You are stopping by Timmy’s to buy $12$ donuts. There are $4$ varieties to choose from. You need to have at least one donut from each variety. How many different ways can you select your $12$ donuts?
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3Affine, abstract, projective and quasi-projective? – Edward Evans Nov 19 '16 at 00:16
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http://math.stackexchange.com/questions/1312337/find-the-number-of-positive-integers-solutions-of-the-equation-3x2y-37 – ET93 Nov 19 '16 at 00:19
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1As the linked post suggests, the stars and bars technique will be helpful. – angryavian Nov 19 '16 at 00:22
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Taking $1$ donut firstly from each variety to fulfill The condition for at least one donut from each variety, so we have $4$ donuts selected. $12-4 = 8$ donuts remain to be selected.
Now we can select $8$ donuts with any number of donuts from each variety(including zero selections), consider the $4$ varieties to be named $a,b,c,d$ then we have the situation as $a+b+c+d = 8$. Here $r = 4$ (four varieties) and $n = 8 $
So, to find the number of ways($w$) in this situation use $w = \binom{n+r-1}{r-1}$. applying this, we get the result as $\binom{11}{3}$ which is equal to $165$.
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It will be
$$\binom{11}{3}=\frac{11\cdot10\cdot9}{3\cdot2\cdot1}=11\cdot5\cdot9=165$$
Consider the $8$ remaining donuts as $8$ points (because you have at least $1$ donut for each different donut type, so quit $4$ donuts of your $12$).
$$\bullet\bullet\bullet\bullet\bullet\bullet\bullet\bullet$$
Then you can separate your remaining $8$ donuts by three bars to identify each of your four types of donuts. For example:
$$\bullet\mid\bullet\mid \bullet\bullet\bullet\mid\bullet\bullet\bullet$$
So if you consider those three bars with the $8$ points all as points again, you will only have to choose three of the new $11$ points to change them by bars and get another arrangement.
iam_agf
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