Question

How many different flags can be made by hoisting 6 differently coloured flags one above the other, when any number of them may be hoisted at once?

Answer 1

To solve this question, we will use the concept of permutation. We will solve for all cases and find the number of ways for hoisting different numbers of flags out of 6 flags and then we will add all 6 cases to get all numbers of signals.

Complete step-by-step answer:
Before we solve this question, let us see what the meaning of permutation and factorial function is
Permutation is an arrangement of members into a sequence or linear order that is we just rearrange the place or order of elements.
Permutation is denoted by $^{n}{{P}_{r}}$ , where n are total objects and r are numbers of objects taken at a time and is equals to $\dfrac{n!}{\left( n-r \right)!}$ .
Factorial function is denoted as x! and is calculated as $x!=x(x-1)(x-2)......3.2.1$

Now, in question it is given that we have 6 differently coloured flags which are being hosted one above the other and also any number of flags can be hoisted at once, which means at once, 1 or 2 or 3 or 4 or 5 or 6 flags can be hoisted.
So, number of ways for hoisting 6 different flags out of 6 given flags will be equals to $^{6}{{P}_{6}}$, which is equals to $\dfrac{6!}{\left( 6-6 \right)!}$
On solving, we get
Number of ways for hoisting 6 different flags out of 6 given flags = 720 ways
number of ways for hoisting 5 different flags out of 6 given flags will be equals to $^{6}{{P}_{5}}$, which is equals to $\dfrac{6!}{\left( 6-5 \right)!}$
On solving, we get
Number of ways for hoisting 5 different flags out of 6 given flags = 720 ways

number of ways for hoisting 4 different flags out of 6 given flags will be equals to $^{6}{{P}_{4}}$, which is equals to $\dfrac{6!}{\left( 6-4 \right)!}$
On solving, we get
Number of ways for hoisting 4 different flags out of 6 given flags = 360 ways

number of ways for hoisting 3 different flags out of 6 given flags will be equals to $^{6}{{P}_{3}}$, which is equals to $\dfrac{6!}{\left( 6-3 \right)!}$
On solving, we get
Number of ways for hoisting 3 different flags out of 6 given flags = 120 ways

number of ways for hoisting 2 different flags out of 6 given flags will be equals to $^{6}{{P}_{2}}$, which is equals to $\dfrac{6!}{\left( 6-2 \right)!}$
On solving, we get
Number of ways for hoisting 2 different flags out of 6 given flags = 30 ways

number of ways for hoisting 1 different flags out of 6 given flags will be equals to $^{6}{{P}_{1}}$, which is equals to $\dfrac{6!}{\left( 6-1 \right)!}$
On solving, we get
Number of ways for hoisting 1 different flags out of 6 given flags = 6 ways
So, total number of signals will be equals to = 720 + 720 + 360 + 120 + 30 + 6 = 1956
Hence, we have total 1956 ways to hoist 6 differently coloured flags one above the other, when any number of them may be hoisted at once

Note: Always remember that permutation is rearranging order of elements and permutation is denoted by $^{n}{{P}_{r}}$, where n are total objects and r are numbers of objects taken at a time and is equals to $\dfrac{n!}{\left( n-r \right)!}$. Evaluate each and every case of different numbers of flags hoisted at once carefully. Try not make any calculation mistakes.