Questions: Chang is arranging 14 cans of food in a row on a shelf. He has 1 can of olives, 6 cans of corn, and 7 cans of beans. In how many distinct orders can the cans be arranged if two cans of the same food are considered identical (not distinct)?

Chang is arranging 14 cans of food in a row on a shelf. He has 1 can of olives, 6 cans of corn, and 7 cans of beans. In how many distinct orders can the cans be arranged if two cans of the same food are considered identical (not distinct)?

Transcript text: Chang is arranging 14 cans of food in a row on a shelf. He has 1 can of olives, 6 cans of corn, and 7 cans of beans. In how many distinct orders can the cans be arranged if two cans of the same food are considered identical (not distinct)? $\square$

Solution

Solution Steps

To solve this problem, we need to determine the number of distinct permutations of the cans, considering that cans of the same type are identical. This is a classic combinatorics problem where we use the formula for permutations of multiset:

\[ \frac{n!}{n_1! \times n_2! \times \ldots \times n_k!} \]

where $ n $ is the total number of items to arrange, and $ n_1, n_2, \ldots, n_k $ are the counts of each distinct item. In this case, $ n = 14 $, $ n_1 = 1 $ (olives), $ n_2 = 6 $ (corn), and $ n_3 = 7 $ (beans).

Step 1: Determine Total Cans and Their Types

We have a total of $ n = 14 $ cans consisting of: