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Question

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A. $^{2n}{{C}_{2r-1}}$

B. $^{\left( n+1 \right)}{{C}_{r}}$

C. $^{n}{{C}_{r+1}}$

D. $^{\left( n+1 \right)}{{C}_{r+1}}$

Answer
Verified

We have to find the value of $^{n}{{C}_{r}}{{+}^{n}}{{C}_{r-1}}$.

The meaning of $^{n}{{C}_{r}}$ in terms of permutation and combination is picking $r$ objects out of $n$ objects.

And in mathematical way $^{n}{{C}_{r}}$ can be represented as,

${{\therefore }^{n}}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}..........................(1)$

Similarly, $^{n}{{C}_{r-1}}$ can also be written as,

${{\Rightarrow }^{n}}{{C}_{r-1}}=\dfrac{n!}{(r-1)!(n-r+1)!}...............(2)$

Also, we know that

$\Rightarrow n!=n\times (n-1)!..................(3)$

Now, from the question, we have

${{\therefore }^{n}}{{C}_{r}}{{+}^{n}}{{C}_{r-1}}$

Substituting values from equation (1) and (2), we get

$\Rightarrow \dfrac{n!}{r!(n-r)!}+\dfrac{n!}{(r-1)!(n-r+1)!}$

Taking $\dfrac{n!}{(r-1)!(n-r)!}$ common from both terms, we get

$\Rightarrow \dfrac{n!}{(r-1)!(n-r)!}\left( \dfrac{1}{r}+\dfrac{1}{n-r+1} \right)$

Simplifying the above equation, we get

$\Rightarrow \dfrac{n!}{(r-1)!(n-r)!}\left( \dfrac{n-r+1+r}{r(n-r+1)} \right)$

Cancelling out $-r$ with $r$, we get

$\Rightarrow \dfrac{n!}{(r-1)!(n-r)!}\times \dfrac{n+1}{r(n-r+1)}$

Rearranging the terms like multiplying $(n+1)$ with $n!$, $r$ with $(r-1)!$ and $(n-r+1)$ with $(n-r)!$ and using (3), we get

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

Also the above expression can be written as

${{\Rightarrow }^{(n+1)}}{{C}_{r}}$

So, we get the value of $^{n}{{C}_{r}}{{+}^{n}}{{C}_{r-1}}{{=}^{(n+1)}}{{C}_{r}}$.