Definitions first. In Mason's terminology a forward path runs from source to sink touching no node more than once; a feedback loop is a closed path returning to its starting node; two loops touch if they share at least one node or branch.

Step 1 — count the forward paths, A. Tracing every route from R to C without repeating a node gives four distinct paths: the upper route through G₁ and G₂ and then G₃; the lower route through G₄ and the unity branch; the route that enters through G₁ and crosses on the upper diagonal into the lower half; and the route that crosses on the other diagonal from the upper half down into G₄. So

\(A=4\)

Step 2 — count the feedback loops, B. The graph has exactly two closed paths, one in each half:

\(L_1=-G_2H_1 \qquad\text{(upper pair)}\)

\(L_2=-G_4H_2 \qquad\text{(lower pair)}\)

No other closed path exists, because the crossing diagonals both run downward — there is no branch carrying a signal back up from the lower half to the upper one, so nothing closes across the two halves. Hence

\(B=2\)

Step 3 — count the touching loops, C. L₁ lives entirely on the two upper nodes and L₂ entirely on the two lower nodes. They share no node and no branch, so they are non-touching, and the number of touching loops is

\(C=0\)

Step 4 — order them.

\(A=4 \gt B=2 \gt C=0\)

which is option 3.

The check through Mason's formula. Because the two loops are non-touching, the determinant carries their product term:

\(\Delta=1-(L_1+L_2)+L_1L_2=1+G_2H_1+G_4H_2+G_2H_1G_4H_2\)

and the overall gain is \(T=\dfrac{1}{\Delta}\sum_k P_k\Delta_k\). The presence of that \(L_1L_2\) term is itself the confirmation that no loops touch — if they did, the product would be absent and ∆ would reduce to \(1+G_2H_1+G_4H_2\).

Hence, the decreasing order is A > B > C.