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Chapter 1: Problem 13

Consider a group of 20 people. If everyone shakes hands with everyone else, how many handshakes take place?

Short Answer

Expert verified
A total of 190 handshakes take place among the group of 20 people. This is calculated using the combination formula, \(C(20,2) = \frac{20!}{2!(18)!}\), which simplifies to \(\frac{20 脳 19}{2 脳 1} = 10 脳 19 = 190\).

Step by step solution

01

Identify the combination parameters

Here, we have to find the number of unique handshakes among 20 people. So, in this case: - \(n = 20\) (total people in the group) - \(k = 2\) (number of people involved in each handshake)
02

Calculate the factorials of n, k, and (n-k)

We need to calculate the factorials of \(n\), \(k\), and \((n-k)\), which are: - \(20! = 20 脳 19 脳 18 脳 \dots 脳 3 脳 2 脳 1\) - \(2! = 2 脳 1\) - \((20 - 2)! = 18!\)
03

Apply the combination formula

Now, we can use the combination formula to find the total number of handshakes: \[C(20, 2) = \frac{20!}{2!(18)!}\]
04

Simplify the expression

To simplify the expression further and calculate the total handshakes, we can cancel out the factorials: \[\frac{20 脳 19 脳 18!}{2 脳 1 脳 18!} = \frac{20 脳 19}{2 脳 1}\]
05

Calculate the total handshakes

Now, we can calculate the final result: \[C(20,2) = \frac{20 脳 19}{2 脳 1} = 10 脳 19 = 190\] So, there are a total of 190 handshakes taking place among the group of 20 people.

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Most popular questions from this chapter

Consider a tournament of \(n\) contestants in which the outcome is an ordering of these contestants, with ties allowed. That is, the outcome partitions the players into groups, with the first group consisting of the players that tied for first place, the next group being those that tied for the next best position, and so on. Let \(N(n)\) denote the number of different possible outcomes. For instance, \(N(2)=3\) since in a tournament with 2 contestants, player 1 could be uniquely first, player 2 could be uniquely first, or they could tie for first. (a) List all the possible outcomes when \(n=3\). (b) With \(N(0)\) defined to equal 1 , argue, without any computations, that $$ N(n)=\sum_{i=1}^{n}\left(\begin{array}{l} n \\ i \end{array}\right) N(n-i) $$ HINT: How many outcomes are there in which \(i\) players tie for last place? (c) Show that the formula of part (b) is equivalent to the following: $$ N(n)=\sum_{i=0}^{n-1}\left(\begin{array}{l} n \\ i \end{array}\right) N(i) $$ (d) Use the recursion to find \(N(3)\) and \(N(4)\).

If 12 people are to be divided into 3 committees of respective sizes 3,4, and 5 , how many divisions are possible?

In how many ways can \(r\) objects be selected from a set of \(n\) if the order of selection is considered relevant?

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Present a combinatorial explanation of why \(\left(\begin{array}{l}n \\\ r\end{array}\right)=\left(\begin{array}{c}n \\ r, n-r\end{array}\right)\)
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