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Math Challenge II-A Combinatorics

 
 
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Math Challenge II-A Combinatorics
by Neo Liang - Sunday, September 1, 2019, 4:29 PM
 

Lecture 2 part 1, Problem 2.2 b

I got $\left({\dbinom{15}{1}\dbinom{10}{1}\dbinom{23}{3}}\right)^2$, but the answer says that it's $\left({\dbinom{25}{5}-\dbinom{15}{5}-\dbinom{10}{5}}\right)^2$. I got this because you need at least one man and one woman, and the rest doesn't matter. The same is for the Party Planning Committee, so you just square the answer.

 
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Re: Math Challenge II-A Combinatorics
by Areteem Professor - Wednesday, September 4, 2019, 1:40 PM
 

Good question! This a common mistake that can be confusing.

The answer if $\binom{15}{1}\binom{10}{1}\binom{23}{3}$ has some hidden overcounting. First, to clarify where this answer comes from, it is three steps:

  1. Choose one of the 15 males to be on the committee, THEN
  2. Choose one of the 10 females to be on the committee, THEN
  3. Fill in the remaining 3 spots from any of the remaining 23 people.

The answer you gave correctly counts the number of ways to do the $3$ steps listed above. As an example suppose our $15$ males are $A,B,\ldots, O$ and $P,Q,\ldots$ are female.

Consider the three steps: 1. Choose $A$, 2. Choose $P$, 3. Choose $B$, $C$, $D$. Note in the end this gives us the group with members $A, B, C, D, P$.

Now consider the different three steps: 1. Choose $D$, 2. Choose $P$, 3. Choose $A$, $B$, $C$. In the end this ALSO gives the group with members $A, B, C, D, P$.

This is the problem, as we only care about the different possible groups, so we are overcounting the above scenario.

This is of course only one example of how the overcounting can be done, but hopefully it shows what can go wrong with the method you used.