We need to take values for this questions.
1. y = 2

Option A: \(\frac{y!}{5(y-2)!} = \frac{2!}{5 \times 0!} = \frac{2}{5} = 0.4\)

Option B: \(\frac{(y+1)!}{3(y-1)!} = \frac{3!}{3 \times 2!} = \frac{6}{6} = 1\)

Option A < Option B

2. y = 5

Option A: \(\frac{y!}{5(y-2)!} = \frac{5!}{5 \times 3!} = \frac{120}{30} = 4\)

Option B: \(\frac{(y+1)!}{3(y-1)!} = \frac{6!}{3 \times 4!} = \frac{720}{72} = 10\)

Option A < Option B

We can see that as we keep increasing the value of y, the gap between Option A and Option B will keep increasing and Option A will always be smaller than Option B.

OA, B