An employee of a grocery store is placing an order for soda. There are 8 varieties of soda and they are sold in cases. Each case contains only one variety of soda. The store will order 50 cases total. How many different ways are there to place the order if the employee orders at least three cases of each variety?
My attempt
I used pigeon hole principle and got the answer as
$3(50-1)+1$ $=148$ $ways$
Can anyone please verify whether this is correct or not
Your answer is incorrect.
Since there must be at least three cases of each of the eight varieties of soda in the order, the employee only has a choice of which $50 - 3 \cdot 8 = 50 - 24 = 26$ additional cases to order. Let $x_k$ be the number of additional cases of variety $k$ that the employee opts to order. Then $$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 = 26$$ is an equation in the nonnegative integers. A particular solution of the equation corresponds to the placement of seven addition signs in a row of $26$ ones. For instance, $$1 1 1 1 + 1 1 1 + 1 1 + + 1 1 1 1 1 + 1 1 1 1 1 1 + 1 1 + 1 1 1 1$$ corresponds to the solution $x_1 = 4$, $x_2 = 3$, $x_3 = 2$, $x_4 = 0$, $x_5 = 5$, $x_6 = 6$, $x_7 = 2$, $x_8 = 4$. The number of such solutions is $$\binom{26 + 7}{7} = \binom{33}{7}$$ since we must choose which $7$ of the $33$ positions required for $26$ ones and $7$ addition signs will be filled with addition signs.