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I'm trying to prove that ${n \choose r}$ is equal to ${{n-1} \choose {r-1}}+{{n-1} \choose r}$ when $1\leq r\leq n$.

I suppose I could use the counting rules in probability, perhaps combination= ${{n} \choose {r}}=\frac{n!}{r!(n-r!)}$.

I want to see an actual proof behind this equation. Does anyone have any ideas?

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BTW This is sometimes called Pascal's rule proofwiki.org/wiki/Pascal%27s_Rule –  Martin Sleziak Dec 10 '11 at 12:52

6 Answers 6

Consider $n$ balls in a basket. Let there be $1$ red ball and $n-1$ blue balls. Now look at the number of ways of choosing $r$ balls in two different ways

One way is to choose $r$ balls out of the $n$ balls. So the number of ways is $C(n,r)$

The other way is to look at the cases when out of the $r$ balls chosen if we have a red ball or not. We have only two options namely out of the $r$ balls we could have one red ball or no red balls

The number of ways of having $1$ red ball is to choose the one red ball which can be done in $C(1,1)$ ways and choose the remaining $(r-1)$ balls from the $(n-1)$ blue balls which can be done in $C(n-1,r-1)$ ways

Similarly, the number of ways of having no red balls is to choose all the balls as blue balls which can be done in $C(n-1,r)$ ways

These are the only two cases and these are mutually exclusive and hence the total number of ways is $C(n-1,r-1)+C(n-1,r)$

Hence, we get $$C(n,r) = C(n-1,r-1) + C(n-1,r)$$

The same idea could be extended to prove a generalization of the above $$C(m+n,r) = \displaystyle \sum_{k=\max(0,r-n)}^{\min(r,m)} C(m,k) C(n,r-k)$$

Consider a basket with $m$ red balls and $n$ blue balls and we want to count the number of ways in which $r$ balls can be drawn. Argue by two different ways to count (same as above) to prove this.

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Thanks! I was having a hard time visualizing the concept, and I figured proving this would help. Your example is brilliant. –  user6639 Feb 4 '11 at 23:36
@Christopher: More generally, any equality involving only natural numbers can be argued purely on the basis of counting. –  user17762 Feb 5 '11 at 0:14

As Sivaram and Chandru1 suggested, a combinatorial argument is often a very good way to understand/prove that kind of identities.

The other way would be, as you said, to use the explicit formula for the Binomial coefficient:

$${{n-1} \choose {r-1}}+{{n-1} \choose r}=\frac{(n-1)!}{(n-k)!(k-1)!}+\frac{(n-1)!}{(n-k-1)!k!}$$

which reduces to $\frac{n!}{(n-k)!k!}={n\choose k}$.

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Since you mentioned that you were having a hard time visualising this and I always seem to find myself visualising it whenever I have the need to write it down here is what goes through my mind as the pen is moving across the paper:

We are placing $r$ identical balls in $n$ boxes (at most one in each) that are in a straight line, so ${ n \choose r}$ ways to do this, now either the last box is empty, that's ${n-1 \choose r}$ ways, or the last box is full, that's ${ n-1 \choose r-1}$ ways. QED

This is equivalent to Sivaram's answer but does away with the colours, which for the purposes of visualising is probably slightly easier.

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Here's a take on this formula from a different direction. Suppose you were given the recurrence relation $R(n,k) = R(n-1,k) + R(n-1,k-1)$, for $0 \leq k \leq n$, and boundary condition $R(0,k) = [k=0]$. How would you derive the solution $R(n,k)$ if you didn't already know what it was? You could use generating functions.

(The following is borrowed from Wilf's Generatingfunctionology, 2nd edition, p. 14). Let $$G_n(x) = \sum_{k=0}^{\infty} R(n,k) x^k.$$ Multiplying the recurrence relation above by $x^k$ and summing $k$ from $1$ to $\infty$ yields $$G_n(x) -1 = G_{n-1}(x) - 1 + x G_{n-1}(x).$$ Thus $G_n(x) = (x+1)G_{n-1}(x)$, with $G_0(x) = 1$. Thus $G_n(x) = (x+1)^n$. Since $R(n,k)$ is the coefficient of $x^k$ in $(x+1)^n$, applying the binomial theorem tells us that $R(n,k) = \binom{n}{k}$.

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See this Wikipedia page:

Under the subsection Recursion formula, HINT for proving this formula is given. Hope you can do it from there.

The formula follows either from tracing the contributions to $X^{k}$ in $(1 + X)^{n−1}(1 + X)$, or by counting k-combinations of {1, 2, ..., n} that contain n and that do not contain n separately

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How about this:

Note that

$${n \choose r} = \frac {n(n-1)(n-2)...(n-r+1)} {1\cdot2\cdot3\cdot...(r-1)r}$$

This can be written as $${n \choose r} = \frac nr \cdot \frac {(n-1)(n-2)...(n-r+1)} {1\cdot2\cdot3\cdot...(r-1)} = \frac nr {n-1 \choose r-1} $$ and also $${n \choose r} = \frac n{n-r} \cdot \frac {(n-1)(n-2)...(n-r+1)(n-r)} {1\cdot2\cdot3\cdot...(r-1)r}= \frac n{n-r} {n-1 \choose r}$$

So $$\begin {align} {n-1 \choose r-1} + {n-1 \choose r} & = \frac rn {n\choose r} + \frac {n-r} n {n\choose r} \\ & = {n \choose r} \end {align} $$

as required.

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