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Needed help for the following question in combination?

For $2 \leq r\leq n$. To what does$^nC_r + 2\times^nC_{r-1} + ^nC_{r-2}$ is equal to ?

Thank's

Akash

peterwhy
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Deiknymi
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  • I really didn't tried to solve it cause I am not getting what the question asking to do if you can tell that also then it would also help – Deiknymi Sep 14 '13 at 17:43
  • what all do you know about $^nC_r$??? –  Sep 14 '13 at 17:45
  • well I know enough just having problem with the language of the ques – Deiknymi Sep 14 '13 at 17:48
  • well, if you know enough, it would not be difficult for you to see at least what is the question.... Do not try to be rude... Ask help and please don't place orders.. –  Sep 14 '13 at 17:53
  • haha no I am not being rude but apologies if it sounded like that and I am not placing order's just came here for help in place I thought I could get one – Deiknymi Sep 14 '13 at 17:56
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    $\TeX$tip: one usually writes \binom{n}{k} to get $\binom{n}{k}$, which is a much more readable notation! – Mariano Suárez-Álvarez Sep 14 '13 at 18:00
  • @MarianoSuárez-Alvarez : Thanks for the suggestion :) –  Sep 14 '13 at 18:08

1 Answers1

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Actual Question is to calculate what does $\binom{n}{r}+2\binom{n}{r-1}+\binom{n}{r-2}$ equal to???

With out assuming much,

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

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

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

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

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

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

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

Thus, we conclude that $\binom{n}{r}+\binom{n}{r-1}=\binom{n+1}{r}$

with similar observation, (replacing $r$ by $r-1$) we have $\binom{n}{r-1}+\binom{n}{r-2}=\binom{n+1}{r-1}$

and so $\binom{n}{r}+2\binom{n}{r-1}+\binom{n}{r-2}=\binom{n}{r}+\binom{n}{r-1}+\binom{n}{r-1}+\binom{n}{r-2}$

As we have already seen that $\binom{n}{r}+\binom{n}{r-1}=\binom{n+1}{r}$ and $\binom{n}{r-1}+\binom{n}{r-2}=\binom{n+1}{r-1}$

we have $\binom{n}{r}+2\binom{n}{r-1}+\binom{n}{r-2}=\binom{n}{r}+\binom{n}{r-1}+\binom{n}{r-1}+\binom{n}{r-2}=\binom{n+1}{r}+\binom{n+1}{r-1}=\binom{n+2}{r}$

So, we have $\binom{n}{r}+2\binom{n}{r-1}+\binom{n}{r-2}=\binom{n+2}{r}$

  • ok question is solved but I am still confused what was the use of the condition $2\leq r \leq n$ – Deiknymi Sep 15 '13 at 01:59
  • please convince yourself that one can not use $r-2$ in binomial coefficient if $r=1$ or $r=0$.... and you know why $r\leq n$ ;) –  Sep 15 '13 at 02:30