1. Calculate the total number of ways to choose 7 hikers from 10 hikers without restrictions

$^{10}C_7 = 120$

1. Identify the possible cases for having at least 4 hikers from team $C$ and set up the expression for the total number of combinations.

Since there are 5 hikers in team $C$ and 7 hikers needed in total, there are two cases:
Case 1: Exactly 4 from $C$, and $7 - 4 = 3$ from the rest (teams $A$ and $B$, which have $2+3=5$ hikers in total).
Number of ways $= {}^{5}C_{4} \times {}^{5}C_{3}$

Case 2: Exactly 5 from $C$, and $7 - 5 = 2$ from the rest.
Number of ways $= {}^{5}C_{5} \times {}^{5}C_{2}$

Total number of ways $= {}^{5}C_{4} \times {}^{5}C_{3} + {}^{5}C_{5} \times {}^{5}C_{2}$

2. Evaluate the combinations to find the total number of ways.

$= (5 \times 10) + (1 \times 10)$
$= 50 + 10$
$= 60$

1. State the unrestricted total and formulate the complement method

Total unrestricted ways to choose 7 hikers from 10 is $\binom{10}{7} = 120$.
To ensure at least one hiker from each team is chosen, we subtract the cases where team A, team B, or team C is completely missing from the total.

2. Calculate the combinations where at least one team is missing

If team A is missing, the 7 hikers must be chosen from the remaining $3 + 5 = 8$ hikers (teams B and C):
$\text{Ways missing team A} = \binom{8}{7} = 8$

If team B is missing, the 7 hikers must be chosen from the remaining $2 + 5 = 7$ hikers (teams A and C):
$\text{Ways missing team B} = \binom{7}{7} = 1$

If team C is missing, the 7 hikers would need to be chosen from teams A and B, which only have $2 + 3 = 5$ hikers. This is impossible ($\binom{5}{7} = 0$).

(Note: Missing two teams simultaneously is also impossible as the largest single team only has 5 hikers.)

3. Subtract the invalid cases from the total to find the final answer

$\text{Valid ways} = 120 - 8 - 1 = 111$