A child received a gift of six different soccer team flags, including

Direct counting approach:

We have 3 cases:

Case #1: Liverpool flag is on the wall which leaves us with only 4 flags to choose from (Arsenal cannot be taken).

\(4C3*4!\)

Case #2: Arsenal flag is on the wall which, like in previous case, leaves us with only 4 flags to choose from (Liverpool cannot be taken).

\(4C3*4!\)

Finally case #3: When neither Arsenal nor Liverpool flags are on the wall leaving us with only 4 possible options to fill 4 positions on the wall.

\(4!\)

In total:

\(4*4! + 4*4! + 4! = 24*(4 + 4 + 1) = 216\)

Answer B.

We are given that a child received 6 different soccer team flags, including Liverpool and Arsenal. We need to determine the number of arrangements that are possible when the flags are displayed 4 at a time and the Arsenal and Liverpool flags are not displayed at the same time.

To start we can create the following equation:

Total number of ways to display the flags = (number of ways to display the flags when both the Arsenal and Liverpool flags are displayed together) + (number of ways to display the flags when the Arsenal and Liverpool flags are not displayed together).

Let’s first determine the total number of ways to display the 4 flags from a choice of 6 flags.

This is a permutation problem because the order in which the flags are displayed is important.

Number of ways to display the 4 flags from 6 flags = 6P4 = 6 x 5 x 4 x 3 = 360

Next we can determine the number of ways to display the flags when both Arsenal and Liverpool are displayed. Since we know that the Arsenal and Liverpool flags are definitely selected, that leaves us with 4 flags for the 2 remaining spots, so there are 4C2 ways to select the two remaining flags, which equals (4 x 3)/2! = 6 ways. Finally, there are 4! ways to arrange those 4 flags, which equals 24 ways. Thus, there are 24 x 6 = 144 ways to select and arrange the flags in which both Arsenal and Liverpool are displayed.

Finally, there are 360 - 144 = 216 ways to select and arrange the 4 flags when Arsenal and Liverpool cannot be displayed at the same time.

Answer: B

Total number of ways to arrange four flags from 6 = 4P6 = 6*5*4*3 = 360

Number of ways with Arsenal AND Liverpool flags: Set Arsenal and Liverpool in the first two spaces and determine the possible arrangements for the remaining two spaces (out of four). 2C4 = (4*3)/2=6. So there are 6 combinations that include Liverpool and Arsenal. Each of these 6 combinations can be ordered in 4! ways. So number of ways with Arsenal AND Liverpool = 6*4! = 144.

Number of desired arrangements = 360 - 144 = 216

OFFICIAL SOLUTION

A helpful way to begin this problem is to consider how many options the boy would have were there no restriction. With 6 flags to choose from and 4 spaces for them, he would then have 6 choices for the first flag, 5 for the second, 4 for the third, and 3 for the fourth, for a total of \(6 * 5 * 4 * 3 = 360\) options. But there is a restriction, so you know that there are fewer options than 360. This at the very least allows you to eliminate choices D and E while you investigate the restriction.

Now you want to determine how many arrangements would violate the restriction. For those arrangements, Liverpool and Arsenal are definitely displayed together. That means that of the 4 other flags, you will need to select 2 to make the group (that's a combination), and then you will have 4 flags to arrange (that's how you'll bring it back to a permutation). selecting 2 out of 4 means that you'll use \(\frac{4!}{{2!2!}}\), which becomes \(\frac{4∗3}{2}=6\) combinations for which the two flags will appear together. Then multiply that by the number of ways to arrange each group of 4, which is 4!. That means that 144 of the 360 arrangements will not work, so that leaves 216 possible arrangements, which is answer choice B.

I really like if the math problems of arrangement or possibilities only concerns with a small number of factors in each group and with a small number of groups.
this question is such an example.

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