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hiuwaitong@gmail.com

 
 
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hiuwaitong@gmail.com
by Tonia Yuan - Monday, 16 January 2023, 9:57 PM
 

Can you please tell me why in Math Challenge I-A   Counting and Probability  Question 5.14  we have to do

 (3 + 2    2- 1) when in the example question 5.2 (a) we do (3 + 3 -1    3)?

 
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Re: hiuwaitong@gmail.com
by Areteem Professor - Tuesday, 17 January 2023, 1:15 PM
 

Good question(s)! I'm not quite sure exactly your question, so let me answer a few just to be sure.

First, let's talk about the general stars and bars trick.

  • In the Example 5.2a), we have 3 balls and 3 boxes, so we have 3 stars (for the balls) and 3-1=2 bars (for the boxes).
  • In 5.14, we have 3 balls and only 2 boxes, so there are still 3 stars (for the balls) but now 2-1=1 bar (for the boxes).

In both cases, our stars and bars trick says that we just need to arrange the stars and bars to count the total number of outcomes.

In Example 5.2a), there are 3+3-1 = 5 total symbols (stars+bars), and we choose where the stars go, giving $\binom{5}{3} = 10$ outcomes. Could we instead have chosen where the bars go? YES! Then we'd have $\binom{5}{2} = 10$ outcomes. Note we get the same answer whether we choose where the stars go or choose where the bars go.

Hence, we could have two ways for 5.14, both giving the same answer:

  • Choosing the stars: $\displaystyle\binom{3+2-1}{3} = \binom{4}{3} = 4$ ways.
  • Choosing the bars: $\displaystyle\binom{3+2-1}{2-1} = \binom{4}{1} = 4$ ways.

In general we have the identity: $\displaystyle\binom{n}{k} = \binom{n}{n-k}$ which we're seeing here.

Hope this helps!

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Re: hiuwaitong@gmail.com
by Tonia Yuan - Sunday, 22 January 2023, 2:35 PM
 

Then why do we need to do 2-1 for 5.14 and 5.2 we do not need to do 3-1 ?

Please help  T^T 

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Re: hiuwaitong@gmail.com
by Areteem Professor - Monday, 23 January 2023, 11:14 AM
 

I think it's a little confusing because while every counting problem should have 1 numerical/number as an answer, there are sometimes different ways to get that answer that might seem different.

With the stars and bars trick, there are two ways (that give the same numerical answer after simplifying). You can have$$(i)\ \binom{\text{# of stars} + \text{# of bars}}{\text{# of stars}}\text{ OR } (ii)\ \binom{\text{# of stars} + \text{# of bars}}{\text{# of bars}}.$$

Anytime you answer a stars and bars question, either way works.

For example, in 5.2a the final numerical answer is 10. But you can get 10 with either (i) or (ii) above:$$(i)\ \binom{3+(3-1)}{3}=10 \text{ OR } (ii)\ \binom{3+(3-1)}{3-1}=10.$$

Similarly for 5.14, the final answer is 4, but$$(i)\ \binom{3+(2-1)}{3}=4 \text{ OR } (ii)\ \binom{3+(2-1)}{2-1}=4.$$

To review the other differences, 5.2a is 3 balls (so 3 stars) and 3 boxes (so 3-1 bars) while 5.14 is 3 balls (so 3 stars) and 2 boxes (so 2-1 bars). In both cases the number of bars is 1 less than the number of boxes.

Stars and bars is definitely tricky, so don't worry if it takes a little time to get use to it! Hope this helps!