Question

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

Answer 1

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 : nPr = n!/n–r

Substitute n = 9

r = 4

9P4 = 9!/(9-4)! = 9×8×7×6×5!/5! = 9×8×7×6 = 3024

=9×8×7×6=3024

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

Answer 2

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

$~~~~~~~~~~\mapsto\sf r× ^{n-1} P_{r-1} = 4× ⁹ P_3$

${~~~~~~~~~~\sf \quad\quad \quad\quad\quad\: = 4 \times 9 \times 8 \times 7}$

${~~~~~~~~~~\sf \quad\quad \quad\quad\quad\: = 2016}$

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

$\: \: \: \: \: \mathcal { ~~~~~\therefore~~~~~⁹P_4 = \dfrac{9!}{(9-4!)}}$

$\: \: \: \: \: \mathcal { ~~~~~~~~~~ \quad \quad = 9 \times 8 \times 7 \times 6}$

$\: \: \: \: \: \mathcal { ~~~~~~~~~~ \quad \quad = 3024}$