Prove that

To Prove: ${ }^{9} \mathrm{P}_{3}+3 \times{ }^{9} \mathrm{P}_{2}={ }^{10} \mathrm{P}_{3}$

Formula Used:

Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by,

${ }_{n} P_{r}=\frac{n !}{(n-r) !}$

The equation given below needs to be proved i.e

${ }^{9} P_{3}+3 \times{ }^{9} P_{2}={ }^{10} P_{3}$

$\frac{9 !}{(9-3) !}_{+}\left(3 \times \frac{9 !}{(9-2) !}\right)=\frac{10 !}{(10-3) !}$

$(9 \times 8 \times 7)+(3 \times 9 \times 8)=10 \times 9 \times 8$

$10 \times 9 \times 8=10 \times 9 \times 8$

Hence, proved.

${ }^{9} \mathrm{P}_{3}+3 \times{ }^{9} \mathrm{P}_{2}={ }^{10} \mathrm{P}_{3}$