imc

Let the ordered triplet $(x,y,z)$ represent the assortment of $x$ chocolate chip cookies, $y$ oatmeal cookies, and $z$ peanut butter cookies. Using casework: Pat selects $0$ chocolate chip cookies: Pat needs to select $6-0=6$ more cookies that are either oatmeal or peanut butter. The assortments are: $\{(0,6,0);(0,5,1);(0,4,2);(0,3,3);(0,2,4);(0,1,5);(0,0,6)\}\to 7$ assortments. Pat selects $1$ chocolate chip cookie: Pat needs to select $6-1=5$ more cookies that are either oatmeal or peanut butter. The assortments are: $\{(1,5,0);(1,4,1);(1,3,2);(1,2,3);(1,1,4);(1,0,5)\}\to 6$ assortments. Pat selects $2$ chocolate chip cookies: Pat needs to select $6-2=4$ more cookies that are either oatmeal or peanut butter. The assortments are: $\{(2,4,0);(2,3,1);(2,2,2);(2,1,3);(2,0,4)\}\to 5$ assortments. Pat selects $3$ chocolate chip cookies: Pat needs to select $6-3=3$ more cookies that are either oatmeal or peanut butter. The assortments are: $\{(3,3,0);(3,2,1);(3,1,2);(3,0,3)\}\to 4$ assortments. Pat selects $4$ chocolate chip cookies: Pat needs to select $6-4=2$ more cookies that are either oatmeal or peanut butter. The assortments are: $\{(4,2,0);(4,1,1);(4,0,2)\}\to 3$ assortments. Pat selects $5$ chocolate chip cookies: Pat needs to select $6-5=1$ more cookies that are either oatmeal or peanut butter. The assortments are: $\{(5,1,0);(5,0,1)\}\to 2$ assortments. Pat selects $6$ chocolate chip cookies: Pat needs to select $6-6=0$ more cookies that are either oatmeal or peanut butter. The only assortment is: $\{(6,0,0)\}\to 1$ assortment. The total number of assortments of cookies that can be collected is $7+6+5+4+3+2+1=28\Rightarrow \boxed{28}$ There is a much faster way to do casework. Case 1: 1 type of cookie - 3 ways Case 2: 2 types of cookies - 3 ways to choose the 2 types of cookies, and 5 ways to choose how of each there are 1 and 5 2 and 4 3 and 3 4 and 2 5 and 1 $3\ast 5=15$ cases for case 2 Case 3: 3 types of cookies - quick examination shows us that the only ways to use all three cookies are the following: 4, 1, 1: this gives us 3!/2!*1! = 3 ways 3, 2, 1: this gives us 3! = 6 ways 2, 2, 2: this gives us 1 way total of 10 cases for this case 10+15+3= $28$ total