
In a class of 22 students, every student had a handshake with every other. How many handshakes will be there in total?
A) 150
B) 200
C) 250
D) 231
Answer
587.7k+ views
Hint:Here, in the given question total no. of students available for handshake are 22. We know, each student will shake hands with all other remaining students i.e. 22-1=21 because he can’t shake hands with himself. So, this means 1 student will shake hands with 21 students $\therefore $ the total handshakes between 22 students will be $22 \times 21$
But this is not the correct answer because here we have counted the no. of handshakes twice. Since, student 1 shaking hands with student 2 and student 2 shaking hands with student 1 are both the same things.And this way we have to find the solution for the given problem.
Complete step-by-step solution:
Let n=total number of students that will shake hands
Since all (total) persons can't shake hands with themselves, hence we subtract one individual to calculate the handshakes done by each person.
$\therefore $(n−1) = total number of handshakes an individual student would do
Hence, we multiply both total number of students with total number of handshakes an individual student would do:
n(n−1)
But this counts every handshake twice, because student 1 shaking hands with student 2 and student 2 shaking hands with student 1 are both the same things.
So, we have to divide by 2.
Therefore;
Total Number of handshakes = $\dfrac{{n\left( {n - 1} \right)}}{2}$
Here, in above question n = 22
Therefore;
Total Number of handshakes among students =
$\dfrac{{22\left( {22 - 1} \right)}}{2} = \dfrac{{22 \times 21}}{2} = 11 \times 21 = 231{\text{ handshakes}}$
Note:Alternative method:
We can simply solve this problem using the formula for combination:
$C\left( {n,r} \right) = {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Where, n = number of items in the set
And r = number of items selected from the set
Here n will be total no. persons available for handshake = 22
And r will be no. of persons required for a handshake = 2
Putting n = 22 and r = 2 in combination formula:
Total number of handshakes =
\[
{}^{22}{C_2} = \dfrac{{22!}}{{2!\left( {22 - 2} \right)!}} = \dfrac{{22!}}{{2!\left( {20} \right)!}} \\
= \dfrac{{22 \times 21 \times 2{0}!}}{{2! \times 2{0}!}} \\
= \dfrac{{22 \times 21}}{2} \\
= 11 \times 21 = 231 \\
\]
But this is not the correct answer because here we have counted the no. of handshakes twice. Since, student 1 shaking hands with student 2 and student 2 shaking hands with student 1 are both the same things.And this way we have to find the solution for the given problem.
Complete step-by-step solution:
Let n=total number of students that will shake hands
Since all (total) persons can't shake hands with themselves, hence we subtract one individual to calculate the handshakes done by each person.
$\therefore $(n−1) = total number of handshakes an individual student would do
Hence, we multiply both total number of students with total number of handshakes an individual student would do:
n(n−1)
But this counts every handshake twice, because student 1 shaking hands with student 2 and student 2 shaking hands with student 1 are both the same things.
So, we have to divide by 2.
Therefore;
Total Number of handshakes = $\dfrac{{n\left( {n - 1} \right)}}{2}$
Here, in above question n = 22
Therefore;
Total Number of handshakes among students =
$\dfrac{{22\left( {22 - 1} \right)}}{2} = \dfrac{{22 \times 21}}{2} = 11 \times 21 = 231{\text{ handshakes}}$
Note:Alternative method:
We can simply solve this problem using the formula for combination:
$C\left( {n,r} \right) = {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Where, n = number of items in the set
And r = number of items selected from the set
Here n will be total no. persons available for handshake = 22
And r will be no. of persons required for a handshake = 2
Putting n = 22 and r = 2 in combination formula:
Total number of handshakes =
\[
{}^{22}{C_2} = \dfrac{{22!}}{{2!\left( {22 - 2} \right)!}} = \dfrac{{22!}}{{2!\left( {20} \right)!}} \\
= \dfrac{{22 \times 21 \times 2{0}!}}{{2! \times 2{0}!}} \\
= \dfrac{{22 \times 21}}{2} \\
= 11 \times 21 = 231 \\
\]
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