Question

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

Explanation:

Step1: Calculate the total number of cans

The total number of cans \(n=1 + 6+7=14\)

Step2: Use the formula for permutations of multi - sets

The formula for the number of permutations of a multi - set with \(n\) objects, where there are \(n_1\) of one kind, \(n_2\) of another kind, \(\cdots\), \(n_k\) of the \(k\)th kind is \(\frac{n!}{n_1!n_2!\cdots n_k!}\)
Here \(n = 14\), \(n_1=1\) (beans), \(n_2 = 6\) (olives), \(n_3=7\) (carrots)
The number of arrangements \(N=\frac{14!}{1!×6!×7!}\)
We know that \(n!=n\times(n - 1)\times\cdots\times1\), so \(14! = 14\times13\times12\times11\times10\times9\times8\times7!\)

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Since \(6!=6\times5\times4\times3\times2\times1 = 720\)

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Answer:

\(12012\)