The new ointment was applied to four locations, and a control was applied to the other four. How many different choices were there for the four locations to apply the new ointment?

there are 70 possible choices for the four locations to apply the new ointment

Step-by-step explanation:

Since we have a total of 8 locations ( 4 to the new ointment and 4 to the control) ,  each one can be chosen and since the order of the locations that are chosen for the new ointment is not relevant  , then we know that the number of choices is given by the number of combinations of 4 elements in 8

number of combinations = 8 possible locations to the first ointment * 7 possible locations to the second ( since the first one was already located) * 6 to the third * 5 locations for the fourth / number of times the same combination is repeated ( the same locations but in different positions) = 8*7*6*5 / (4 possible positions for the first ointment* 3 possible positions to the second ointment (since the first one was already located * 2 possible positions of the third * 1 possible position of the fourth)

therefore

number of combinations = 8*7*6*5/(4*3*2*1 ) = 8!/((8-4)!*4!) = 70 possible combinations

thus there are 70 possible choices for the four locations to apply the new ointment

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length