# Calculate the amount of different combinations with rules [closed]

I am wondering how to calculate the amount of possible combinations for this scenario: Say a party bag contains 6 different sweets, and that these are a random selection from 8 different types of sweets. However, no more than 2 of the same type of sweet can be put into the party bag and no fewer than 4 different types of sweet can be put into the party bag. How many different party bags are possible, and how would I calculate it?

-

## closed as off-topic by ᴡᴏʀᴅs, Hakim, Cookie, studiosus, hardmathJul 4 '14 at 1:49

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – ᴡᴏʀᴅs, Hakim, Cookie, studiosus, hardmath
If this question can be reworded to fit the rules in the help center, please edit the question.

The numbers are quite small, so we need not be very clever in our counting. Divide into cases.

(a) Maybe the bag has $6$ different types of sweet. These can be chosen in $\binom{8}{6}=28$ ways.

(b) Maybe the bag has $5$ different types of sweet. Then we will have $2$ of one type, and $1$ each of $4$ other types. The type we have $2$ of can be chosen in $\binom{8}{1}$ ways. For each such choice, there are $\binom{7}{4}$ ways to select the other types, for a total of $\binom{8}{1}\binom{7}{4}=280$.

(c) Maybe the bag has $4$ different types only. Since $3$ of one type is not allowed, we must have $2$ each of $2$ types, and $1$ each of $2$ other types.

The two types we have $2$ each of can be chosen in $\binom{8}{2}$ ways, and then the $2$ singletons in $\binom{6}{2}$ ways, for a total of $\binom{8}{2}\binom{6}{2}=420$.

Add up the $3$ numbers we have obtained.

-
Sorry, I forgot to put "no fewer than 4 different types of sweet can be put in". This means we cannot have the numbers obtained in (c), I do not think, because we cannot have 3 of one type. We can only have 2 of one type and no less than 4 of different types. Basically, we can have them all different or we can have two of one type with the 4 other ones different, in a bag. So I think that (a) and (b) from you are still valid. Thank you for your response. – user64266 Feb 27 '13 at 21:51
Next, I will use the two numbers that you gave me as n and r respectively, and use the formula n! / (r!(n-r)!) – user64266 Feb 27 '13 at 21:58
I have made the change, now there is only one entry under (c), marginally simpler. Note that $2$ of sweet A, $2$ of sweet C, and $1$ each of sweets D and F is allowed by the wording. So some of (c) survives and has to be counted. – André Nicolas Feb 27 '13 at 21:58
I do not agree with your interpretation of the wording. You will need to use also $\binom{8}{2}\binom{6}{2}$. – André Nicolas Feb 27 '13 at 22:00
Oh right, yes. However, will \binom{8}{1} not have to be multiplied by two, seeing as we are getting it twice? – user64266 Feb 27 '13 at 22:19