Bill is sent to a donut shop to purchase exactly six donuts. If the shop has four kinds of donuts and Bill is to get at least one of each kind, how many combinations will satisfy Bill's order requirements?
Bill is required to get at least 1 of each of the 4 kinds. Once he has done that, he has two donuts left to buy with no restrictions. He can do this by buying 2 of the same kind, which can be done in 4 ways, or he can do this by buying two donuts which are different kinds. If he buys donuts of different kinds, there are 4 options for the type of the first donut and 3 options for the second donut, but since the order that he selects them in doesn't matter we need to divide by two to get to a final count of $\dfrac{4\cdot3}{2}=6$ ways to buy two distinct donuts. This gives us a total of $6+4=10$ ways to buy the last 2 donuts once he has bought one of each kind, so $\boxed{10}$ is our answer.
10
Level 5
Prealgebra
Problem: Bill is sent to a donut shop to purchase exactly six donuts. If the shop has four kinds of donuts and Bill is to get at least one of each kind, how many combinations will satisfy Bill's order requirements?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the problem, we first recognize that Bill needs to buy six donuts with the condition that he must have at least one of each of the four kinds available. This means that we can start by allocating one donut of each kind to satisfy the requirement of having at least one of each. 

After giving one donut of each kind, Bill has already used up four donuts (one from each of the four kinds). This leaves him with two more donuts to choose freely among the four kinds. 

Now, we need to determine how to distribute these remaining two donuts. This is a classic combinatorial problem where we can use the "stars and bars" theorem. The problem can be rephrased as finding the number of non-negative integer solutions to the equation:

x1 + x2 + x3 + x4 = 2

where x1, x2, x3, and x4 represent the additional donuts of each kind that Bill can choose. 

According to the stars and bars theorem, the number of solutions to the equation x1 + x2 + x3 + x4 = n (where n is the total number of indistinguishable items to distribute) is given by the formula:

C(n + k - 1, k - 1)

where n is the number of items to distribute (in this case, 2 donuts), and k is the number of categories (in this case, 4 kinds of donuts). 

Substituting the values into the formula, we have:

C(2 + 4 - 1, 4 - 1) = C(5, 3)

Calculating C(5, 3):

C(5, 3) = 5! / (3! * (5 - 3)!) = 5! / (3! * 2!) = (5 * 4) / (2 * 1) = 10

Thus, the total number of combinations that satisfy Bill's order requirements is 10.<|END_OF_LATENT|> The final answer is $10$.