Binomial Number proof

**Stonehambey** · August 20th 2009, 09:21 AM

Hi,

I'm required to show that

${n \choose r} + 2{n \choose r-1} + {n \choose r-2} = {n+1 \choose r}$

Now I know that ${n \choose r} + {n \choose r-1} = {n+1 \choose r}$

and we can write the LHS as

$\left[{n \choose r} + {n \choose r-1}\right] + {n \choose r-1} + {n \choose r-2}$

$={n+1 \choose r} + {n \choose r-1} + {n \choose r-2}$

Which means for the equality to hold I would have to show that ${n \choose r-1} + {n \choose r-2} = 0$ , which I don't think is right.

So where have I messed up?

Thanks

**Moo** · August 20th 2009, 09:33 AM

Hello,

I'd say it's rather ${n \choose r} + 2{n \choose r-1} + {n \choose r-2} = {n+{\color{red}2} \choose r}$

Because using the formula, we have ${n \choose r-1} + {n \choose r-2}={n+1 \choose r-1}$

And using the formula again, it results that :
${n \choose r} + 2{n \choose r-1} + {n \choose r-2}={n+1 \choose r} +{n+1 \choose r-1}={n+2 \choose r}$

So could there possibly be a typo ?

**Stonehambey** · August 20th 2009, 09:39 AM

Originally Posted by Moo

So could there possibly be a typo ?

Double checked and I definitely copied the question out correctly, so I guess it must be a typo.

Thanks

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