Stochastic_Jimmy
New member
- Joined
- Jan 10, 2013
- Messages
- 27
Hi all,
I'm curious if I'm approaching this question correctly:
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?
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!
I'm curious if I'm approaching this question correctly:
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?
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!