Recall that, if all objects in a set are distinct, then they can be arranged in $n!$ possible permutations, where $n$ represents the number of objects. The quantity $n!$ is given by: 

$\displaystyle n\cdot (n-1)\cdot (n-2)\cdots 2\cdot 1$

It is easy enough to use this formula to count the number of possible permutations of a set of distinct objects; for example, the number of permutations of three differently-colored balls. However, consider a situation where not all of the elements in a set of distinct objects are used in each permutation. For example, what if $7$ cards are chosen from among a full deck of $52$? In this case, not all of the cards from the deck are chosen for each possible permutation. There exists a formula for solving permutation problems such as this one, which would otherwise be nearly impossible to determine. 

Permutations of a Partial Set

If not all of the objects in a set of unique elements are chosen, the following formula is used. This formula determines the number of possible permutations of $k$ elements selected from the set of $n$ elements:

$\displaystyle \frac {n!}{(n-k)! }$

To understand the application of this concept, consider a race in which $3$ different prizes are awarded to the top $3$ fastest competitors. If $25$ competitors participate in the race, in how many distinct orders could the $3$ prizes be awarded?

To solve this problem, we want to evaluate the number of possible permutations of  $3$elements from the set of $25$ elements; in other words, $k = 3$ and $n=25$. Plugging these values into the formula, we have:

$\displaystyle \frac {25!}{(25-3)! } = \frac {25!}{22!} $

Remember that both $25!$ and $22!$ contain the terms $22 \cdot 21 \cdots 2 \cdot 1$. Thus, these values cancel from the numerator and denominator, and the equation can be simplified:

$$ $ \displaystyle \begin{aligned} \frac{25!}{22!} &= \frac{25 \cdot 24 \cdot 23 \cdot 22 \cdot 21 \cdots 2 \cdot 1}{22 \cdot 21 \cdots 2 \cdot 1} \\ &= 25 \cdot 24 \cdot 23 \\ &= 13,800 \end{aligned}$

There are $13,800$ possible permutations in which the $3$ top prizes may be awarded to the $25$ race competitors.

General Considerations

It is worth noting that this formula does not exclude what we might call 'duplicate' permutations. In other words, the order of the elements selected does matter. Consider $3$ cards drawn from a pack: an ace of spades, a $10$ of diamonds, and a $3$ of clubs. The hand is exactly the same as the following: a $10$ of diamonds, an ace of spaces and a $3$ of clubs. If you are playing a game in which the order of your cards does not matter, you will only want to count each permutation of cards once. The formula introduced here does not apply to such situations. 

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