Question

Tim Horton’s coffee shop has three types of “Timbits” (little doughnuts): plain, chocolate, and glazed. When purchasing a box of twelve Timbits, a customer can decide how many of each type is put into the box. How many ways are there to do this? (Hint: there are less than 100 ways) Show your work.

Answer 1

Assuming there is no restriction on number of each type of timbits. Let x1, x2 and x3 represent the number of timbits 9f each type of timbits put in a box.

Example : (2,4,6) means there are 2 Number of type1 timbits ,4 number of type 2 timbits and 6 number of type 3 timbits in a box.

Let X1 + x2 + x3 = 12 where each of x1,x2,x3 lie between 0 and 12 (including).

If X1= 0 , then x2+x3= 12 and then there are 13 possible solutions are (0,12),(1,11),(2,10),(3,9),(4,8),(5,7)(6,6)(7,5)(8,4)(9,3)(10,2)(11,1)(12,0)

If X1 = 1, then x2+x3 =11 has 12 possible solutions

(0,11)(1,10)(2,9)(3,8)(4,7)(5,6)(6,5)(7,4)(8,3)(9,2)(10,1)(11,0)

Similarly, we can predict that

When X1 = 2 , x2+x3 = 10 has 11 solutions

When x1 = 3 , x2 +x3 = 9 has 10 solutions

When x1 =4 , x2+x3 =8 has 9 solutions

When x1 = 5, x2 + x3 = 7 has 8 solutions

When x1 =6, x2 + x3 = 6 has 7 solutions

When x1 =7 , x2 + x3 =5, has 6 solutions

When x1 =8, x2+x3 =4 has 5 solutions

When x1=9, x2+x3=3 has 4 solutions (3,0)(2,1)(1,2)(0,3)

When x1 =10, x2+x3 =2 has 3 solutions (2,0)(1,1)(0,2)

When x1 =11, x2+x3 = 1 has 2 solutions (1,0)(0,1)

When x1 = 12, x2 +x3 =0 has one solution which is (0,0)

So, total number of ways to put 12 timbits in a box = number of solutions of x1 +x2+x3 =12

= 13+12+11+10+9+8+7..+1

= 1+2+3+4+5+..+13

= 13*14/2 = 13*7 =91

Hence there are 91 ways to put 12 timbits of 3 types in a box.