Question

the number of ways in which we can select 3 numbers from 1 to 30 so as to exclude every selection of all even numbers​

Answer 1

Answer:

3605

Step-by-step explanation:

from 1 to 30, there are 15 even numbers. The total number of combinations of three even numbers out of 15 even numbers is: ¹⁵C₃ = $\frac{15!}{(15 - 3)!*3!}$

= 455

Now number of ways of choosing any 3 numbers out of 30 numbers is: ³⁰C₃

=  $\frac{30!}{(30 - 3)!*3!}$ = 4060

Since we need to avoid all even possibilities, we subtract 455 from 4060.

Therefore, answer is 3605

Hope it helps!

Forgive any silly mistakes :P

P.S. I see that you too are part of the anime culture :D

Answer 2

Step-by-step explanation:

3605 correct as above explanation is excellent do go or search for other answer thanks you make above one as brainlist and me to