Question

Find the number of different signals that can be generated by arranging at least 2 flags in order (one below the other) on a vertical staff, if five different flags are available.

Answer 1

Hint: To solve this question, we have to consider different cases in which we are taking different numbers of flags to arrange. Also, to select r things from n objects given, (where each thing is considered different). We do it by the formula $\Rightarrow {}^{n}{{C}_{r}}$ .

Complete step-by-step answer:
So, let us start taking up the cases considering different numbers of flags.
Case I: - When there are only two flags, the number of ways we can arrange the flag at first position is ${}^{5}{{C}_{1}}$ . Now, coming to second position we have only four flags left. So, at second position there can be ${}^{4}{{C}_{1}}$ choices. So, total ways are: - ${}^{5}{{C}_{1}}\times {}^{4}{{C}_{1}}=5\times 4=20$
Case II: - When there are three flags, the first position has 5 choices. At second position, there are 4 choices and third position there are 3 choices. So, total number of arrangements are $=5\times 4\times 3=60$
Case III: - When the number of flags are four, the first position has 5 choices. Now after putting the flag there, four flags are left. Out of four flags one flag is placed at second position. It is done by ${}^{4}{{C}_{1}}=4$ ways. Now only three flags are left. At third position, one flag is arranged is ${}^{3}{{C}_{1}}$ ways. Now at fourth position, out of 2 flags left we have to put 1 flag. It is done in ${}^{2}{{C}_{1}}$ ways. So, the total number of ways to arrange flags are: -
${}^{5}{{C}_{1}}\times {}^{4}{{C}_{1}}\times {}^{3}{{C}_{1}}\times {}^{2}{{C}_{1}}=120$
Case IV: - We have to arrange five flags in five positions. This arrangement can be done in 5! Ways. So, total ways $=120$ in this case.
Now, to get the overall number of ways, we will add the number of ways from each case. Thus, total ways
$=20+60+120+120$
$=320$

Note: The shortcut method of arranging these flags is given by ${}^{n}{{P}_{r}}$ where ${}^{n}{{P}_{r}}={}^{n}{{C}_{r}}\times r!$ . So, the total ways of arranging two flags is ${}^{5}{{P}_{2}}$ , three flags is ${}^{5}{{P}_{3}}$ and so on.