4.7 Counting Rules

There are times when the sample space is very large and is not feasible to write out. In that case, it helps to have mathematical tools for counting the size of the sample space. These tools are known as counting techniques or counting rules.

Example 3

How many ways can the three letters a, b, and c be arranged with no letters repeating?

Show solution

Three tasks must be done in this case. The tasks are to pick the first letter, then the second letter, and then the third letter. The first task can be done 3 ways since there are 3 letters. The second task can be done 2 ways, since the first task took one of the letters (repeats are not allowed). The third task can be done 1 way, since the first and second task took two of the letters.

There are $\frac{3}{1^{st} letter} \cdot \frac{2}{2^{nd} letter} \cdot \frac{1}{3^{rd} letter}$

You can also look at this example in a tree diagram, see Figure 4-17. There are 6 different arrangements of the letters. The solution was found by multiplying 3·2·1 = 6.

Figure 4-17

If we have 10 different letters for, say, a password, the tree diagram would be very time-consuming to make because of the length of options and tasks, so we have some shortcut formulas that help count these arrangements.

Many counting problems involve multiplying a list of decreasing numbers, which is called a factorial. The factorial is represented mathematically by the starting number followed by an exclamation point, in this case 3! = 3·2·1 = 6. There is a special symbol for this and a special button on your calculator or computer.

TI-84: On the home screen, enter the number of which you would like to find the factorial. Press [MATH]. Use cursor keys to move to the PRB menu. Press 4 (4:!) Press [ENTER] to calculate.

TI-89: On the home screen, enter the number of which you would like to find the factorial. Press [2^nd] [Math] > 7:Probability > 1:!. Press [ENTER] to calculate.

Excel: In an empty cell type in =FACT(n) where n is the number so 4! would be =FACT(4).

Sometimes we do not want to select the entire group but only select r objects from n total objects. The number of ways to do this depends on if the order you choose the r objects matters or if it does not matter. As an example, if you are trying to call a person on the phone, you have to have the digits of their number in the correct order. In this case, the order of the numbers matters. If you were picking random numbers for the lottery, it does not matter which number you pick first since they always arrange the numbers from the smallest to largest once the numbers are drawn. As long as you have the same numbers that the lottery officials pick, you win. In this case, the order does not matter.

A permutation is an arrangement of items with a specific order. You use permutations to count items when the order matters.

When the order does not matter, you use combinations. A combination is an arrangement of items when order is not important. When you do a counting problem, the first thing you should ask yourself is “are repeats allowed,” then ask yourself “does order matter?”

TI-84: Enter the number “trials” (n) on the home screen. Press [MATH]. Use cursor keys to move to the PRB menu. Press 2 for permutation (2: _nP_r), 3 for combination (3: _nC_r). Enter the number of “successes” (r). Press [ENTER] to calculate.

TI-89: Press [2nd] Math > 7:Probability > Press 2 for permutation (2: _nP_r), 3 for combination (3: _nC_r). Enter the sample size on the home screen, then a comma, then enter the number of “successes,” then end the parenthesis. Press [ENTER] to calculate.

Excel: In a blank cell type in the formula =COMBIN(n, r) or =PERMUT(n, r) where n is the total number of objects and r is the smaller number of objects that you are selecting out of n. For example =COMBIN(8, 3).

The following flow chart in Figure 4-18 may help with deciding which counting rule to use.

Start on the left; ask yourself if the same item can be repeated. For instance, a person on a committee cannot be counted as two distinct people; however, a number on a car license plate may be used twice. If repeats are not allowed, then ask, does the order in which the item is chosen matter? If it does not then we use the combinations, if it does then ask are you ordering the entire group, use factorial, or just some of the group, use permutation.

Decision flowchart for counting rules: if repeats are allowed, use the Fundamental Counting Rule (multiply the total number of outcomes in each event); if not and order matters, ordering all of n gives n! while ordering r out of n gives permutations nPr = n!/(n−r)!; if order does not matter, use combinations nCr = n!/(r!(n−r)!).

Figure 4-18

Example 5

Critical Miss, PSU's Tabletop Gaming Club, has 15 members this term. How many ways can a slate of 3 officers consisting of a president, vice-president, and treasurer be chosen?

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In this case, repeats are not allowed since we don’t want the same member to hold more than one position. The order matters, since if you pick person 1 for president, person 2 for vice-president, and person 3 for treasurer, you would have different members in those positions than if you picked person 2 for president, person 1 for vice-president, and person 3 for treasurer. This is a permutation problem with n = 15 and r = 3.

$
{ }_{15} \mathrm{P}_{3}=\frac{15 !}{(15-3) !}=\frac{15 !}{12 !}=2730
$
There are 2,730 ways to elect these three positions.

In general, if you were selecting items that involve rank, a position title, 1^st, 2^nd, or 3^rd place or prize, etc. then the order in which the items are arranged is important and you would use permutation.

Example 6

Critical Miss, PSU's Tabletop Gaming Club, has 15 members this term. They need to select 3 members to have keys to the game office. How many ways can the 3 members be chosen?

Show solution

In this case, repeats are not allowed, because we don’t want one person to have more than one key. The order in which the keys are handed out does not matter. This is a combination problem with n = 15 and r = 3.

$_{15} C_{3} = \frac{15!}{(3! (15 - 3)!)} = \frac{15!}{(3! \cdot 12!)} = 455 There are 455 ways to hand out the three keys.$

We can use these counting rules in finding probabilities. For instance, the probability of winning the lottery can be found using these counting rules.

Example 7

What is the probability of winning the jackpot in the Pick-4 Lottery? To play Pick-4, you choose 4 numbers from 0 to 9. This will give you a number between 0000 and 9999. You win the jackpot if you match your 4 numbers in the exact order they are drawn.

https://www.oregonlottery.org/jackpot/pick-4/ Solution There is only one winning number, so the numerator for the probability will just be 1. The denominator will be all the ways to select the 4 numbers. Repeat numbers are allowed, for example you can have 4242 with repeating 4s and 2s. The order in which the balls are selected does matter by the rules of the game. Use the fundamental counting rule combined with the fundamental counting rule and we would get 10·10·10·10 = 10,000. Thus, the probability of winning the jackpot would be $\frac{1}{10000} = 0.0001$

Example 8

What is the probability of getting a full house if 5 cards are randomly dealt from a standard deck of cards?

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A full house is a combined three of a kind and pair, for example, QQQ22. There are ₁₃C₁ ways to choose a card between Ace, 2, 3, …, King. Once a number is chosen, there are 4 cards with that rank and there are ₄C₃ ways to choose a three of kind from that rank. Once we use up one of the ranks, such as the three queens, there are ₁₂C₁ ways to choose the rank for the pair. Once the pair is chosen there are ₄C₂ ways to choose a pair from that rank. All together there are ₅₂C₅ ways to randomly deal out 5 cards. The probability of getting a full house with then be

$\frac{_{13} C_{1} \cdot_{4} C_{3} \cdot_{12} C_{1} \cdot_{4} C_{2}}{_{52} C_{5}}$ = $\frac{3744}{2598960}$ = 0.00144

Adapted from Mostly Harmless Statistics by Rachel Webb (Portland State University), hosted on LibreTexts (stats.libretexts.org) and licensed under CC BY-SA 4.0. Changes were made. License: CC-BY-SA-4.0.