Four people in a room. Each shakes hands with each of the others once. How many handshakes are there?

Mathematics

2 answers:

storchak [24]2 years ago

7 0

<span>The correct answer is 6.

The clearest way to determine this is by creating a table of possible hand shaking. If we label the people A-D, the following are the ways they can be combined:
AB
AC
AD
BC
BD
CD

There are no other combinations. </span>

slavikrds [6]2 years ago

3 0

There are 6 handshakes between four people in the room.

<h3>Further explanation</h3>

The probability of an event is defined as the possibility of an event occurring against sample space.

$\large { \boxed {P(A) = \frac{\text{Number of Favorable Outcomes to A}}{\text {Total Number of Outcomes}} } }$

<h2>Permutation ( Arrangement )</h2>

Permutation is the number of ways to arrange objects.

$\large {\boxed {^nP_r = \frac{n!}{(n - r)!} } }$

<h2>Combination ( Selection )</h2>

Combination is the number of ways to select objects.

$\large {\boxed {^nC_r = \frac{n!}{r! (n - r)!} } }$

Let us tackle the problem.

This problem is about Combination.

If there are 4 people in a room , then the number of handshaking between 2 people is analogy as selecting 2 people from 4 people available. We will use combination formula in this problem.

$^4C_2 = \frac{4!}{2! (4-2)!}$