Without-Replacement Draws Counted as Combinations

Why Unordered Samples Need a Different Count

Suppose several objects are drawn from a collection, and the question is about which objects were selected, not the order in which they appeared. If an object is not returned before the next draw, the selection is without replacement. The possible samples are then groups of distinct objects, and combinations count these groups without counting different orders separately.

For example, drawing a red ball and then a blue ball gives the same unordered sample as drawing that blue ball and then that red ball. Counting these as two outcomes would be inappropriate if the event of interest concerns only the colors in the sample. The central steps are to count all possible samples of the required size, count the samples that meet the condition, and take the ratio of those counts.

The objects must be treated as distinct for this count, even when some share a visible feature. For example, two red balls are still different objects if one is labeled R1 and the other R2. A sample containing R1 is a different sample from one containing R2. The labels are useful for counting possible selections; a probability question may then ask about a feature such as color.

Why the Possible Samples Are Equally Likely

Assume that each draw selects uniformly from the objects still available, with no object returned between draws. A particular ordered sequence of \(k\) distinct objects has probability \(1/[N(N-1)\cdots(N-k+1)]\). Every ordered sequence of \(k\) distinct objects has the same probability.

An unordered sample of \(k\) distinct objects can occur in exactly \(k!\) orders. Since each of those orders has the same probability, every unordered sample has the same probability as every other unordered sample. There are \(\binom{N}{k}\) such samples, so each has probability \(1/\binom{N}{k}\). Thus a count of favorable samples can be used as a probability only when the samples being counted are equally likely.

This is the favorable-counts-over-total-counts rule applied to a sample space whose outcomes are unordered groups. The combination \(\binom{N}{k}\) counts the groups of size \(k\) chosen from \(N\) distinct objects, without treating different orders as different groups.

Counting a Specified Number From a Category

Suppose a collection contains \(A\) objects of one category and \(N-A\) objects of another. To draw a sample of size \(k\) containing exactly \(r\) objects from the first category, choose \(r\) of its \(A\) objects and \(k-r\) of the remaining objects. Each pair of choices determines one sample, and every sample with exactly \(r\) from the first category is formed in exactly one such way.

The numerator counts favorable samples: choose \(r\) objects from the category and \(k-r\) from outside it. The denominator counts every possible sample of size \(k\). The formula applies when the requested counts are possible: \(0\le r\le A\) and \(0\le k-r\le N-A\). If those conditions fail, there are no favorable samples and the probability is zero.

Worked Examples

A Short Justification of the Counting Method

The ratio works because equally likely outcomes give an event probability equal to the number of outcomes in that event divided by the total number of outcomes. Here the outcomes are the unordered samples. To see why they have equal probability under uniform sequential drawing, fix one particular sample of \(k\) objects. Its \(k!\) possible orders are mutually exclusive sequences. Each sequence has probability \(1/[N(N-1)\cdots(N-k+1)]\), so the probability of obtaining that unordered sample is

The equality follows from \(\binom{N}{k}=N!/[k!(N-k)!]\): the product \(N(N-1)\cdots(N-k+1)\) is \(N!/(N-k)!\). The resulting probability is the same for every sample, which justifies dividing the favorable sample count by \(\binom{N}{k}\).

When This Method Applies

The combinations method is appropriate when the question concerns a subset, the selection is without replacement, and the objects are selected uniformly. It is especially useful when many different draw orders would produce the same event. Before calculating, identify the objects, the sample size, what counts as one outcome, and whether those outcomes are equally likely.

If the order matters to the question, an unordered count may not distinguish the outcomes the question treats as different. If objects can be selected more than once, the possible samples and their probabilities also change. And if some objects are more likely to be selected than others, equal counting of samples does not generally produce their probabilities. The probability rule depends on both the counting and the sampling mechanism.

Common Mistake: Counting Orders in Only Part of the Calculation

A tempting approach is to count the draw sequences that meet a condition but divide by the number of unordered samples. That mixes two different sample spaces, so the ratio need not be a probability. For instance, an ordered count distinguishes red-then-blue from blue-then-red, while an unordered count treats them as the same sample.

The correction is to count consistently. Either count unordered samples in both numerator and denominator, using combinations, or count ordered sequences in both. For a question about the composition of a sample rather than its order, the unordered approach is usually the more direct one.