Question

A cafeteria offers 5 types of sandwiches. Moreover, for each type of sandwich, a customer can choose one of 4 breads and opt for either small or large sized sandwich. Optionally, the customer may also add up to 2 out of 6 available sauces. The number of different ways in which an order can be placed for a sandwich, is:

• A. \(600\)
• B. \(840\)
• C. \(880\)
• D. \(800\)

Hint:

When a question says “up to $k$ items” from $n$ options, sum the combinations: \[ \sum_{r=0}^{k} \binom{n}{r}. \] Then multiply by the number of ways for all other independent choices.

Answer 1

The Correct Option is C

Approach: Every part of the order is an independent choice, so multiply the counts. The only twist is the sauces, where "up to 2 out of 6" means you separately count taking 0, 1 or 2 sauces.

Step 1: Independent fixed choices. Type of sandwich: \(5\) ways. Bread: \(4\) ways. Size (small/large): \(2\) ways. Together \[ 5 \times 4 \times 2 = 40. \]

Step 2: Sauce choices. "Up to 2" allows 0, 1, or 2 sauces, and order does not matter: \[ \binom{6}{0} + \binom{6}{1} + \binom{6}{2} = 1 + 6 + 15 = 22. \]

Step 3: Combine. \[ 40 \times 22 = 880. \] Note why we add for sauces but multiply elsewhere: the sauce sub-cases (0/1/2) are alternatives for the same slot, so they add; bread, size and sauces are different slots, so they multiply.

\[ \boxed{880} \] which is option (3).

Answer 2

Approach: Instead of adding \(\binom{6}{0}+\binom{6}{1}+\binom{6}{2}\) directly, use the symmetry of Pascal's Triangle: choosing \(k\) sauces out of 6 has the same count as choosing \(6-k\) sauces, and all 7 possible subset sizes (0 through 6) must add up to \(2^6 = 64\).

Step 1: Fixed choices. Sandwich type, bread, size: \(5 \times 4 \times 2 = 40\) ways.

Step 2: Sauce count via symmetry. By symmetry, the number of subsets of size 0, 1, 2 equals the number of subsets of size 6, 5, 4 respectively, and the middle size, 3, is left over. So
\[ 2^6 = 2\big[\text{(subsets of size 0, 1 or 2)}\big] + \binom{6}{3}. \]
Since \(\binom{6}{3} = 20\), this gives
\[ 64 = 2(\text{sauce count}) + 20 \Rightarrow \text{sauce count} = \frac{64-20}{2} = 22. \]

Step 3: Combine. \(40 \times 22 = 880\).

\[ \boxed{880} \] option (3).