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Question: If \(n\) and \(r\) are positive integers such as \(r < n\), then \(^{n}{{C}_{r}}{{+}^{n}}{{C}_{r-1}}...

If nn and rr are positive integers such as r<nr < n, then nCr+nCr−1=^{n}{{C}_{r}}{{+}^{n}}{{C}_{r-1}}=
A. 2nC2r−1^{2n}{{C}_{2r-1}}
B. (n+1)Cr^{\left( n+1 \right)}{{C}_{r}}
C. nCr+1^{n}{{C}_{r+1}}
D. (n+1)Cr+1^{\left( n+1 \right)}{{C}_{r+1}}

Explanation

Solution

In order to solve this question, we have to know about what nCr^{n}{{C}_{r}} stands for and then substitute nCr=n!r!(n−r)!^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!} in the equation and then solve it. After solving, rearrange the value that we got according to the options and get the answer.

Complete step by step answer:
We have to find the value of nCr+nCr−1^{n}{{C}_{r}}{{+}^{n}}{{C}_{r-1}}.
The meaning of nCr^{n}{{C}_{r}} in terms of permutation and combination is picking rr objects out of nn objects.
And in mathematical way nCr^{n}{{C}_{r}} can be represented as,
∴nCr=n!r!(n−r)!..........................(1){{\therefore }^{n}}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}..........................(1)
Similarly, nCr−1^{n}{{C}_{r-1}} can also be written as,
⇒nCr−1=n!(r−1)!(n−r+1)!...............(2){{\Rightarrow }^{n}}{{C}_{r-1}}=\dfrac{n!}{(r-1)!(n-r+1)!}...............(2)
Also, we know that
⇒n!=n×(n−1)!..................(3)\Rightarrow n!=n\times (n-1)!..................(3)
Now, from the question, we have
∴nCr+nCr−1{{\therefore }^{n}}{{C}_{r}}{{+}^{n}}{{C}_{r-1}}
Substituting values from equation (1) and (2), we get
⇒n!r!(n−r)!+n!(r−1)!(n−r+1)!\Rightarrow \dfrac{n!}{r!(n-r)!}+\dfrac{n!}{(r-1)!(n-r+1)!}
Taking n!(r−1)!(n−r)!\dfrac{n!}{(r-1)!(n-r)!} common from both terms, we get
⇒n!(r−1)!(n−r)!(1r+1n−r+1)\Rightarrow \dfrac{n!}{(r-1)!(n-r)!}\left( \dfrac{1}{r}+\dfrac{1}{n-r+1} \right)
Simplifying the above equation, we get
⇒n!(r−1)!(n−r)!(n−r+1+rr(n−r+1))\Rightarrow \dfrac{n!}{(r-1)!(n-r)!}\left( \dfrac{n-r+1+r}{r(n-r+1)} \right)
Cancelling out −r-r with rr, we get
⇒n!(r−1)!(n−r)!×n+1r(n−r+1)\Rightarrow \dfrac{n!}{(r-1)!(n-r)!}\times \dfrac{n+1}{r(n-r+1)}
Rearranging the terms like multiplying (n+1)(n+1) with n!n!, rr with (r−1)!(r-1)! and (n−r+1)(n-r+1) with (n−r)!(n-r)! and using (3), we get
⇒(n+1)!(r)!(n−r+1)!=(n+1)!(r)!(n+1−r)!\Rightarrow \dfrac{(n+1)!}{(r)!(n-r+1)!}=\dfrac{(n+1)!}{(r)!(n+1-r)!}
Also the above expression can be written as
⇒(n+1)Cr{{\Rightarrow }^{(n+1)}}{{C}_{r}}
So, we get the value of nCr+nCr−1=(n+1)Cr^{n}{{C}_{r}}{{+}^{n}}{{C}_{r-1}}{{=}^{(n+1)}}{{C}_{r}}.

So, the correct answer is “Option B”.

Note: This question tests the understanding of expression solving and rearranging knowledge of students. In these type of question students often do two mistakes, first one is that they put the value of nCr^{n}{{C}_{r}} and other terms in factorial and then solve it but they get struck at the last step i.e. converting back the final expression in nCr^{n}{{C}_{r}}, so they leave it. Second mistake it that they get confused in nCr^{n}{{C}_{r}} and nPr^{n}{{P}_{r}}, they mix these two things and end up getting wrong answer. So, carefully read the question and substitute the correct value and then rearrange the final expression to get the correct answer.