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How do you solve this?

Find out which recurrence relation involving $\dbinom{n}{r}$ is valid, and thus prove that we can compute $\dbinom{n}{r}$ in a recursive manner.

I appreciate any help. Thank You

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    @J.M. - Yes, but I must solidify my friendship with that neat triangle2012-05-09

3 Answers 3

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To prove: $\dbinom{n}{k}=\dbinom{n-1}{k-1}+\dbinom{n-1}{k}$

We have: $\dbinom{n-1}{k-1}=\frac{(n-1)!}{(k-1)!(n-k)!}=\frac{(n-1)!}{(k-1)!(n-k-1)!}\frac{1}{n-k}$

$\dbinom{n-1}{k}=\frac{(n-1)!}{(k)!(n-k-1)!}=\frac{(n-1)!}{(k-1)!(n-k-1)!}\frac{1}{k}$

Adding the two together:

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

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

$=\dbinom{n}{k}$

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From the familiar $\binom{n}{r}=\frac{n!}{r!(n-r)!}$, you can readily derive the formula $\binom{n}{r}=\frac{n}{r}\binom{n-1}{r-1}.$ This can be computationally and theoretically useful, particularly when we are interested in the expression $\binom{n}{r}p^r(1-p)^{n-r}$ that comes up when we are working with the Binomial Distribution.

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    Or $\binom{n}{r}=\frac{n}{n-r}\binom{n-1}{r}$ which may be quicker if $2r \gt n$2012-05-09
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There are many recurrence relations for $\dbinom{n}{r}$. One of the most commonly used one is the following. $\binom{n+1}{r} = \binom{n}{r} + \binom{n}{r-1}.$ There are many ways to prove this and one simple way is to look at $\displaystyle (1+x)^{n+1}$. We know that $(1+x)^{n+1} = (1+x)^{n} (1+x).$ Now compare the coefficient of $x^r$ on both sides.

The left hand the coefficient of $x^r$ is $\dbinom{n+1}{r}.$ On the right hand side, the $x^r$ term is obtained by multiplying the $x^r$ term of $(1+x)^n$ with the $1$ from $(1+x)$ and also by multiplying the $x^{r-1}$ term of $(1+x)^n$ with the $x$ from $(1+x)$. Hence, the coefficient of $x^r$ is given by the sum of these two terms from the right hand side i.e. $\dbinom{n}{r} + \dbinom{n}{r-1}$ enter image description here As Rob and J.M have pointed out, these recurrences define the celebrated Pascal's triangle. A pictorial representation is shown above. Each row represents the value of $n$ starting from $0$. On a row, as we proceed from left to right, the different values of $r$ from $0$ to $n$ are hit. The arrows indicate how the $r^{th}$ value on the $(n+1)^{th}$ row is obtained as the sum of the $(r-1)^{th}$ and $r^{th}$ value on the $n^{th}$ row.

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    No worries; I learned it rather recently, too.2012-05-09