Prove that $\dfrac{^nC_r}{^nC_{r-1}} = \dfrac{n-r+1}{r}$

> [!TLDR] Expanding both combinations by their factorial definitions and cancelling proves $\dfrac{^nC_r}{^nC_{r-1}}=\dfrac{n-r+1}{r}$. > [!FORMULA] $^nC_r=\dfrac{n!}{r!\,(n-r)!}$. 1. Write out both terms: $$^nC_r=\dfrac{n!}{r!\,(n-r)!}, \qquad {}^nC_{r-1}=\dfrac{n!}{(r-1)!\,(n-r+1)!}$$ 2. Form the ratio (dividing by a fraction = multiplying by its reciprocal): $$\dfrac{^nC_r}{^nC_{r-1}}=\dfrac{n!}{r!\,(n-r)!}\times\dfrac{(r-1)!\,(n-r+1)!}{n!}$$ 3. The $n!$ in numerator and denominator cancels: $$=\dfrac{(r-1)!\,(n-r+1)!}{r!\,(n-r)!}$$ 4. Since $r!=r\times(r-1)!$, we have $\dfrac{(r-1)!}{r!}=\dfrac{1}{r}$. …

Prove that dfracⁿCrⁿCr-1 = (n-r+1)/(r)

Worked Examples · Example 42

\dfrac{^nC_r}{^nC_{r-1}} = \dfrac{n-r+1}{r}

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Concept understanding — Combination

Combination: Choosing Without Ordering

Imagine you have three friends: A, B, and C. You need to pick two of them to form a committee. How many different committees can you make?

You could pick A and B, or A and C, or B and C. That's three committees. Notice something crucial: the committee {A, B} is exactly the same as {B, A}. Who gets picked first doesn't matter — only who is on the committee matters. This is the core idea of a combination: a selection where the order of selection does not matter.

The Intuition: "How many groups?"

A combination answers the question: "From a set of $n$ distinct items, how many ways can I choose $r$ of them, ignoring the order in which I pick them?"

Think of it as forming a team, a committee, a hand of cards, or a set of ingredients for a recipe. In all these cases, swapping two members around doesn't give you a new team — it's still the same group.

The Precise Statement

A combination is a selection of $r$ objects from a set of $n$ distinct objects, where the order of selection does not matter. The number of such combinations is denoted by:

\binom{n}{r} \quad \text{or} \quad ^nC_r

and is given by the formula:

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

where $n!$ (read "n factorial") means $n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1$ , and $0! = 1$ by definition.

Why the $r!$ in the denominator?

This is the key to understanding combinations. If order did matter, the number of ways to pick $r$ objects from $n$ would be $n \times (n-1) \times \cdots \times (n-r+1)$ , which is written as $\frac{n!}{(n-r)!}$ . That's called a permutation.

But in a combination, every group of $r$ objects can be rearranged in $r!$ different orders. Since we don't care about order, we divide by $r!$ to "collapse" all those orderings into one single group.

So:

\binom{n}{r} = \frac{\text{number of ordered selections}}{\text{number of ways to order each selection}} = \frac{n!}{(n-r)!} \div r! = \frac{n!}{r! \, (n-r)!}

A Quick Example

From a class of 10 students, how many ways can you choose a team of 3?

Here $n=10$ , $r=3$ .

\binom{10}{3} = \frac{10!}{3! \, 7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120

So there are 120 different teams possible. …

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Expanding both combinations by their factorial definitions and cancelling proves $\dfrac{^nC_r}{^nC_{r-1}}=\dfrac{n-r+1}{r}$ .

^nC_r=\dfrac{n!}{r!\,(n-r)!}

Write out both terms:

^nC_r=\dfrac{n!}{r!\,(n-r)!}, \qquad {}^nC_{r-1}=\dfrac{n!}{(r-1)!\,(n-r+1)!}

Form the ratio (dividing by a fraction = multiplying by its reciprocal):

\dfrac{^nC_r}{^nC_{r-1}}=\dfrac{n!}{r!\,(n-r)!}\times\dfrac{(r-1)!\,(n-r+1)!}{n!}

The $n!$ in numerator and denominator cancels:

=\dfrac{(r-1)!\,(n-r+1)!}{r!\,(n-r)!}

Since $r!=r\times(r-1)!$ , we have $\dfrac{(r-1)!}{r!}=\dfrac{1}{r}$ . …

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Q.Prove that nCrnCr−1=n−r+1r\dfrac{^nC_r}{^nC_{r-1}} = \dfrac{n-r+1}{r}nCr−1​nCr​​=rn−r+1​

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Q.Prove that $\dfrac{^nC_r}{^nC_{r-1}} = \dfrac{n-r+1}{r}$

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