Ordered Counting with Restrictions Problem

[kendang]

5 Boys and 8 Girls go to the Halloween dance. Lori and Julie will not dance with Bert or Tyler. Charles will not dance with Diana or Sara. How many boys/girls couples can be made?

So I know that between Lorie & Julie and Bert & Tyler, there are 4 restricted possibilities.
Diana & Sara and Charles, there are 2 restricted possibilities.

Total possibilities to match up boy/girl: P(8,2) = 56.

Thus, 56 - 6 is the solution?

Thanks in advance.

 

[JeffM]

5 Boys and 8 Girls go to the Halloween dance. Lori and Julie will not dance with Bert or Tyler. Charles will not dance with Diana or Sara. How many boys/girls couples can be made?

So I know that between Lorie & Julie and Bert & Tyler, there are 4 restricted possibilities.
Diana & Sara and Charles, there are 2 restricted possibilities.

Total possibilities to match up boy/girl: P(8,2) = 56.

Thus, 56 - 6 is the solution?

Thanks in advance.

Your general approach looks good: calculate the total number of possibilities and then deduct the prohibited ones.

I agree with 6 prohibited possibilities.

I disagree with the total.
Boy A can be paired 8 distinct ways.
For each distinct pairing for Boy A, Boy B can then be paired 7 distinct ways.
For each of the 56 distinct pairings for Boys A and B, Boy C can then be paired 6 distinct ways.
Etc

So I think the answer is (8! / 3!) - 6 = (8 * 7 * 6 * 5 * 4) - 6 = 6,720 - 6 = 6,714.