Question

In a cafeteria, there are 5 breads. One can choose 1 bread from the available breads, either small or large sized, and can choose up to 2 sauces from 6 available sauces. What is the number of different ways one can place an order?

Hint:

Treat the bread choice and the sauce choice as two separate, independent decisions and multiply their counts at the end. For the sauces, "up to 2" includes the option of choosing none at all -- do not forget to count that case too.

Answer 1

Correct Answer: 220

Approach: An order is two independent decisions $-$ the bread (with its size) and the sauce bundle. Count each separately, then multiply.

Step 1 (bread): 5 breads, each available small or large, so \(5 \times 2 = 10\) bread choices.

Step 2 (sauces): "Up to 2 sauces" from 6 means choose 0, 1, or 2: \[ \binom{6}{0} + \binom{6}{1} + \binom{6}{2} = 1 + 6 + 15 = 22. \]

Step 3: The two choices are independent, so multiply: \[ 10 \times 22 = 220. \]

Final Answer: Number of possible orders \(= \boxed{220}\).

Answer 2

Approach: Treat "bread type + size" as a single combined choice, and "sauce selection" as choosing a subset of size \(0\), \(1\), or \(2\) from \(6\) sauces, then multiply independent choices.

Bread choice: \(5\) breads \(\times\ 2\) sizes (small/large) \(=10\) options.
Sauce choice ("up to 2" means \(0\), \(1\), or \(2\) sauces): \[ \binom{6}{0}+\binom{6}{1}+\binom{6}{2}=1+6+15=22. \]
Since the bread choice and the sauce choice are independent, total orders \[ =10\times22=\boxed{220} \]

Answer 3

Concept:

• Total number of orders equals (number of bread choices) times (number of sauce choices), since the two decisions are made independently.
• Instead of adding up the sauce combinations for 0, 1 and 2 sauces one by one, the same count can be found by removing the "too many sauces" cases from the full power set of all possible sauce subsets.

Step 1: Count the bread choices.
5 breads, each available in 2 sizes, giving $5 \times 2 = 10$ bread choices.

Step 2: Count every possible sauce subset, of any size.
Each of the 6 sauces is independently included or excluded, so the total number of subsets (sizes 0 through 6) is $2^6 = 64$.

Step 3: Remove the subsets that use more than 2 sauces.
$\binom{6}{3}+\binom{6}{4}+\binom{6}{5}+\binom{6}{6} = 20+15+6+1 = 42$

Step 4: Subtract to get the valid sauce selections (0, 1, or 2 sauces).
$64 - 42 = 22$

Step 5: Multiply the two independent choices.
$10 \times 22 = 220$

Final Answer: $220$