Partial Permutation, Combination question for buying dogs at the pet store

[Memo]

TL;DR Summary: A pet store has 5 Chihuahua, 3 Fox and 4 Poodle. A person wants to buy 7 dogs. How many ways for the person to choose and buy with at least 3 Chihuahua and 2 Poodle.

There're C(5,3)*C(4,2) ways to buy dogs with at least 3 Chihuahua and 2 Poodle.
There're C(7,2) ways to buy dogs of any breeds.
In conclusion, there're C(5,3)*C(4,2)*C(7,2)=1260 ways for the person to pick and buy dogs.
Could someone check if my solution is correct? Thank you very much.

 

[PeroK]

Your solution is wrong.

 

[Memo]

Your solution is wrong.

Could you be more specific? thanks.

 

[PeroK]

Could you be more specific? thanks.

You are double/multiple counting cases where more than three Chihuahua's and/or more than 2 Poodles are bought.

Note: it's not clear to me whether the question is at least 2 Poodles or exactly 2 Poodles.

 

[Memo]

You are double/multiple counting cases where more than three Chihuahua's and/or more than 2 Poodles are bought.

Note: it's not clear to me whether the question is at least 2 Poodles or exactly 2 Poodles.

I still don't get what you mean, could you point out whether it's the first or second part, or the whole part.
I think the question implies at least 2 Poodles

 

[PeroK]

I still don't get what you mean, could you point out whether it's the first or second part, or the whole part.
I think the question implies at least 2 Poodles

How many ways are there to choose 7 dogs from 12 total?

 

[Memo]

How many ways are there to choose 7 dogs from 12 total?

Ok, I see that my answer exceed the total possible ways. If 5 dogs are certain then they're 7 dogs left, and the person picks 2 out of 7 regardless of their breeds. How do I fix it?

 

[PeroK]

Ok, I see that my answer exceed the total possible ways. If 5 dogs are certain then they're 7 dogs left, and the person picks 2 out of 7 regardless of their breeds. How do I fix it?

If we have at least 3 C's and at least 2 P's, then there are a number of different combinations that must be calculated separately. I don't see a quick way. I put the calculations on an Excel spreadsheet.

In any case, the possibilities are:

C	P	F
5	2	0
4	3	0
4	2	1
3	4	0
3	3	1
3	2	2

You can then, if you want, calculate all the other possibilities where there either $C < 3$ or $P < 2$ and check that the sum of all the possibilities is $\binom{12} 7$.

 

[PeroK]

PS that's why I thought the question might mean exactly two P's. Because that's simpler, as there are only three cases.

 

[erobz]

I think you can group three Chihuahua's $C_3$, and group 2 Poodles $P_2$, leaving 2 dogs to select in any way.

 

[PeroK]

I think you can group three Chihuahua's $C_3$, and group 2 Poodles $P_2$, leaving 2 dogs to select in any way.

Not easily without multiple counting. That was the OP's initial error.

 

[erobz]

Not easily without multiple counting. That was the OP's initial error.

Yeah...never mind, not exactly straight forward.

 

[Memo]

If we have at least 3 C's and at least 2 P's, then there are a number of different combinations that must be calculated separately. I don't see a quick way. I put the calculations on an Excel spreadsheet.

In any case, the possibilities are:

C	P	F
5	2	0
4	3	0
4	2	1
3	4	0
3	3	1
3	2	2

You can then, if you want, calculate all the other possibilities where there either $C < 3$ or $P < 2$ and check that the sum of all the possibilities is $\binom{12} 7$.

So would the answer be 6^3=216 ways? Like 5C 2P and 2P 5C would be considered the same but still in 2 options

 

[erobz]

So would the answer be 6^3=216 ways? Like 5C 2P and 2P 5C would be considered the same but still in 2 options

You have to sum combinations for each specific case ( each row in the table ), I don't get 216 when I do that.

 

[PeroK]

So would the answer be 6^3=216 ways? Like 5C 2P and 2P 5C would be considered the same but still in 2 options

Not at all. There are six possibilities that are all different. You have to calculate them all.

Do you know how to use a spreadsheet?

 

[Memo]

Not at all. There are six possibilities that are all different. You have to calculate them all.

Do you know how to use a spreadsheet?

No, I can't use a spreadsheet while in exams. Is there really not a quick way since there were much harder questions waiting and if I solve it in the conventional ways then I just lose points

 

[erobz]

May I suggest solving it as suggested (you don’t actually need a spreadsheet), then you can spend as much time as you like looking for a efficient method once you have the result.

 

[PeroK]

No, I can't use a spreadsheet while in exams

You won't have me in an exam.

. Is there really not a quick way since there were much harder questions waiting

I can imagine. The sooner you realise not all problems can be solved by a single calculation the better.

 

[WWGD]

I can see an example of multiple counting here( though can't think of how to deal with it).
So we have 12 dogs, 3 of type A: A1, A2, A3; similarly 4 of type B , 5 of type C. We want to select 7 dogs from the 12, so that we have 3 of type A, 2 of type B.
Consider this:

1)Choose A1,A2,A3 ;B1,B2; A4,A5

2) Choose A1,A4,A5; B1,B2; A2,A3

Both represent the choice , as a set:
{A1,A2,A3 A4,A5, B1,B2}

That's just one such duplicate.

 

[scottdave]

Are the Chihuahuas, Poodles, and Foxes distinct? That would make a difference. Then you have 5 slots spoken for - you only have 2 slots to fill with the remaining animals. If you have to distinguish which Poodles that you are using, it is more complex.