Question

How many different ways are there to paint 'n' balls with 4 colors [A,B,C,D] the number of balls in color A must be odd. The number of balls in color B must be even. [includes 0]

Answer 1

Answer:

There are infinitely many ways. You increase the value of n, the ways to color also increases.

Step-by-step explanation:

Plz mark this as brainliest if it helped you

Answer 2

Answer: This is referred to as the coupon collector's dilemma.

The average number of attempts required to view each of the four hues is four.

$44+43+42+41=813$

The terms are $4/4$ for the time to take one ball; then $4/3$ for the average time it takes after the first ball until you see another code; $4/2$ for the average time it takes after the first time you see the second color until you see a third, and finally $4/1$ for the time you then have to wait until you see the last color. (These can just be added due to the additivity of expectations).  

#SPJ2