Question

Austin is arranging 14 cans of food in a row on a shelf. He has 5 cans of corn, 1 can of olives, and 8 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)?

Answer 1

Answer:

18018

Step-by-step explanation:

Given:

Number of cans = 14

Number of cans of corn = 5

Number of cans of olives = 1

Number of cans of beans = 8

When there are $n$ objects to be arranged in order, out of which  $r_{1}$ objects are of one kind, $r_{2}$ objects are of another and so on, then the number of different arrangements is given as:

$\frac{n!}{r_{1}!\times r_{2}!\times ...r_{k}!}$

Here, $n = 14, r_{1}=5,r_{2}=8$

Therefore, the distinct orders in which the cans can be arranged is $\frac{14!}{5! \times 8!} = 18018$