The question asks us to determine the total quantity of different signals that can be created by arranging flags on a vertical staff. We are given 5 different flags, and the condition is that at least 2 flags must be used in sequence (one below the other) to form a signal.

This problem involves the concept of permutations because the order in which the flags are arranged matters. For example, a signal with flag A on top and flag B below it is different from a signal with flag B on top and flag A below it.

Flags and Permutations Explained

In this scenario, we have a total of 5 different flags available. We need to create signals by organizing these flags in a sequence on a vertical staff. The key condition is "at least 2 flags," which means we can use 2 flags, 3 flags, 4 flags, or all 5 flags to form a signal.

The formula for permutations of 'n' distinct items taken 'r' at a time is given by:

$$ P(n, r) = \frac{n!}{(n-r)!} $$

Where:

• $\mathbf{n}$ is the total number of distinct items available (here, 5 different flags).
• $\mathbf{r}$ is the number of items being chosen and arranged (here, the number of flags used in a signal).
• $\mathbf{!}$ denotes the factorial function, e.g., $5! = 5 \times 4 \times 3 \times 2 \times 1$.

Calculating Signals for Each Case

We need to calculate the number of possible signals for each case based on the number of flags used (r) and then sum these quantities to find the total number of different signals.

Signals Using Exactly 2 Flags

In this case, we choose and arrange 2 flags out of the 5 available flags. Here, $n=5$ and $r=2$.

$$ P(5, 2) = \frac{5!}{(5-2)!} = \frac{5!}{3!} $$

$$ P(5, 2) = \frac{5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1} = 5 \times 4 = 20 $$

So, there are 20 different signals that can be created using exactly 2 flags.

Signals Using Exactly 3 Flags

Next, we consider using exactly 3 flags out of the 5 available. Here, $n=5$ and $r=3$.

$$ P(5, 3) = \frac{5!}{(5-3)!} = \frac{5!}{2!} $$

$$ P(5, 3) = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = 5 \times 4 \times 3 = 60 $$

Thus, 60 different signals can be created using exactly 3 flags.

Signals Using Exactly 4 Flags

Now, let's calculate the signals using exactly 4 flags out of the 5 available. Here, $n=5$ and $r=4$.

$$ P(5, 4) = \frac{5!}{(5-4)!} = \frac{5!}{1!} $$

Since $1! = 1$, we have:

$$ P(5, 4) = \frac{5 \times 4 \times 3 \times 2 \times 1}{1} = 120 $$

This means there are 120 different signals that can be created using exactly 4 flags.

Signals Using Exactly 5 Flags

Finally, we calculate the signals using all 5 flags available. Here, $n=5$ and $r=5$.

$$ P(5, 5) = \frac{5!}{(5-5)!} = \frac{5!}{0!} $$

It's important to remember that $0! = 1$. So:

$$ P(5, 5) = \frac{5 \times 4 \times 3 \times 2 \times 1}{1} = 120 $$

Therefore, 120 different signals can be created using all 5 flags.

Total Quantity of Different Signals

To find the total quantity of different signals, we sum the number of signals from each case where "at least 2 flags" are used:

Total Signals = Signals with 2 flags + Signals with 3 flags + Signals with 4 flags + Signals with 5 flags

Total Signals = $20 + 60 + 120 + 120 = 320$

Thus, the total quantity of different signals that can be created is 320.