Question

The number of signals that can be made with 3 flags each of different colour by hoisting 1 or 2 or 3 above the other is:
A) $3$
B) $7$
C) $15$
D) $16$

Answer 1

We will use the method of permutations to calculate the total number of signals given by the formula: ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$ where n is the total number of flags and r is the number of flags to be chosen (in this case). Here, we can either consider only 1 flag to hoist or 2 or all three of them. So, we will add up all the possible ways to give signals by calculating the number of ways individually for 1 or 2 or 3 flags’ arrangement.

Complete step by step solution:

We are given that there are 3 flags and we are required to calculate the number of signals that can be produced by hoisting the flags either 1 or 2 or 3 above the other flag.

Now, we can observe that there are a total of three cases to calculate the possible ways.

First of all, we can either hoist any one flag or we can hoist 2 of them or all three flags.

Considering one flag to give a signal, we can say that any of the three colours can be used for the same.

So, we can say that we need to choose 1 flag out of 3. So, the number of ways to do so will be ${}^3{P_1}$ .

$\Rightarrow {}^3{P_1} = \dfrac{{3!}}{{\left( {3 - 1} \right)!}} = \dfrac{{3!}}{{2!}} = \dfrac{{3 \times 2 \times 1}}{{2 \times 1}} = 3$

So, there are 3 ways to choose any one of the flags to give signal.

Now, for any two flags, we can say that we need to choose any two out of three flags.

So, the possible ways are ${}^3{P_2} = \dfrac{{3!}}{{\left( {3 - 2} \right)!}} = \dfrac{{3!}}{{1!}} = \dfrac{{3 \times 2 \times 1}}{1} = 6$

Therefore, there are 6 ways to signal by hoisting 2 flags.

Now, for giving signal using 3 flags can be given by ${}^3{P_3}$.

$\Rightarrow {}^3{P_3} = \dfrac{{3!}}{{\left( {3 - 3} \right)!}} = \dfrac{{3!}}{{0!}} = \dfrac{{3 \times 2 \times 1}}{1} = 6$

Therefore, there are 6 ways to hoist 3 flags above each other to give signal.

So, we can say that the total possible ways to signal using 3 different flags can be calculated by adding total ways to signal by individual flags (1 at a time + 2 at a time + all 3 at a time).

$\Rightarrow$ Total possible ways = $3 + 6 + 6 = 15$

So, there are a total of 15 different ways.

Hence, option (C) is correct.

Note:

In this question, you may go wrong in the final step where we added all the individual ways to signal as in the question it was mentioned either 1 or 2 or 3 flags above the other. If it was asked 1 and 2 and 3 flags, then we should have multiplied all the possible individual ways to calculate the number of signals that can be made using 3 different flags.