# 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?

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## closed as off-topic by 1999, Hakim, les guimauves, studiosus, hardmathJul 4 '14 at 1:49

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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.

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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