Question
If $$\pi \left( n \right)$$ denotes product of all binomial coefficients in $${\left( {1 + x} \right)^n},$$ then ratio of $$\pi \left( {2002} \right)$$ to $$\pi \left( {2001} \right)$$ is
A.
$$2002$$
B.
$$\frac{{{{\left( {2002} \right)}^{2001}}}}{{\left( {2001} \right)!}}$$
C.
$$\frac{{{{\left( {2001} \right)}^{2002}}}}{{\left( {2002} \right)!}}$$
D.
$$2001$$
Answer :
$$\frac{{{{\left( {2002} \right)}^{2001}}}}{{\left( {2001} \right)!}}$$
Solution :
$$\eqalign{
& \frac{{\pi \left( n \right)}}{{\pi \left( {n + 1} \right)}} = \frac{{^n{C_0} \cdot {\,^n}{C_1} \cdot {\,^n}{C_2}{{.....}^n}{C_n}}}{{^{n + 1}{C_0} \cdot {\,^{n + 1}}{C_1} \cdot {\,^{n + 1}}{C_2}{{.....}^{n + 1}}{C_{n + 1}}}} \cr
& = \frac{1}{{^{n + 1}{C_0}}}\left( {\frac{{^n{C_0}}}{{^{n + 1}{C_1}}}} \right)\left( {\frac{{^n{C_1}}}{{^{n + 1}{C_2}}}} \right).....\left( {\frac{{^n{C_n}}}{{^{n + 1}{C_{n + 1}}}}} \right) \cr
& = \frac{1}{1}\left( {\frac{1}{{n + 1}}} \right)\left( {\frac{2}{{n + 1}}} \right).....\left( {\frac{{n + 1}}{{n + 1}}} \right)\left[ {\because \frac{{^n{C_r}}}{{^{n + 1}{C_{r + 1}}}} = \frac{{r + 1}}{{n + 1}}} \right] \cr
& = \frac{{\left( {n + 1} \right)!}}{{{{\left( {n + 1} \right)}^{n + 1}}}} = \frac{{n!}}{{{{\left( {n + 1} \right)}^n}}} \cr
& \therefore \frac{{\pi \left( {2002} \right)}}{{\pi \left( {2001} \right)}} = \frac{{{{\left( {2002} \right)}^{2001}}}}{{\left( {2001} \right)!}} \cr} $$