Lizzie has a strange deck of 24 number cards that contains cards labeled 2 through 9 in three different colors (red, blue, and green).

(a) In how many ways can she draw one card from each color so that the sum of all the cards is 10?

(b) In how many ways can she draw one card from each color so that the sum of all the cards is 20?

Note that the colors matter, so 234 is different from 234 if the digits are different colors

Guest Jul 30, 2019

#1**+2 **

\(a) \text{ The partitions of 10 are }\\ (2,2,6), ~(2,3,5),~(2,4,4),~(3,3,4),\\ \text{ and all arrangements of these among the colors}\\ \text{there will be $\dbinom{3}{1}+3! + \dbinom{3}{1}+\dbinom{3}{1} = 15$ arrangements}\)

\(b) \text{ Is basically the same deal but now we find the partition of 20 rather than 10}\\ (2,9,9),~(3,8,9),~(4,7,9),~(4,8,8),~(5,6,9),~(5,7,8),~(6,6,8),~(6,7,7),~\text{and all arrangements by color}\\ \text{as before this will be}\\ 3 + 3! + 3! + 3 + 3!+3! +3+3= 36\)

.Rom Jul 31, 2019

#2**+1 **

Hi, I your answer is wrong I was able to solve the problem and verify my solution.

A: 24

B: 48

Guest Jul 31, 2019

#5**+2 **

I have been asked to discuss this question further.

I have only looked at part A again. I assume the problem with part A is the same as that of part B.

Lizzie has a strange deck of 24 number cards that contains cards labeled 2 through 9 in three different colors (red, blue, and green).

(a) In how many ways can she draw one card from each color so that the sum of all the cards is 10?

The only triads that add to 10 are

6,2,2

5,2,3

4,4,2

4,3,3

If you think there is a problem with this so far then let me know.

Consider 6,2,2

The 6 can be red blue or green 3 choices. Once the 6 is allocated a colour, both the other colours must be 2.

SO there are 3 ways to allocate 6,2 and 2 if each one must be a different colour.

The same logic can be used for 4,4,2 and for 4,3,3

So far there is 3+3+3=9 ways

The only triad not considered so far is 5,2,3

The 5 can be one of 3 colours, the 2 can be one of 2 remaining colours and the 3 must be the last colour. So that is 3*2=6 possibilities.

9+6 = 15 ways that this can be done.

-----------------

Think about it, see if you can get 36 for the second one by yourself.

Melody Oct 7, 2019

#6**+1 **

Hi Melody,

I’ve read your post in detail and I understand your logic and methodology. It’s consistent with Rom’s view and solution process.

It still seems odd to me that the output sets, which are distinguishable by observation (because of the color), are factored out and discounted when the numeric element is redundant. I know that sometimes certain elements that make a set unique are ignored because it’s irrelevant to the question. Similarly, some sets that are truly indistinguishable are treated as distinguishable because it affects the probabilities of their incidence.

For most counting problems, it seems straightforward for when to count or not to count redundancies. For counting questions of this type, I suppose it’s a matter of practiced experience to understand the nuances for when to include or exclude redundancies. Considering that both you and Rom have solved these problems for years and years, long before I was born, I accept the consensus that the general solution should factor out redundancies, (only counting them if an explicit condition in the problem requires it).

Thank you for your input on the solution to this question.

GA

GingerAle
Oct 7, 2019

#7**+1 **

I sometimes find it difficult to get my head around whether to include or not include redundancies too.

In this case I have no problem (it is a listing problem - the lists look the same then they are the same)

But having difficulties often means that you are on the way to a full understanding.

If something that other people find difficult seems easy to you then it is likely that you are not really understanding.

Melody
Oct 8, 2019