# Difference operation on factorials

Please how is the combination addition formula ${{t}\choose{r}}={{t-1}\choose{r}}+{{t-1}\choose{r-1}}$ useful in proving the difference equation $\Delta_{t}{{r+t}\choose{t}}={{r+t}\choose{t+1}}$?

Secondly, does ${{t}\choose{k}}={{t}\choose{t-k}}$ need verification? I thought definition only testifies this.

The difference equation is an immediate consequence of the addition formula:

$$\Delta_t\binom{r+t}t=\binom{r+t+1}t-\binom{r+t}t=\binom{r+t}{t-1}\;,$$

since

$$\binom{r+t+1}t=\binom{r+t}t+\binom{r+t}{t-1}\;.$$

Whether the identity $\binom{t}k=\binom{t}{t-k}$ needs proof depends on how you defined the binomial coefficient. If you defined it in terms of factorials, virtually no proof is required. If you defined it combinatorially, or as

$$\binom{t}k=\frac{t^{\underline k}}{k!}\;,$$

then some argument is required.

• Thanks so much Brian M. Scott.
– YYG
Nov 13 '13 at 22:27
• @YYG: You’re very welcome. Nov 13 '13 at 22:27
• But Brian what is that argument that you have mentioned about because I consider the binomial coefficient in terms of the falling factorial like you have written there.
– YYG
Nov 13 '13 at 22:30
• @YYG: That identity is valid only for integer $t\ge 0$ and integer $k$, so I would simply show that if $0\le k\le t$, then $$\frac{t^{\underline k}}{k!}=\frac{t!}{k!(t-k)!}=\frac{t^{\underline{t-k}}}{(t-k)!}\;,$$ and if $k$ is an integer outside that range, then both binomial coefficients are $0$. Nov 13 '13 at 22:33
• Okay, but sorry, how do you manage to 'kill' that $k!$ and get $t$ to the falling $t-k$ at the numerator on the last part? And by the way, the equation can be defined even for non-integer and/or negative values of $t$ if we utilize the definition of Gamma function.
– YYG
Nov 13 '13 at 22:41

Just write your difference equation explicitely: $$\Delta_t \binom{r+t}{t}=\binom{r+t+1}{t+1}-\binom{r+t}{t} = \binom{r+t}{t+1}.$$

The identity $\binom{x}{y}=\binom{x}{x-y}$ is obvious if you defined binomial coefficients via factorials. If you have a combinatoric definition, then it's an evident exercise.

• Thanks so much TZakrevskiy.
– YYG
Nov 13 '13 at 22:26
• @YYG you're welcome! Nov 13 '13 at 22:28