# You are stopping by Timmy’s to buy $12$ donuts. There are $4$ varieties to choose from.

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?

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 $12$).

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Then you can separate your remaining $8$ donuts by three bars to identify each of your four types of donuts. For example:

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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.

• The correct answer is 165. – Tom Sadan Nov 19 '16 at 0:51
• Thanks I didn't read that consideration. – MonsieurGalois Nov 19 '16 at 0:57
1. 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.
2. 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 = (n+r-1)C(r-1). applying this, we get the result as 11C3 which is equal to 165.