { - 1} \right)}^2} - 4\left( 1 \right)\left( {132} \right)} }}{{2 \cdot 1}}\]
Simplify the equation.
\[\begin{array}{l}n = \dfrac{{1 + \sqrt {1 + 528} }}{2},\dfrac{{1 - \sqrt {1 + 528} }}{2}\\n = \dfrac{{1 + \sqrt {529} }}{2},\dfrac{{1 - \sqrt {529} }}{2}\end{array}\]
Simplifying the expression, we get
\[\begin{array}{l}n = \dfrac{{1 + 23}}{2},\dfrac{{1 - 23}}{2}\\n = 12, - 11\end{array}\]
The number of people cannot be negative, so \[n = 12\].
\[\therefore\] Total number of people in the room is 12.
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