Suppose that you own a fancy bakery and there are 6 muffins

Suppose that you own a fancy bakery, and there are 6 muffins and 7 donuts waiting to be eaten. No one\'s around, which means you get to have them all to yourself. How many groups of 5 pastries can you eat at one time How many groups of 4 pastries can you eat at once, if there must be at least one muffin in each group How many groups of (i pastries can you eat at once, if two specific donuts refuse to be eaten together

Solution

(a)

This is done in 6 ways

0 muffins 5 donuts   in C(7,5) =21 groups

1 muffins 4 donuts   in C(6,1)C(7,4)=210 groups

2 muffins 3 donuts   in C(6,2)C(7,3)=525 groups

3 muffins 2 donuts   in   C(6,3)C(7,2)=420 groups

4 muffins 1 donuts    in   C(6,4)C(7,1)=105 groups

5 muffins 0 donuts    in    C(6,5) = 6 groups

SO total ,

6+105+420+525+210+21=1287 groups

An easy way to compute this is to simply choose 5 pastries from total 13 (6+7)

in :C(13,5)=1287 groups

(b)

At least one muffin in each group. This one muffin is chosen in C(6,1)=6 ways. Now 3 positions to fill

in 4 ways

0 muffins 3 donuts   in C(7,3) =35 groups

1 muffins 2 donuts   in C(5,1)C(7,2)=105 groups

2 muffins 1 donuts   in C(5,2)C(7,1)=70 groups

3 muffins 0 donuts   in   C(5,3)C(7,0)=10 groups

So total: 6*(10+70+105+35)=1320 groups

Easy way to do this is: C(6,1)C(12,3)=1320 groups

(c)

Two cases

Case 1. The two donuts which refuse to be eaten together are not selected

This is done in: C(11,6) =462

Case 2: One of the two donuts is selected

This is done as follows

Select 1 of the those two pastries and then select 5 from remaining 11 pastries

C(2,1)C(11,5)=924 groups

Hence total number of groups are:

924+462=1386