# Math Help - 4 balls & 4 cases & probability

1. ## 4 balls & 4 cases & probability

Hello,

i did not anderstand this exercise.
May you help me ?

We have 4 balls numbered 1,2,3,4 and a box of 4 cases.

a) what is the number of possibility to arrange the balls in the box in order to put one ball per case?
b)find this possibility : 2 boxes allied never contain 2 number allied.

2. (a) This is the same as the number of possible arrangements of the numbers 1, 2, 3, and 4.

(b) Since there are few possibilities, you can count them and verify that the only two are 3, 1, 4, 2 and 2, 4, 1, 3.

3. hello,

thanks you !

Is it right if i do that ?

4. Yes, your method is correct. You can also calculate that number this way: There are four choices for the first number, three choices for the second number, two choices for the third number, and one choice for the fourth number. Hence the total number of possibilities is

$4 \cdot 3 \cdot 2 \cdot 1 = 4! = 24$.

5. Ok.