BdMO Online Forum

Problem 8:
At a conference, flags of $5$ countries are to be hoisted on $2$ poles so that no pole is left empty and all the flags get hoisted. More than one flag must not be placed at the same height of the same pole. Rather, they can be placed above or below an already placed flag. In this manner, more than one flag can be placed serially on a single pole. How many ways can the flags be hoisted?

5 flag can be arranged in two poles having no empty poles in 4 ways.and 5 flages arranged in 5! ways.(how).so the ans is 4.5!=480

My answer is 480, please check it
there are four ways, the flags can be flagged like this
1st pole 2nd pole
a. 4 1
b. 3 2
c. 2 3
d. 1 4

case a- the total number of permutations is 4! 5C4= 120(since 4 out of 5 can be selected in 5C4 ways)(all of them in the first pole)
case b- the total number of permutations is 3!2!5C3= 120
case c- the total number of permutations is 2!3!5C2=120
case 4- the total number of permutations is 4!5C1=120

so the total is 480??
I think I got the question right....

There are four ways in which the five flags can be hoisted...
1st 2nd pole
4 1
3 2
2 3
1 4
In every possibilities of the 4,the five flags can be hoisted in 5! ways...
So the answer is 4*5!=480 ...