A new airlines company is planning to start operations in a country. The company has identified ten different cities which they plan to connect through their network to start with. The flight duration between any pair of cities will be less than one hour. To start operations, the company has to decide on a daily schedule.

The underlying principle that they are working on is the following:

Any person staying in any of these 10 cities should be able to make a trip to any other city in the morning and should be able to return by the evening of the same day.

Suppose three of the ten cities are to be developed as hubs. A hub is a city which is connected with every other city by direct flights each way, both in the morning as well as in the evening. The only direct flights which will be scheduled are originating and/or terminating in one of the hubs. Then the minimum number of direct flights that need to be scheduled so that the underlying principle of the airline to serve all the ten cities is met without visiting more than one hub during one trip is:

The correct option is C.

Let the total number of cities be 10, out of which 3 cities are developed as hubs, and the remaining 7 cities are non-hubs.

Let the hubs be denoted as $H_1,H_2,H_3$ and the non-hubs as $N_1,N_2,...,N_7$.

We analyze the two required types of travel routes:
1. Hub to Hub flights:
Since there are 3 hubs, each hub must be connected to every other hub. For any pair of hubs, say $H_i$ and $H_j$:
• There must be a direct flight from $H_i$ to $H_j$ in the morning and from $H_j$ to $H_i$ in the evening.
• Conversely, there must be a direct flight from $H_j$ to $H_i$ in the morning and from $H_i$ to $H_j$ in the evening.
Thus, each pair of hubs requires 4 direct flights. With 3 hubs, the number of pairs is $C_2^3=3$ pairs.
Total flights between hubs = $3\times 4=12$ flights.

2. Non-hub to Hub flights:
For any non-hub city $N_a$ to reach any other non-hub city $N_b$ without visiting more than one hub during the trip, the route must be:
• In the morning: $N_a\to H_x\to N_b$
• In the evening: $N_b\to H_y\to N_a$
where $H_x$ and $H_y$ are hubs (which could be the same hub or different hubs).
For this to work for any pair of non-hubs, every non-hub city must have a morning flight to all hubs and an evening flight from all hubs, and also a morning flight from all hubs and an evening flight to all hubs.
Specifically, for each non-hub city $N_a$ and each hub $H_x$, we need:
• Morning: $N_a\to H_x$ and $H_x\to N_a$
• Evening: $H_x\to N_a$ and $N_a\to H_x$
This means we need 4 flights for every (non-hub, hub) pair.
With 7 non-hubs and 3 hubs, there are $7\times 3=21$ pairs.
Total flights between non-hubs and hubs = $21\times 4=84$ flights.

3. Total scheduled flights:
Summing the two types of flights gives:
Total = $12+84=96$ flights.