Answer:
- 220 options, 84 options, 20 options, 1 option
Step-by-step explanation:
<h3>The first dinner</h3>
Combinations of 3 out of 12 items
- 12C3 = 12!/(12-3)!3! = 12!/9!3! = 12*11*10/2*3 = 220
<h3>The second dinner</h3>
Combinations 3 out of remaining 9 items
- 9C3 = 9!/(9 - 3)!3! = 9!/6!3! = 9*8*7/2*3 = 84
<h3>The third dinner</h3>
Combinations of 3 out of remaining 6 items
- 6C3 = 6!/(6 - 3)!3! = 6!/3!3! = 6*5*4/2*3 = 20
<h3>The fourth dinner</h3>
Combinations of 3 out of 3 remaining items
- 3C3 = 3!/(3 - 3)!3! = 3!/0!3! = 1
Answer:
Step-by-step explanation:
The picture of the question in the attached figure
we know that
In the right triangle EFG
----> by TOA (opposite side divided by the adjacent side)
substitute
X false -10 cg(centigrams)+8x-3
