Question

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

Answer 1

Hint: In this question it is given that we have to find the value of $${}^{n}C_{r+1}+\ ^{n} C_{r-1}+2\times \ ^{n} C_{r}$$. So for this we have to know that
$${}^{n}C_{p}+\ ^{n} C_{p-1}=\ ^{n+1} C_{p}$$.......(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,
$${}^{n}C_{r+1}+\ ^{n} C_{r-1}+2\times \ ^{n} C_{r}$$
$$= {}^{n}C_{r+1}+\ ^{n} C_{r-1}+\ ^{n} C_{r}+\ ^{n} C_{r}$$
$$= {}^{n}C_{r+1}+\ ^{n} C_{r}+\ ^{n} C_{r}+\ ^{n} C_{r-1}$$
$$= ({}^{n}C_{r+1}+\ ^{n} C_{r})+(\ ^{n} C_{r}+\ ^{n} C_{r-1})$$
$$= ({}^{n}C_{r+1}+\ ^{n} C_{\left( r+1\right) -1})+(\ ^{n} C_{r}+\ ^{n} C_{r-1})$$........(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,
$${}^{n}C_{r+1}+\ ^{n} C_{\left( r+1\right) -1}=\ ^{n+1} C_{r+1}$$.......(3)
And in the second term considering p=r, we get,
$${}^{n}C_{r}+\ ^{n} C_{r-1}=\ ^{n+1} C_{r}$$........(4)
Now by putting the values of (3) and (4) in (2), we get,
$$ ({}^{n}C_{r+1}+\ ^{n} C_{\left( r+1\right) -1})+(\ ^{n} C_{r}+\ ^{n} C_{r-1})$$
$$={}^{n+1}C_{r+1}+\ ^{n+1} C_{r}$$
$$={}^{n+1}C_{r+1}+\ ^{n+1} C_{(r+1)-1}$$
Now again applying formula (1) where p=r+1 and taking n+1 as n, we get,
$${}^{n+1}C_{r+1}+\ ^{n+1} C_{(r+1)-1}$$
$$={}^{\left( n+1\right) +1}C_{r+1}$$
$$={}^{n+2}C_{r+1}$$
Hence the correct option is option B.

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