Why Order Changes the Count
Suppose several distinct objects are available, and we select some of them to fill ordered positions. The same objects in a different order make a different result: assigning Maya to first place and Luis to second is not the same outcome as assigning Luis to first and Maya to second. To count such outcomes, we can count the choices for each position in sequence. Because an object cannot be selected twice, the number of available choices decreases after every selection.
This tutorial derives a compact formula for that count. The key idea is not the formula itself, but why its factors have the values they do: \(n\) choices for the first position, \(n-1\) for the next, and so on until \(r\) positions have been filled.
Here \(n\) and \(r\) are whole numbers with \(0\le r\le n\). The \(n\) objects are distinct, and the positions being filled are distinguishable. For example, president and treasurer are distinguishable positions, as are first, second, and third place. These conditions matter: the formula counts ordered selections without replacement, not selections in which the same object can fill multiple positions.
Counting One Position at a Time
Imagine filling \(r\) ordered positions from \(n\) distinct objects. For the first position, all \(n\) objects are available. After choosing one, only \(n-1\) remain for the second position. After two choices, \(n-2\) remain for the third. This continues until the \(r\)th position.
There are \(r\) factors in this product. The first factor is \(n\), and each later factor is one smaller than the factor before it. Therefore the last factor is \(n-(r-1)\), which equals \(n-r+1\). By the multiplication principle for sequential choices, the total number of ordered selections is the product of the numbers of choices at each step:
For instance, if 8 distinct finalists fill 3 ordered places, the factors are \(8\), \(7\), and \(6\). The first person can occupy first place in 8 ways; for each such choice, a different person can occupy second place in 7 ways; and for each pair, a third person can occupy third place in 6 ways. Multiplying these stage counts includes every possible ordered result exactly once.
Turning the Product Into a Factorial Formula
The product above is often convenient for direct counting. A factorial expression is a more compact way to write the same product and can be easier to calculate. Recall that for a nonnegative integer \(m\), \(m!\) is the product of the positive integers from \(m\) down to 1, with \(0!=1\). In particular,
The part of \(n!\) through \(n-r+1\) is exactly the product for \({}_nP_r\). The remaining factors are \((n-r)(n-r-1)\cdots 1\), which together equal \((n-r)!\). Dividing \(n!\) by \((n-r)!\) cancels those remaining factors, leaving just the desired product.
This derivation also explains why the denominator is \((n-r)!\): it removes from \(n!\) the factors corresponding to the objects left after the \(r\) selected positions have been filled. The formula does not come from treating each position as having \(n\) choices. Its decreasing factors reflect the fact that a chosen object is no longer available.
Worked Examples
Worked Example: Awarding Three Places
Eight distinct runners finish a race. In how many ways can first, second, and third place be awarded?
Check the counting conditions. The runners are distinct, there are three ordered places, and each runner can receive at most one place. Thus this is an ordered selection without replacement, with \(n=8\) and \(r=3\).
Count the choices in sequence. There are 8 choices for first place, 7 remaining choices for second, and 6 remaining choices for third:
Use the formula as a check.
There are 336 possible assignments of the three places. For example, a result with runner A first, runner B second, and runner C third is counted separately from a result with B first, A second, and C third.
Worked Example: Assigning Distinct Roles
A group has 6 distinct members. How many ways can it assign a chair, a secretary, a treasurer, and a recorder, with one different person in each role?
Check the counting conditions. The roles are distinguishable, so changing who is chair or secretary changes the assignment. Each person can fill only one role. There are \(n=6\) members and \(r=4\) roles.
The chair can be selected in 6 ways. After that, 5 members remain for secretary; then 4 for treasurer; and 3 for recorder. Therefore:
The factorial formula gives the same count:
There are 360 possible assignments of members to these four roles. The count concerns who fills each particular role, not merely which four people are selected.
Worked Example: Arranging Every Object
Five distinct books are placed in a row on a shelf. How many different orders are possible?
All five books are used, so \(n=5\) and \(r=5\). The first place has 5 choices, the second has 4, and the choices continue down to 1 for the final place:
The formula confirms this result:
Because \(0!=1\), the formula works when all available objects are arranged. There is one empty arrangement if \(r=0\) as well: the formula gives \({}_nP_0=n!/n!=1\). That convention is consistent with making no choices in zero positions.
How to Decide Whether the Formula Fits
Before calculating \({}_nP_r\), identify what counts as one outcome. Ask whether the problem fills ordered positions and whether the same object may appear more than once. If the positions are distinguishable and no object can be reused, the decreasing-choice product applies. If either condition fails, this formula does not describe the requested count without adjustment.
Count the \(n\) distinct objects available. Check that the objects are treated as different possible choices.
Determine how many positions or stages are filled. This number is \(r\); each position must be distinguishable from the others for order to matter.
Confirm that a selected object cannot be selected again. The permutation product assumes the available choices decrease after each selection.
Use \(n(n-1)\cdots(n-r+1)\) or \(n!/(n-r)!\), then state what the resulting count represents in the original situation.
Common Mistakes and Why the Formula Matters
A frequent error is to use \(n^r\), reasoning that each of the \(r\) positions has \(n\) possible choices. That count assumes the number of choices stays at \(n\) at every step, so it allows an object to be selected repeatedly. For selections without replacement, the correct choices decrease: \(n\), then \(n-1\), and so on. For example, in the race with 8 runners, \(8^3=512\) counts outcomes that could assign the same runner to more than one place. The valid count is \(8\cdot7\cdot6=336\).
Another mistake is to overlook whether order matters. If a problem asks who fills chair, secretary, treasurer, and recorder, swapping two people between roles gives a different assignment. If it asks only which people are selected, the roles are absent and that reasoning does not apply as stated. The permutation formula is specifically for ordered selections; recognizing that distinction is essential before substituting values.
The formula is useful because many counting problems describe a sequence of choices rather than a single choice. It translates that description into a transparent product, and the factorial form makes the product compact. In either form, the result is meaningful only when the objects, positions, and no-reuse condition match the situation.
Check Your Understanding
For each question, focus first on the objects, the positions, and whether reuse is allowed.
- There are 9 distinct finalists. In how many ways can first and second place be assigned? Write both the decreasing-choice product and the factorial expression.
- Explain why \({}_7P_3\) has three factors and identify the value of its final factor.
- Five distinct speakers are assigned to five different time slots. How many schedules are possible? Express the answer using a factorial.
- A student writes \(6^4\) to count four ordered selections from six distinct objects without replacement. What assumption in this calculation is incorrect, and what product should be used instead?
- What does \({}_nP_0=1\) mean in terms of filling positions?