Question:
Prove that $1+1 .{ }^{1} P_{1}+2 .{ }^{2} P_{2}+3 .{ }^{3} P_{3}+\ldots . n .{ }^{n} P_{n}={ }^{n+1} P_{n+1}$
Solution:
To Prove: $1+1 .{ }^{1} \mathrm{P}_{1}+2 .{ }^{2} \mathrm{P}_{2}+3 .{ }^{3} \mathrm{P}_{3}+\ldots . . \mathrm{n} .{ }^{n} \mathrm{P}_{\mathrm{n}}={ }^{\mathrm{n}+1} \mathrm{P}_{\mathrm{n}+1}$
Formula Used:
Total number of ways in which $\mathrm{n}$ objects can be arranged in $\mathrm{r}$ places (Such that no object is replaced) is given by,
${ }_{n}{ }_{P_{r}}=\frac{n !}{(n-r) !}$
$1+1 .{ }^{1} P_{1}+2 .{ }^{2} P_{2}+3 .{ }^{3} P_{3}+\ldots . . n .{ }^{n} P_{n}={ }^{n+1} P_{n+1}$
$1+(2 !-1 !)+(3 !-2 !)+(4 !-3 !)+\ldots \ldots((n+1) !-n !)=(n+1) !$
$1+((n+1) !-1 !)=(n+1) !$
$(n+1) !=(n+1) !$
Hence proved.