Question

Rachel is arranging 12 cans of food in a row on a shelf. She has 8 cans of beets, 3 cans of corn, and 1 can 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

Given:

Total number of food cans = 12

Cans of Beets = 8

Cans of corn = 3

Can of beans = 1

To find:

How many distinct orders can the cans be arranged if two cans of the same food are considered identical.

Solution:

To find the distinct ways arrangement, we have a formula:

$\text{Number of distinct ways}=\dfrac{n!}{r_1!r_2!...r_k!}$        ...(i)

Where, n is the number of objects and $r_1,r_2,...,r_k$ are repeated objects.

Total number of food cans is 12. So, $n=12$.

She has 8 cans of beets. So, $r_1=8$

She has 3 cans of corns. So, $r_2=3$

She has 1 can of beans. So, $r_3=1$

Substituting these values in (i), we get

$\text{Number of distinct ways}=\dfrac{12!}{8!3!1!}$

$\text{Number of distinct ways}=\dfrac{12\times 11\times 10\times 9\times 8!}{8!\times 3\times 2\times 1\times 1}$

$\text{Number of distinct ways}=\dfrac{11880}{6}$

$\text{Number of distinct ways}=1980$

Therefore, the number of distinct orders to arrange the cans is 1980.