0

Prove that,

$$\binom{3n}{r}=\binom{3n-1}{r} + \binom{3n-1}{r-1}$$

Here is my solution below, this is my first time working through a pascal identity problem and I just wanted to see if this made sense:

By the definition $\binom{n}{r}$ we have,

$$\binom{3n-1}{r-1}=\frac{(3n-1)!}{(3n-1-(r-1))!(r-1)!} = \frac{(3n-1)!}{(3n-r)!(r-1)!}$$

and,

$$\binom{3n-1}{r}=\frac{(3n-1)!}{(3n-1-r)!r!}$$

Thus, starting with the right hand side of the equation,

$$\binom{3n-1}{r-1} + \binom{3n-1}{r} = \frac{(3n-1)!}{(3n-r)!(r-1)!} + \frac{(3n-1)!}{(3n-1-r)!r!} \\ = \frac{(3n-1)!r}{(3n-r)r!} + \frac{(3n-1)!(3n-r)}{(3n-r)!r!} \\ = \frac{(3n-1)!(r+3n-r)}{(3n-r)!r!} \\ = \frac{3n!}{(3n-r)!r!} \\ = \binom{3n}{r}$$ \

mt12345
  • 315
  • You can replace 3n with any number m – nonuser Jan 26 '18 at 14:52
  • 1
    But...I think his idea is still good and the OP did not know about duplicate, so as far as learning experience is concerned, it is fruitful and that is important! – imranfat Jan 26 '18 at 14:52
  • 1
    Not quite a duplicate because it's a proof verification question... the first question linked is not proof verification, and the second is proof verification of a different proof... – BallBoy Jan 26 '18 at 14:54
  • @imranfat Yes, I agree. But it is useful to compare first with other posts, which have attempted exactly the same proof. This kind of question is really very frequent here. And no, the other proof attempts and proof verifications are really more or less the same. – Dietrich Burde Jan 26 '18 at 14:54
  • 1
    @DietrichBurde In the second linked question, the proof being verified is combinatorial. I wouldn't call that "more or less the same." – BallBoy Jan 26 '18 at 14:56
  • 1
    @DietrichBurde I'm sure that's true! But for the sake of future people who come across this page, we shouldn't lead them on a wild goose chase by marking a duplicate and linking to the wrong question. – BallBoy Jan 26 '18 at 14:58
  • The first duplicate is not so wrong; the OP could really compare the proof there. I have the feeling that this is very similar, indeed. – Dietrich Burde Jan 26 '18 at 15:25

1 Answers1

1

The proof looks perfect.

Why use $3n$ though? Note that the proof doesn't depend on using the $3$ at all. If you substituted $n$, you'd have the same proof for the general form of Pascal's identity.

BallBoy
  • 14,740