Question
If $$P_n$$ denotes the product of the binomial coefficients in the expansion of $${\left( {1 + x} \right)^n},\,$$ then $$\frac{{{P_{n + 1}}}}{{{P_n}}}$$ equals
A.
$$\frac{{n + 1}}{{n!}}$$
B.
$$\frac{{n }}{{n!}}$$
C.
$$\frac{{{{\left( {n + 1} \right)}^{n }}}}{{\left( {n + 1} \right)!}}$$
D.
$$\frac{{{{\left( {n + 1} \right)}^{n + 1}}}}{{\left( {n + 1} \right)!}}$$
Answer :
$$\frac{{{{\left( {n + 1} \right)}^{n + 1}}}}{{\left( {n + 1} \right)!}}$$
Solution :
$$\eqalign{
& {\text{Here, }}{P_n} = {\,^n}{C_0} \cdot {\,^n}{C_1} \cdot {\,^n}{C_2}.....{\,^n}{C_n} \cr
& {\text{and }}{P_{n + 1}} = {\,^{n + 1}}{C_0} \cdot {\,^{n + 1}}{C_1} \cdot {\,^{n + 1}}{C_2}.....{\,^{n + 1}}{C_{n + 1}} \cr
& \therefore \frac{{{P_{n + 1}}}}{{{P_n}}} = \frac{{^{n + 1}{C_0} \cdot {\,^{n + 1}}{C_1} \cdot {\,^{n + 2}}{C_2}.....{\,^{n + 1}}{C_{n + 1}}}}{{^n{C_0} \cdot {\,^n}{C_1} \cdot {\,^n}{C_2}.....{\,^n}{C_n}}} \cr
& = \left( {\frac{{^{n + 1}{C_1}}}{{^n{C_0}}}} \right)\left( {\frac{{^{n + 1}{C_2}}}{{^n{C_1}}}} \right)\left( {\frac{{^{n + 1}{C_3}}}{{^n{C_0}}}} \right).....\left( {\frac{{^{n + 1}{C_{n + 1}}}}{{^n{C_n}}}} \right) \cr
& = \left( {\frac{{n + 1}}{1}} \right)\left( {\frac{{n + 1}}{2}} \right)\left( {\frac{{n + 1}}{3}} \right).....\left( {\frac{{n + 1}}{{n + 1}}} \right) \cr
& = \frac{{{{\left( {n + 1} \right)}^{n + 1}}}}{{\left( {n + 1} \right)!}} \cr} $$