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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

We have 20 thousand dollars that must be invested among 4 possible opportunities. Each investment must be integral in units of 1 thousand dollars, and there are minimal investments that need to be made if one is to invest in these opportunities. The minimal investments are \(2,2,3\), and 4 thousand dollars. How many different investment strategies are available if (a) an investment must be made in each opportunity; (b) investments must be made in at least 3 of the 4 opportunities?

A student is to answer 7 out of 10 questions in an examination. How many choices has she? How many if she must answer at least 3 of the first 5 questions?

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

Consider three classes, each consisting of \(n\) students. From this group of \(3 n\) students, a group of 3 students is to be chosen. (a) How many choices are possible? (b) How many choices are there in which all 3 students are in the same class? (c) How many choices are there in which 2 of the 3 students are in the same class and the other student is in a different class? (d) How many choices are there in which all 3 students are in different classes? (e) Using the results of parts (a) through (d), write a combinatorial identity.

From a set of \(n\) people a committee of size \(j\) is to be chosen, and from this committee a subcommittee of size \(i, i \leq j\), is also to be chosen. (a) Derive a combinatorial identity by computing, in two ways, the number of possible choices of the committee and subcommittee-first by supposing that the committee is chosen first and then the subcommittee, and second by supposing that the subcommittee is chosen first and then the remaining members of the committee are chosen. (b) Use part (a) to prove the following combinatorial identity: $$ \sum_{j=i}^{n}\left(\begin{array}{l} n \\ j \end{array}\right)\left(\begin{array}{l} j \\ i \end{array}\right)=\left(\begin{array}{l} n \\ i \end{array}\right) 2^{n-i} \quad i \leq n $$ (c) Use part (a) and Theoretical Exercise 13 to show that $$ \sum_{j=i}^{n}\left(\begin{array}{l} n \\ j \end{array}\right)\left(\begin{array}{l} j \\ i \end{array}\right)(-1)^{n-j}=0 \quad i \leq n $$
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