Hi could somebody explain me how the linked genes Aa and Cc

Hi, could somebody explain me how the linked genes A/a and C/c affect the probability in the following problem?

You are running a dihybrid cross involving 4 different genes. Three different chromosomes are involved and only the A/a and C/c genes are linked. With diploid parents AaBbccDd x AaBbcCdd (Ac and aC are linked in the second parent) and no crossing over, what are the odds of progeny with genotypes (a) aaBbccdd, (b) aabbCcDd, (c) AABbccdd?

Here are the solutions but I do not understand how they obtiened the calculations for A/a and C/c

Answers: (a) Can’t happen (second parent provides gametes Ac and aC only; (b) ¼ (bb part) * ½ (Dd) * ¼ (aaCc) = 1/32; (c) ½ * ½ * ¼ (same order as with (b)) = 1/16

Solution

For genotype aa Bbcc dd to occur the gamates from the parents should be a, B/b, c and d. Since a and c are linked as given in the question as Ac and aC the combination of ac does not exist in the provided genotype. Hence the genotype could not happen.

For genotype aabbCcDd, the probability for obtaining bb is 1/4; Because a cross between Bb and Bb will yield BB, bb, Bb and bB - Hence it is 1/4; Like wise for Dd (Dd x dd = Dd Dd dd and dd= 2/4 = 1/2) the probability is 1/2 and since a and C are linked it is 1/4 (Aa cc x AacC = AAcc; AaCc; aacC and aacc) = 1/4x1/2 x 1/4 = 1/32.

For genotype AA Bb cc dd = The probability for each gamate is 1/2 x 1/2 x 1/4 like the option B - Hence the answer is 1/16.