Examples

Counting teams with Cooper

A team of 3 student volunteers is formed at random from a class of 40. What is the probability that Cooper is on the team?

Solution

There are $\left(\begin{array}{c}39\\ 2\end{array}\right)$ teams that include Cooper, and $\left(\begin{array}{c}40\\ 3\end{array}\right)$ teams in total. So we have:

$$P=\frac{39!}{2!37!}/\frac{40!}{3!37!}=\frac{3}{40}$$

Haley and Hugo from 2 groups of 3

A UVA class has 40 students. Suppose the professor chooses 3 students on Wednesday at random, and again 3 on Friday. What is the probability that Haley is chosen today and Hugo on Friday?

Solution

(1) Count total outcomes:

• We have $\left(\genfrac{}{}{0px}{}{40}{3}\right)$ possible groups chosen Wednesday.
• We have $\left(\genfrac{}{}{0px}{}{40}{3}\right)$ possible groups chosen Friday.

Therefore $\left(\genfrac{}{}{0px}{}{40}{3}\right)\times \left(\genfrac{}{}{0px}{}{40}{3}\right)$ possible groups in total. (Product of possibilities.)

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(2) Count desired outcomes:

The possible groups of 3 that include Haley can be counted by counting the subgroups of 2 formed of the other students in Haley’s group.

• Therefore we have $\left(\genfrac{}{}{0px}{}{39}{2}\right)$ groups that contain Haley.
• Similarly, we have $\left(\genfrac{}{}{0px}{}{39}{2}\right)$ groups that contain Hugo.

Therefore we count $\left(\genfrac{}{}{0px}{}{39}{2}\right)\times \left(\genfrac{}{}{0px}{}{39}{2}\right)$ total desired outcomes.

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(3) Compute probability:

Let $E$ label the desired event. By the counting rule:

$$P[E]=\frac{|E|}{|S|}$$

Evaluate using our data:

$$\begin{array}{c}P[E]\gg \gg \frac{\left(\genfrac{}{}{0px}{}{39}{2}\right)\times \left(\genfrac{}{}{0px}{}{39}{2}\right)}{\left(\genfrac{}{}{0px}{}{40}{3}\right)\times \left(\genfrac{}{}{0px}{}{40}{3}\right)}\\ \\ \gg \gg {\left(\frac{\frac{39\cdot 38}{2!}}{\frac{40\cdot 39\cdot 38}{3!}}\right)}^2\gg \gg {\left(\frac{3}{40}\right)}^2\end{array}$$

Counting VA license plates

VA license plates have three letters (with no I, O, or Q) followed by four numerals. A random plate is seen on the road.

(a) What is the probability that the numerals occur in strictly increasing order?

(b) What is the probability that at least one number is repeated?

Solution

(a) Numerals in increasing order.

(1) Count total plates:

• Have $23\cdot 23\cdot 23$ options for letters.
• Have $10\cdot 10\cdot 10\cdot 10$ options for numbers.

Thus ${23}^3\cdot {10}^4$ possible plates.

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(2) Count ways to have 4 numerals that occur in increasing order:

There are $\left(\genfrac{}{}{0px}{}{10}{4}\right)$ ways to choose 4 distinct numerals from 10 options.

For each choice of four distinct numerals (i.e. for each combination), there is exactly one ordering that’s increasing.

Therefore, there are $\left(\genfrac{}{}{0px}{}{10}{4}\right)$ ways to have 4 numerals that occur in increasing order.

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(3) Count ways to have a list of 3 letters (excluding I, O, Q).

There are 26 total letters, 3 are excluded, thus 23 options for each letter.

Repetition is allowed, thus we have $23\cdot 23\cdot 23={23}^3$ total ways.

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(3) Compute probability:

Total count of desired plates (taking the product of possibilities):

$${23}^3\cdot \left(\genfrac{}{}{0px}{}{10}{4}\right)$$

Let $E$ label the event that a plate has increasing numerals. The counting formula for probability is:

$$P[E]=\frac{|E|}{|S|}$$

Evaluate using our data:

$$P[E]\gg \gg \frac{{23}^3\cdot \left(\genfrac{}{}{0px}{}{10}{4}\right)}{{23}^3\cdot {10}^4}\gg \gg \frac{\frac{10!}{4!6!}}{10000}\gg \gg \frac{21}{1000}$$

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(b) At least one number repeated.

“At least” is hard to work with! Lots of ways that can happen.

Try the complement event, which is much simpler: “no repetition”

Let $E^c$ be event that no numbers are repeated. All are distinct. Then $E$ is our desired event.

Count the possibilities:

$$|E^c|=23\cdot 23\cdot 23\cdot 10\cdot 9\cdot 8\cdot 7$$

The total number of plates is still ${23}^3\cdot {10}^4$.

Therefore, license plates with at least one number repeated:

$$\begin{array}{c}|E|=|S|-|E|\\ \\ \gg \gg {23}^3\cdot {10}^4-{23}^3\cdot 10\cdot 9\cdot 8\cdot 7\gg \gg 60348320\end{array}$$

Desired outcomes over total outcomes:

$$\frac{|E|}{|S|}\gg \gg \frac{60348320}{{23}^3\cdot {10}^4}\gg \gg 0.496$$

Counting out 4 teams

A board game requires 4 teams of players. How many configurations of teams are there out of a total of 17 players if the number of players per team is 4, 4, 4, 5, respectively.

Solution

This is just the multinomial coefficient with this data:

$n$	$r_1$	$r_2$	$r_3$	$r_4$
17	4	4	4	5

So we have:

$$\#\text{teams}=\frac{n!}{r_1!r_2!r_3!r_4!}\gg \gg \frac{17!}{4!4!4!5!}\gg \gg 214414200$$