0

New to the community. Excited to dive into this world this weekend.

I have a problem that I am having trouble thinking through:

Group A has 4 members with 2 slots to fill Group B has 6 members with 2 slots to fill

If I were to stop here I understand the calc to figure out how many combinations there are: $(4\cdot 3\cdot 2\cdot 1/2\cdot1) \cdot (6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1/2\cdot 1)=(4\cdot 3) \cdot (6\cdot 5\cdot 4\cdot 3) = (12) \cdot (360) = 4320$

Where I get lost is the next step:

If I add in a 3rd group Called "Combined Group" that has 1 slot to fill and can only select from Group Members A&B. Does this change the overall amount of combinations involved and if so, what is the math behind it?

From what I worked through, I can't see how it would decrease the combinations. I can see in very rudimentary examples that the combinations don't change, but I can't work through the math when the amount of members and/or slots change. Intuitively I could see it changing from constant to increasing the combinations given the different combinations of members and slots, but again, I can't see how to work through it with math.

Any help would be greatly appreciated.

As a first-time poster - apologize if any community guidelines weren't followed, I'll be sure to not make the same mistake(s) next time!

Thanks!

Kevin Long
  • 5,159
CuriousTR3
  • 1
  • 1
    To clarify... does it matter what order the slots are filled? With group $A$ consisting of ${a,b,c,d}$ does it matter if the first slot was $a$ and the second slot was $b$ versus if the first slot was $b$ and the second slot was $a$? You appear to have an error with the first part still. – JMoravitz Jan 22 '21 at 16:01
  • The end result should be that this follows from the rule of product and/or binomial coefficients. Look to see how many ways you can fill the first slots from group A, given such a selection of how to fill for group A look to see how many ways you can fill the second set of slots from group B, the finally given a selection for each of these first two steps look to see how many ways you can fill your final slot(s). Multiply the number of options for each step together for final total. – JMoravitz Jan 22 '21 at 16:04
  • If order doesn't matter for these slots, then it should have been $\binom{4}{2}=\dfrac{4!}{2!2!}=\dfrac{4\cdot 3\cdot 2\cdot 1}{2\cdot 1\cdot 2\cdot 1}=6$ ways for the first. If order did matter, then it would have been $4\cdot 3 = 12$ as you had. If order doesn't matter for the second calculation it would have been $\binom{6}{2}=\dfrac{6!}{4!2!}=15$ while if it did it would have been $6\cdot 5=30$. If you insist on calling it a fraction, $\dfrac{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{\color{red}{4\cdot 3\cdot}2\cdot 1} = \dfrac{6!}{(6-2)!}$, not $\dfrac{6!}{2!}$. – JMoravitz Jan 22 '21 at 16:07
  • As you have only one slot for the last group, it can be selected from the remaining $6$ members and that should be multiplied to the combinations for the first two. – Math Lover Jan 22 '21 at 16:37
  • @JMoravitz Thanks for your input - I did not see an email until this am that people had responded. To your first comment & question the order does not matter. Additionally the sets within Groups A & Groups B are separate.

    Combined Group (or Group C) pulls from both sets of A & B.

     – CuriousTR3 Jan 23 '21 at 17:44
  • Thanks for point out my errors - it seems denominator should be (# of members not selected!*# of slots!)

    To @MathLover's post - with the Combination Group (Group C) added in, since it is a combined set of A&B's sets in order to calculate, I would simply assume A & B were filled & determine how many total applicants were left over & multiply the A & B combinations by that (6 in this case). Am I thinking about that correctly?

     – CuriousTR3 Jan 23 '21 at 17:55
  • Next question for clarification, is being "chosen for the third group" distinguishable from being chosen for the first? For instance, with group A consisting of ${a,b,c,d}$, does $a,b$ being chosen for the first slots and then later on $c$ being chosen for the "combined slot", is this necessarily different than $b,c$ being chosen for the first slots and later on $a$ being chosen for the "combined slot"? In both cases we have $a,b,c$ all being chosen in some way... the only difference being whether it was that they were chosen for the first slots or the combined. – JMoravitz Jan 24 '21 at 18:42
  • If it does matter which it was, that those were different outcomes, then yes... you just multiply by $6$... having applied the rule of product. In the event that it didn't matter, then you need to be more careful about it since you will have overcounted – JMoravitz Jan 24 '21 at 18:43
  • @JMoravitz Sounds like it might be a bit more complicated. It does not matter if a,b are in Group A & c is in Group C (combined group) or if b,c are in Group A & a is in Group C (combined group) ... here I'm at a loss. I can't think of how I would address the overcount? Assuming I would divide the 6 by something? another factorial equation?... that doesn't intuitively seem right. Any direction for my further thinking would be much appreciated. – CuriousTR3 Jan 25 '21 at 20:53
  • For much larger generalizations, this becomes quite convoluted, but for this smaller problem at least you can break into cases based on how many total people from $A$ are used and how many people total from $B$ are used. So, in the case of three people from $A$ (two used in $A$'s designated 2 slots and one used in the slot for $C$) and two from $B$ that would be $\binom{4}{3}\binom{6}{2}$. In the case of two from $A$ and three from $B$ that would be $\binom{4}{2}\binom{6}{3}$. Adding these together. – JMoravitz Jan 25 '21 at 20:59
  • @JMoravitz Thank you. I can see how this could get out of hand if more groups were added. It seems that in order to solve you have to assume a "completed state", whereas you have picked all slots, and then figure out how many different ways you can derive at a "completed state" to complete your calculations.

    To make sure I have it down, the total combinations in this example are 180? In case that's incorrect, I used your two expressions above and added them together = (4!/3!)(6!/(4!2!))+(4!/2!)(6!/(3!3!) = 180 Hopfully I got it, and thank you so much for taking your time to walk thru!

     – CuriousTR3 Jan 27 '21 at 02:35

0 Answers0