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Question: If \[{}^{n}C_{r}\] denotes the number of combinations of n things taken r at a time, then the expres...

If nCr{}^{n}C_{r} denotes the number of combinations of n things taken r at a time, then the expression nCr+1+ nCr−1+2× nCr{}^{n}C_{r+1}+\ ^{n} C_{r-1}+2\times \ ^{n} C_{r} equals
A. n+2Cr{}^{n+2}C_{r}
B. n+2Cr+1{}^{n+2}C_{r+1}
C. n+1Cr{}^{n+1}C_{r}
D. n+1Cr+1{}^{n+1}C_{r+1}

Explanation

Solution

Hint: In this question it is given that we have to find the value of nCr+1+ nCr−1+2× nCr{}^{n}C_{r+1}+\ ^{n} C_{r-1}+2\times \ ^{n} C_{r}. So for this we have to know that
nCp+ nCp−1= n+1Cp{}^{n}C_{p}+\ ^{n} C_{p-1}=\ ^{n+1} C_{p}.......(1)
n!=n⋅(n−1)!n!=n\cdot \left( n-1\right) !.........(2)
So by using this we are able to find the solution.

Complete step-by-step solution:
Here given,
nCr+1+ nCr−1+2× nCr{}^{n}C_{r+1}+\ ^{n} C_{r-1}+2\times \ ^{n} C_{r}
=nCr+1+ nCr−1+ nCr+ nCr= {}^{n}C_{r+1}+\ ^{n} C_{r-1}+\ ^{n} C_{r}+\ ^{n} C_{r}
=nCr+1+ nCr+ nCr+ nCr−1= {}^{n}C_{r+1}+\ ^{n} C_{r}+\ ^{n} C_{r}+\ ^{n} C_{r-1}
=(nCr+1+ nCr)+( nCr+ nCr−1)= ({}^{n}C_{r+1}+\ ^{n} C_{r})+(\ ^{n} C_{r}+\ ^{n} C_{r-1})
=(nCr+1+ nC(r+1)−1)+( nCr+ nCr−1)= ({}^{n}C_{r+1}+\ ^{n} C_{\left( r+1\right) -1})+(\ ^{n} C_{r}+\ ^{n} C_{r-1})........(2)
Now we are going to use the formula (1) in the above two terms.
In the first term considering p=r+1, then we get,
nCr+1+ nC(r+1)−1= n+1Cr+1{}^{n}C_{r+1}+\ ^{n} C_{\left( r+1\right) -1}=\ ^{n+1} C_{r+1}.......(3)
And in the second term considering p=r, we get,
nCr+ nCr−1= n+1Cr{}^{n}C_{r}+\ ^{n} C_{r-1}=\ ^{n+1} C_{r}........(4)
Now by putting the values of (3) and (4) in (2), we get,
(nCr+1+ nC(r+1)−1)+( nCr+ nCr−1)({}^{n}C_{r+1}+\ ^{n} C_{\left( r+1\right) -1})+(\ ^{n} C_{r}+\ ^{n} C_{r-1})
=n+1Cr+1+ n+1Cr={}^{n+1}C_{r+1}+\ ^{n+1} C_{r}
=n+1Cr+1+ n+1C(r+1)−1={}^{n+1}C_{r+1}+\ ^{n+1} C_{(r+1)-1}
Now again applying formula (1) where p=r+1 and taking n+1 as n, we get,
n+1Cr+1+ n+1C(r+1)−1{}^{n+1}C_{r+1}+\ ^{n+1} C_{(r+1)-1}
=(n+1)+1Cr+1={}^{\left( n+1\right) +1}C_{r+1}
=n+2Cr+1={}^{n+2}C_{r+1}
Hence the correct option is option B.

Note: You might be thinking how we obtained
nCp+ nCp−1= n+1Cp{}^{n}C_{p}+\ ^{n} C_{p-1}=\ ^{n+1} C_{p}.
So let us try to establish this formula,
LHS,
nCp+ nCp−1{}^{n}C_{p}+\ ^{n} C_{p-1}
=n!p!⋅(n−p)!+n!(p−1)!⋅(n−p+1)!=\dfrac{n!}{p!\cdot \left( n-p\right) !} +\dfrac{n!}{\left( p-1\right) !\cdot \left( n-p+1\right) !}
=n!p!⋅(n−p)!+n!⋅p(p−1)!⋅p⋅(n−p+1)!=\dfrac{n!}{p!\cdot \left( n-p\right) !} +\dfrac{n!\cdot p}{\left( p-1\right) !\cdot p\cdot \left( n-p+1\right) !} [multiplying p on the both side of the second fraction]
=n!p!⋅(n−p)!+n!⋅pp⋅(p−1)!(n−p+1)!=\dfrac{n!}{p!\cdot \left( n-p\right) !} +\dfrac{n!\cdot p}{p\cdot \left( p-1\right) !\left( n-p+1\right) !}
=n!p!⋅(n−p)!+n!⋅pp!(n−p+1)!=\dfrac{n!}{p!\cdot \left( n-p\right) !} +\dfrac{n!\cdot p}{p!\left( n-p+1\right) !}
=n!p!⋅(n−p)!+n!⋅pp!(n−p+1)⋅(n−p)!=\dfrac{n!}{p!\cdot \left( n-p\right) !} +\dfrac{n!\cdot p}{p!\left( n-p+1\right) \cdot \left( n-p\right) !} [since, n!=n(n-1)!]
=n!p!⋅(n−p)!+n!⋅pp!(n−p)!⋅(n−p+1)=\dfrac{n!}{p!\cdot \left( n-p\right) !} +\dfrac{n!\cdot p}{p!\left( n-p\right) !\cdot \left( n-p+1\right) }
=n!p!⋅(n−p)!(1+pn−p+1)=\dfrac{n!}{p!\cdot \left( n-p\right) !} \left( 1+\dfrac{p}{n-p+1} \right) [by taking common]
=n!p!⋅(n−p)!(n−p+1+pn−p+1)=\dfrac{n!}{p!\cdot \left( n-p\right) !} \left( \dfrac{n-p+1+p}{n-p+1} \right)
=n!p!⋅(n−p)!(n+1n−p+1)=\dfrac{n!}{p!\cdot \left( n-p\right) !} \left( \dfrac{n+1}{n-p+1} \right)
=n!⋅(n+1)p!⋅(n−p)!⋅(n−p+1)=\dfrac{n!\cdot \left( n+1\right) }{p!\cdot \left( n-p\right) !\cdot \left( n-p+1\right) }
=(n+1)!p!⋅(n−p)!⋅(n+1−p)=\dfrac{\left( n+1\right) !}{p!\cdot \left( n-p\right) !\cdot \left( n+1-p\right) }
=(n+1)!p!⋅(n−p)!⋅(n+1−p)=\dfrac{\left( n+1\right) !}{p!\cdot \left( n-p\right) !\cdot \left( n+1-p\right) }
= n+1Cp=\ ^{n+1} C_{p} =RHS.