Question

In how many of permutations of $10$ things taken $4$ at a time will (i) one thing always occurs, (ii) never occur?

Answer 1

(i) Total permutation of n different things taken r at a time when a particular item is always included in the arrangement is

$r{\times }^{n-1}{P}_{r-1}=4{\times }^9{P}_3$

$=4\times 9\times 8\times 7$

$=2016$

(ii) Out of 10 different thing one thing never occur then 4 things will be occur out of 9 things

therefore ${}^9{P}_4=\frac{9!}{(9-4!)}$

$=9\times 8\times 7\times 6$

$=3024$