At a local fast-food restaurant in Oregon (no sales tax), fries, soda, hamburgers, cherry pie, and sundaes cost \$1 each. Chicken sandwiches cost \$2 each. You have five dollars. How many different meals can you order?
Let's assign two groups A and B. Let A consist of \$1 items and B consist of \$2 items.
Group A: \$1 items: Fries, soda, hamburgers, cherry pie, sundaes = 5 items
Group B: \$2 items: Chicken sandwich = 1 item
I'm assuming this is a combinatorics problem which is unordered and with replacement (meaning more than one of the same item can be selected). Hence there are 3 possible scenarios because of the \$5 constraint:
(I) AAAAA: Here we have 5 objects for group A's
n=5 obj + 4 dividers = 9, r=5 obj
(II) BAAA: Since there is only one B item here, I thought I could leave it out and only calculate the placement of 3 objects in AAA. This is because I can have only one object in B, but am free to choose the distribution among the other A's.
n= 3 obj + 2 dividers = 5, r = 3 obj
(III) BBA: Again since B's have only one item, and A is only 5 values, this group is simply 5.
So my approach is to find the combinations of (I)-(III) and add them together:
(I) $\binom{9}{5}=126$
(II) $\binom{5}{3}=10$
(III) $\binom{5}{1}=5$
This sums to 141 but the answer is 166. Can anyone see what I am doing wrong or suggest a better method? I am using the following proposition:
The number of unordered samples of r objects, with replacement from, n distinguishable objects is: $C(n+r-1,r)= \binom{n+r-1}{r}$. This is equivalent to the number of ways to distribute r indistinguishable balls into n distinguishable urns without exclusion.
Thank you!
Best Answer
For choice II, $5 \choose 3$ assumes you cannot order two of the same. For sampling with replacement, it should be ${7 \choose 3}=35$ by the same logic you used to get $9 \choose 5$. That increases the count to $166$