Question 611286: I'm really confused about permutations, and especially this problem!
"Flags of 6 countries are to be displayed in a row. The flags of two countries, A and B, must be next to each other.
a) In how many ways can this be done if the flag of country A must be to the left of the flag of country B?
(The answer key says 120, but I don't understand how)
b) In how many orders can the flags be displayed if the 2 flags can be in either order?
(The answer key says 240 but I don't understand how)"
Thank you so much!
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Note: I'm assuming that all 6 flags are different and that order matters.
a)
Since A and B must be next to each other AND A must be to the left of B, this means that we can effectively "combine" the two flags to make a third flag (say AB). This is not a single flag, but because the two can't be separated or ordered differently, we can group the two like this.
So we have the 5 flags: AB, C, D, E, F
Where AB is considered one flag.
From here, there are 5! = 5*4*3*2*1 = 120 different ways to order these five flags.
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b)
This is a lot like part A, but now we consider the case that we're dealing with the "flag" BA". This will double the number in part a) to give you 2*120 = 240
Let me know if this makes sense. Thanks.
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