Understanding Combinations in the Binomial Formula

What Does a Combination Count?

In The Binomial Probability Formula, the coefficient \(\binom{n}{k}\) accounts for the different arrangements that can produce exactly \(k\) successes in \(n\) trials. This tutorial focuses on what that coefficient means, why its calculation avoids counting the same selection more than once, and how to evaluate it with a calculator.

Imagine labeling the trials 1 through \(n\). To get exactly \(k\) successes, we need to choose which \(k\) trial positions will contain successes. Once those positions are chosen, the remaining \(n-k\) positions contain failures. The question is therefore: how many different groups of \(k\) positions can be selected from \(n\) positions?

For example, suppose the successes in eight trials occur in positions 2, 5, and 7. Listing those positions as 2, 5, 7 or as 7, 2, 5 does not create a different outcome. The same three positions are selected. A combination counts that selection once, not once for every order in which the positions could be listed.

Why the Factorial Formula Works

A factorial is the product of a positive whole number and every positive whole number below it. For example, \(5!=5\cdot4\cdot3\cdot2\cdot1=120\). By convention, \(0!=1\). Factorials provide a compact way to count selections.

Here is the reasoning. If order mattered, we could select and arrange \(k\) different positions from \(n\) in

ways. But a particular group of \(k\) positions can be listed in \(k!\) different orders. Since a combination treats all those listings as the same selection, divide by \(k!\). This gives \(\frac{n!}{k!(n-k)!}\). The division removes the repeated counts created by listing the same selected positions in different orders.

In a binomial probability, the trial positions are labeled, so choosing positions 2, 5, and 7 is different from choosing positions 1, 5, and 7. But within one chosen group, the order used to name those positions does not matter. That distinction is why combinations count the arrangements needed in the binomial formula.

The factorial expression often simplifies before you calculate it. Expand only enough of \(n!\) to cancel the \((n-k)!\) below it. This avoids entering very large factorials and makes the calculation easier to check.

Worked Example: Choosing Three Success Positions

This count does not depend on the probability of success. It depends only on the number of trial positions and how many are selected. The probability factors in the binomial formula account separately for how likely each arrangement is.

Calculating nCr with a Calculator

A calculator’s \(nCr\) function evaluates \(\binom{n}{k}\) directly. On a TI-84-style calculator, enter \(n\), open the MATH menu, select the PRB probability menu, choose nCr, enter \(k\), and evaluate. Menu locations or labels can vary slightly across calculator models, so check the model’s menu if needed.

For instance, to evaluate \(\binom{8}{3}\), the entry is equivalent to \(8\ nCr\ 3\). The calculator returns 56. Use the original values \(n\) and \(k\) in the function: the first value is the total number of positions, and the second is the number being selected.

Worked Example: Using nCr for a Larger Selection

A combination and a permutation answer different kinds of questions. If the order of selected items matters, the arrangements are distinct and a combination alone does not count them. Here, only which days are chosen matters, so each set is counted once.

A Useful Symmetry

Choosing \(k\) positions to be successes is equivalent to choosing the other \(n-k\) positions to be failures. Each choice of success positions determines exactly one matching group of failure positions, and vice versa. Therefore, a combination has a useful symmetry:

For example, choosing 11 success positions out of 15 is equivalent to choosing the four failure positions:

This identity is also a practical calculation check. If one side of a calculation differs from the other, review the values entered or the factorial simplification.

Worked Example: Using a Combination in a Binomial Probability

The coefficient 36 counts the possible locations of the two successes. The factor \((0.30)^2(0.70)^7\) is the probability of any one particular arrangement with two successes and seven failures. Multiplying them accounts for all 36 arrangements.

Common Mistakes and AP Exam Tip

When a binomial formula includes \(\binom{n}{k}\), make clear what the coefficient counts in the context. A correct numerical coefficient is more meaningful when connected to the selected trial positions.

• Counting order when it does not matter. Multiplying \(n(n-1)\cdots\) counts ordered lists. To count groups of selected positions, divide out the \(k!\) possible orders, as the combination formula does.
• Reversing the inputs. In \(\binom{n}{k}\), \(n\) is the total number of available positions and \(k\) is the number selected. Swapping them can produce an invalid expression or a different count.
• Confusing the coefficient with a probability. A value such as \(\binom{9}{2}=36\) is a number of selections. It is not the probability of exactly two successes. In the binomial formula, it must be multiplied by \(p^k(1-p)^{n-k}\).
• Expanding every factorial. Writing every factor in \(n!\) can create large numbers and make errors more likely. Cancel \((n-k)!\) first, then simplify.
• Ignoring the meaning of “exactly.” For exactly \(k\) successes, choose \(k\) positions and leave the other \(n-k\) positions as failures. The coefficient does not include arrangements with a different number of successes.

For a full-credit explanation, state that \(\binom{n}{k}\) counts the ways to choose the \(k\) success positions from the \(n\) trials. If showing the factorial calculation, include the substitution and enough cancellation to make the arithmetic verifiable. If using a calculator, show the nCr expression and then use the result in the complete probability formula when a probability is requested.

Key Takeaway

A combination counts selections when the order of selected positions does not matter. In a binomial setting, \(\binom{n}{k}\) counts the different choices of \(k\) success positions among \(n\) trials. Calculate it with \(\frac{n!}{k!(n-k)!}\) or a calculator’s nCr function, then use that count as the coefficient in the binomial probability formula.