Different way of solving this problem
-
Suppose we choose the groups one at a time, without replacement. For the first group, we have \(\dbinom{8}{2}\) ways to choose the first group. For the second group, we have \(8-2=6\) people left, so there are \(\dbinom{6}{2}\) ways to choose the second group. For the third group, we have \(6-2=4\) people left, so there are \(\dbinom{4}{2}\) ways to choose the third group. Finally, the last group is already fixed for us. So is the answer \(\dbinom{8}{2}\dbinom{6}{2}\dbinom{4}{2}\)? Nope... but that is close! We notice that we are overcounting the groups \(4!\) times, so we must divide by \(4!\).
Therefore, the answer is \(\dfrac{\binom{8}{2}\binom{6}{2}\binom{4}{2}}{4!}=\boxed{105}.\)
-
@professionalbronco bump, mods, can you confirm please? it has been 2 months later...
-
@professionalbronco mods it has now been another 2-3 months. could you please respond??
-
So sorry about that @professionalbronco !!
But yes, your solution is absolutely right. Another way of looking at is the "algorithm" Professor Loh mentioned in the video: by sorting alphabetically, you'll get that there are 7 options for person A to make a pair, 5 options for the second pair, 3 for the third, and 1 for the last. $$7\cdot5\cdot3\cdot1=\boxed{105}$$ as well.
Also from the lesson, using the expression you had but expanding out the 4!:
$$\frac{\binom{8}{2}\cdot\binom{6}{2}\cdot\binom{4}{2}\cdot\binom{2}{2}}{4\cdot3\cdot2\cdot1}$$
and then expanding the binomial coefficients and cancelling:
$$\frac{\frac{8\cdot7}{2}\cdot\frac{6\cdot5}{2}\cdot\frac{4\cdot3}{2}\cdot\frac{2\cdot1}{2}}{4\cdot3\cdot2\cdot1} \implies \frac{(\cancel{4}\cdot7)\cdot(\cancel{3}\cdot5)\cdot(\cancel{2}\cdot3)\cdot(\cancel{1}\cdot1)}{\cancel{4}\cdot\cancel{3}\cdot\cancel{2}\cdot\cancel{1}}$$which leaves us the same expression for \( 7!! \)
also !!
did you know that the exclamation in the factorial is kind of like the "index" of how much you're shifting each number (double factorial shifts by 2, triple by 3, and so on). Soooo... technically we could get to triple factorials (or higher) maybe:
\(8!!!=8\cdot5\cdot2=80\)
\(11!!!!=11\cdot7\cdot3=231\)
