Combination of n things taken r at a time when k particular objects occur (C(n-k, r-k))

When dealing with combinations in mathematics, we often encounter problems that involve selecting a subset of items from a larger set. A common type of problem is finding the number of ways to choose r items from a set of n distinct items, which is denoted as $C(n, r)$ or $\binom{n}{r}$ and is called a combination.

However, there are scenarios where we want to count combinations that include a specific subset of items. In this case, we are interested in the number of ways to choose r items from n items, but with the condition that k particular items must always be included.

Understanding the Formula: C(n-k, r-k)

The formula $C(n-k, r-k)$ or $\binom{n-k}{r-k}$ is used in situations where we have n total items, and we are required to choose r items such that k particular items are always included in the selection.

Here's the reasoning behind the formula:

1. Since k particular items must always be included, we have already chosen k items, and we need to choose the remaining r-k items from the remaining n-k items.
2. The number of ways to choose r-k items from n-k items is given by the combination formula $C(n-k, r-k)$.

Formula

The general formula for combinations is:

$$ C(n, r) = \binom{n}{r} = \frac{n!}{r!(n-r)!} $$

When k particular items must be included, the formula becomes:

$$ C(n-k, r-k) = \binom{n-k}{r-k} = \frac{(n-k)!}{(r-k)!(n-r)!} $$

Table of Differences and Important Points

Aspect	General Combination $C(n, r)$	Combination with k Particular Items $C(n-k, r-k)$
Total Items	n distinct items	n distinct items, with k specific items always included
Items to Choose	r items	r items, but k are predetermined
Remaining Choices	All n items are available for selection	Only n-k items are available for selection
Formula	$\binom{n}{r} = \frac{n!}{r!(n-r)!}$	$\binom{n-k}{r-k} = \frac{(n-k)!}{(r-k)!(n-r)!}$
Example	Choosing 3 books from a shelf of 5	Choosing 3 books from a shelf of 5, but 2 specific books must be included

Examples

Example 1: Basic Scenario

Suppose you have 5 different books on a shelf, and you want to choose 3 of them to take on a trip. The number of ways to do this without any restrictions is:

$$ C(5, 3) = \binom{5}{3} = \frac{5!}{3!(5-3)!} = 10 $$

Example 2: Including Specific Items

Now, let's say that out of the 5 books, there are 2 particular books that you definitely want to take with you. How many ways can you choose the 3 books now?

Since 2 books are already chosen, you only need to choose 1 more book from the remaining 3 books:

$$ C(5-2, 3-2) = C(3, 1) = \binom{3}{1} = \frac{3!}{1!(3-1)!} = 3 $$

So, there are 3 ways to choose the books when 2 specific ones must be included.

Example 3: Real-world Application

Imagine you are organizing a committee of 5 members from a group of 10 people, and there are 2 individuals who must be on the committee. How many different committees can be formed?

Here, n = 10, r = 5, and k = 2. The number of ways to form the committee is:

$$ C(10-2, 5-2) = C(8, 3) = \binom{8}{3} = \frac{8!}{3!(8-3)!} = 56 $$

Thus, there are 56 different ways to form the committee with the 2 specific individuals included.

Conclusion

The concept of $C(n-k, r-k)$ is a powerful tool in combinatorics that allows us to count the number of combinations where certain items are guaranteed to be included. It is important to understand the difference between general combinations and combinations with specific items included, as this can significantly alter the number of possible outcomes. By mastering this concept, you can solve a wide range of problems in probability, statistics, and other fields that involve combinatorial analysis.