There are 15 persons in a party and each person shakes hands with another then the total number of handshakes is?
(a) ${}^{15}{{P}_{2}}$
(b) ${}^{15}{{C}_{2}}$
(c) $15!$
(d) $2\left( 15! \right)$
Answer
522.9k+ views
Hint: Assume the total number of persons in the party as n and the number of persons required for one handshake as r. Using the fact that for one handshake two persons are required, use the formula of combinations given as ${}^{n}{{C}_{r}}$ to select any two persons from a total of 15 persons at the party for the required number of handshakes.
Complete step-by-step solution:
Here we have been told that 15 persons are present at a party and each of them shakes hands with one another. We are asked to determine the total number of handshakes that will occur.
We know that for one handshake two persons will be required. Since there are 15 people at the party, we have to select two people for a handshake. Therefore, the total number of handshakes will be equal to the total number of ways in which any two persons can be selected.
Now, we know that if we have to select “r” number of things from a total of n number of things then we use the formula of combination given as ${}^{n}{{C}_{r}}$. So let us consider the total number persons present at the party as n and the number of persons to be selected for a handshake as r, so we have n = 15 and r = 2, substituting these values in the formula we get,
$\Rightarrow $ Number of handshakes = ${}^{n}{{C}_{r}}$
$\therefore $ Number of handshakes = ${}^{15}{{C}_{2}}$
Hence, option (b) is the correct answer.
Note: You can also think of the solution in a different way. Since there are 15 people in the party, each person will shake hands with 14 others because one cannot shake hands with himself. Since, each person has to shake hand with the other person only once and not twice therefore the total number of handshakes will become $\dfrac{15\times 14}{2}$, which is nothing but the expansion of ${}^{15}{{C}_{2}}$ given by the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
Complete step-by-step solution:
Here we have been told that 15 persons are present at a party and each of them shakes hands with one another. We are asked to determine the total number of handshakes that will occur.
We know that for one handshake two persons will be required. Since there are 15 people at the party, we have to select two people for a handshake. Therefore, the total number of handshakes will be equal to the total number of ways in which any two persons can be selected.
Now, we know that if we have to select “r” number of things from a total of n number of things then we use the formula of combination given as ${}^{n}{{C}_{r}}$. So let us consider the total number persons present at the party as n and the number of persons to be selected for a handshake as r, so we have n = 15 and r = 2, substituting these values in the formula we get,
$\Rightarrow $ Number of handshakes = ${}^{n}{{C}_{r}}$
$\therefore $ Number of handshakes = ${}^{15}{{C}_{2}}$
Hence, option (b) is the correct answer.
Note: You can also think of the solution in a different way. Since there are 15 people in the party, each person will shake hands with 14 others because one cannot shake hands with himself. Since, each person has to shake hand with the other person only once and not twice therefore the total number of handshakes will become $\dfrac{15\times 14}{2}$, which is nothing but the expansion of ${}^{15}{{C}_{2}}$ given by the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Master Class 9 Maths: Engaging Questions & Answers for Success

Master Class 9 Science: Engaging Questions & Answers for Success

Master Class 9 English: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Two of the body parts which do not appear in MRI are class 11 biology CBSE

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

10 examples of friction in our daily life

Name the Largest and the Smallest Cell in the Human Body ?

