Answer:
1/21.
Step-by-step explanation:
There are 9 digits and the total number of permutations of 3 from 9 is 9P3
= 9!/6! = 504.
There are 4 even digits so the number of permutations of 3 from these 4 is 4! (4-3)! = 4*3*2 = 24.
So the required probability = 24/504
= 1/21.
The distance is 8 units
Hope this helps!!
Plz tell me if im right
Isn't 2,351 the standardized form already?
there is word form: two thousand, three hundred fifty one
there is expanded form: (2* 1000)+ (3*100)+ (5* 10)+ (1* 1)= 2,351
