Combinations (Unordered Sampling Without Replacement)

1.

In this case the order does not matter, which means each distribution of courses is a combination. That means the number of ways your first semester can look like is given by $$\begin{array}{llll}\hfill \left(\genfrac{}{}{0.0pt}{}{1700}{5}\right)& =1700C5& \hfill & \\ \hfill & =\frac{1700!}{5!\cdot (1700-5)!}& \hfill & \\ \hfill & =117626840087840& \hfill & \end{array}$$

2.

The probability of getting each course is equal, so you can use uniform probability. That gives you

$$25C5=51530$$

different ways all your courses are among the courses you like. You found the total number of possible outcomes in Exercise 1, so the probability of all the five courses being the ones you like becomes

$$\begin{array}{llll}\hfill P\left(\text{favorites}\right)& =\frac{51530}{117626840087840}& \hfill & \\ \hfill & =4.38\cdot 10^{-10}& \hfill & \end{array}$$

You could say that it’s a good idea to let the students choose their own courses, and not get their education based on the results of a lottery!

Because we’re talking about the composition of groups, and not what order they’re chosen in, this is an unordered set.

1.

In this case, you’re not interested in specific boys or girls, only that there are three girls and two boys. That means you have a case where the order does not matter, which tells you that each outcome is a combination. That makes the number of outcomes that have three boys and two girls

$$\begin{array}{llll}\hfill \left(\genfrac{}{}{0.0pt}{}{13}{3}\right)\cdot \left(\genfrac{}{}{0.0pt}{}{7}{2}\right)& =13C3\cdot 7C2& \hfill & \\ \hfill & =\frac{13!}{3!\cdot (13-3)!}\cdot & \hfill & \\ \hfill & \phantom{=}\frac{7!}{2!\cdot (7-2)!}& \hfill & \\ \hfill & =6006& \hfill & \end{array}$$

$$\begin{array}{llll}\hfill \left(\genfrac{}{}{0.0pt}{}{13}{3}\right)\cdot \left(\genfrac{}{}{0.0pt}{}{7}{2}\right)& =13C3\cdot 7C2& \hfill & \\ \hfill & =\frac{13!}{3!\cdot (13-3)!}\cdot \frac{7!}{2!\cdot (7-2)!}& \hfill & \\ \hfill & =6006& \hfill & \end{array}$$

2.

Since there’s an equally large probability of a random girl being chosen as a random boy being chosen, just with different numbers of girls and boys in each group, you can use uniform probability. The number of favorable outcomes becomes

$$\begin{array}{llll}\hfill 13C1\cdot 7C4& =\frac{13!}{1!\cdot (13-1)!}\cdot & \hfill & \\ \hfill & \phantom{=}\frac{7!}{4!\cdot (7-4)!}& \hfill & \\ \hfill & =13\cdot 35& \hfill & \\ \hfill & =455& \hfill & \end{array}$$

$$\begin{array}{llll}\hfill 13C1\cdot 7C4& =\frac{13!}{1!\cdot (13-1)!}\cdot \frac{7!}{4!\cdot (7-4)!}& \hfill & \\ \hfill & =13\cdot 35& \hfill & \\ \hfill & =455& \hfill & \end{array}$$

The number of possible groups of five is

$$\begin{array}{llll}\hfill \text{Possible outcomes}& =20C5& \hfill & \\ \hfill & =15504& \hfill & \end{array}$$

$$\begin{array}{lll}\hfill \text{Possible outcomes}=20C5=15504& & \hfill \end{array}$$

That makes the probability

$$\begin{array}{llll}\hfill P\left(\text{one girl, four boys}\right)& =\frac{455}{15504}& \hfill & \\ \hfill & =0.029& \hfill & \\ \hfill & =2.9\text{\%}& \hfill & \end{array}$$