counting with replacement

Hi all,

I'm curious if I'm approaching this question correctly:

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There are 11 flavors at an ice cream shop. 6 are frozen yogurt and 5 are traditional ice cream. Among the 6 frozen yogurt, 2 are low-fat and 4 are regular. And among the 5 traditional ice creams 3 are low-fat and 2 are regular.

You're with a large group and all together you want to order 10 frozen yogurts (4 low-fat and 6 regular) and 12 ice creams (7 low-fat and 5 regular). If the ice-cream shop has plenty of stock (i.e. they won't run out of any flavors), then in how many ways can you make your choices?

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Here's what I've come up with:

First, let's assume that it doesn't matter in what order we make the choices, just that we end up with the totals given above. Then, I think this requires combinations (not permutations), but because the store has more than enough of each kind I think I need to count with replacement.

What I've done is split the problem up into 4 cases: 1) choose the 4 low-fat yogurts, 2) choose the 6 regular yogurts, 3) choose the 7 low-fat ice creams, 4) choose the 5 regular ice creams. Then, if I'm not mistaken, for each of the 4 cases I can use n+k-1 Choose k to get the number of ways to make the choices. By the multiplication rule, the total number of ways to make the choices will be the product of the results of cases 1 through 4.

Here are my calculations:
\begin{align*}
&\binom{2+4-1}{4} \times \binom{4+6-1}{6} \times\binom{3+7-1}{7} \times\binom{2+5-1}{5} \\[5pt]
&= \binom{5}{4} \times \binom{9}{6} \times \binom{9}{7} \times \binom{6}{5} \\[5pt]
&= 5 \times 84 \times 36 \times 6 \\[5pt]
&= \mathbf{90720} \text{ ways we can place our order at the ice cream shop.}
\end{align*}

Does that seem like the right approach??

Thanks in advance for your help. I really appreciate it!

Although I did not do the calculations, your reasoning is correct.

pka said:

Although I did not do the calculations, your reasoning is correct.

Click to expand...

Thanks, pka!

I get a different result using the fundamental counting principal as follows:

i) For each of the 4 low-fat yogurts, we have 2 choices, so there are \(\displaystyle 2^4\) ways to choose these.

ii) For each of the 6 regular yogurts, we have 4 choices, so there are \(\displaystyle 4^6\) ways to choose these.

iii) For each of the 7 low-fat ice-creams, we have 3 choices, so there are \(\displaystyle 3^7\) ways to choose these.

iv) For each of the 5 regular ice-creams, we have 2 choices, so there are \(\displaystyle 2^5\) ways to choose these.

Thus, the total number of ways N to place the order is:

\(\displaystyle N=2^4\cdot4^6\cdot3^7\cdot2^5=2^{21}\cdot3^7=4586471424\)

MarkFL said:

I get a different result using the fundamental counting principal as follows:
i) For each of the 4 low-fat yogurts, we have 2 choices, so there are \(\displaystyle 2^4\) ways to choose these.
Thus, the total number of ways N to place the order is:
\(\displaystyle N=2^4\cdot4^6\cdot3^7\cdot2^5=2^{21}\cdot3^7=4586471424\)

Click to expand...

The trouble with that solution is that these are multi-selections
We are finding the number of non-negative integer solutions to
\(\displaystyle \begin{align*}a+b &=4\\ c+d+e+f &=6 \\x+v &=7\\ x+y+z&=5 \end{align*}\)

Where \(\displaystyle a~\&~b\) are the number of choices of low-fat yogurts; \(\displaystyle c,~d,~e~\&~f\) are the numbers each type of regular yogurts; etc.

There are \(\displaystyle \dbinom{6+4-1}{6}\) ways to make six selections from four flavors of regular yogurts.

Thank you kindly!

I figured my reasoning was incorrect (as is frequently the case with counting problems ), but now I understand why. Much appreciated. :cool: