Question

the number of permutations of 10 different things taken 4 at a time in which one particular thing never occurs is

Answer 1

Out of 10 different things one thing never occur then 4 things will be occur out of 9 things.
Therefore, 9p4=(9!)/(9-4)!
=(9*8*7*6)
=3024

Answer 2

Answer:

The number of permutations of 10 different things taken 4 at a time in which one particular thing never occurs is 3024

Step-by-step explanation:

Total no. of items = 10

No. of items to be taken at a time = 4

Now we are given that one particular thing never occurs

So, We will pick 4 items from 9

Formula : $^nP_r=\frac{n!}{(n-r)!}$

Substitute n = 9

r = 4

$^9 P_4=\frac{9!}{(9-4)!}=\frac{9!}{5!}=\frac{9 \times 8 \times 7 \times 6 \times 5!}{5!}=9 \times 8 \times 7 \times 6=3024$

Hence the number of permutations of 10 different things taken 4 at a time in which one particular thing never occurs is 3024 .