0.2 is rounded to the nearest tenth
Solution-
First 9 letters of alphabet are- A,B,C,D,E,F,G,H,I
Total number of ways of selection of 4 letters from 9 alphabets = 9C4
=9!÷((4!).(9-4)!) = 9!÷(4!×5!) = (9×8×7×6)÷(24) = 126
The number of ways of arranging these 4 numbers = 4! = 24
∴ Total number of possible permutations = 9C4×4! = 126×24 = 3024
∴ option number 2 is correct.
Answer:
Hey!
Step-by-step explanation:
Your answer is 2409725.0
Best of luck ♥
Answer:
1)B
2)C
3)B
4)A
5)C
Step-by-step explanation:
BrainLiest please
Probably A I am not really sure I am trying sorry if it is wrong
