Question

1. (2 points) in a group there are three professors and three doctors. if they are seated in a row, find the probability that those of the same profession sit together.

Explanation:

Step1: Calculate total number of arrangements

The total number of people is \(3 + 3=6\). The total number of ways to arrange \(n\) distinct objects in a row is \(n!\). So the total number of arrangements of 6 people in a row is \(n = 6\), and the total arrangements \(N=6! = 720\).

Step2: Treat groups of same - profession as single entities

Treat the 3 professors as one unit and the 3 doctors as one unit. Then the number of ways to arrange these two units is \(2!\). And the number of ways to arrange the 3 professors within their unit is \(3!\), and the number of ways to arrange the 3 doctors within their unit is also \(3!\). So the number of favorable arrangements \(M = 2!\times3!\times3!\)
\[M=2\times6\times6= 72\]

Step3: Calculate the probability

The probability \(P\) is the number of favorable arrangements divided by the total number of arrangements. So \(P=\frac{M}{N}=\frac{72}{720}=\frac{1}{10}\)

Answer:

\(\frac{1}{10}\)