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:
[tex]\text{Number of distinct ways}=\dfrac{n!}{r_1!r_2!...r_k!}[/tex] ...(i)
Where, n is the number of objects and [tex]r_1,r_2,...,r_k[/tex] are repeated objects.
Total number of food cans is 12. So, [tex]n=12[/tex].
She has 8 cans of beets. So, [tex]r_1=8[/tex]
She has 3 cans of corns. So, [tex]r_2=3[/tex]
She has 1 can of beans. So, [tex]r_3=1[/tex]
Substituting these values in (i), we get
[tex]\text{Number of distinct ways}=\dfrac{12!}{8!3!1!}[/tex]
[tex]\text{Number of distinct ways}=\dfrac{12\times 11\times 10\times 9\times 8!}{8!\times 3\times 2\times 1\times 1}[/tex]
[tex]\text{Number of distinct ways}=\dfrac{11880}{6}[/tex]
[tex]\text{Number of distinct ways}=1980[/tex]
Therefore, the number of distinct orders to arrange the cans is 1980.
