Question 1200728: A restaurant offers it's customers a choice of 3 side dishes with each meal. the side dishes can be chosen from a list of fifteen possibilities with duplication allowed. For instance a customer can order two sides of mashed potatoes and one side of string beans. show that there are 680 possible choices for the three side dishes.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Method 1
We have 3 cases which I'll denote as
AAA
AAB
ABC
AAA refers to picking 3 of the same side dish.
There are 15 ways to do this case.
AAB is when you pick 2 of the same dish, and the third dish is something different.
Case AAB has 15*14 = 210 different possibilities.
This is because there are 15 choices for the "AA" portion and 14 choices for the B portion. The order doesn't matter so AAB is the same as ABA and also BAA.
Case ABC is when all three dishes are different.
The order doesn't matter.
We use the nCr combination formula.
n = 15
r = 3
n C r = (n!)/(r!(n-r)!)
15 C 3 = (15!)/(3!*(15-3)!)
15 C 3 = (15!)/(3!*12!)
15 C 3 = (15*14*13*12!)/(3!*12!)
15 C 3 = (15*14*13)/(3!)
15 C 3 = (15*14*13)/(3*2*1)
15 C 3 = (2730)/(6)
15 C 3 = 455
Many calculators have the nCr function built into them.
This is one online calculator that does such a task.
https://www.omnicalculator.com/statistics/combination
To summarize, we have these counts:
AAA = 15
AAB = 210
ABC = 455
Adding them up gets us: 15+210+455 = 680
This shows why we have 680 ways to select 3 dishes from a pool of 15 such that repeat dishes are allowed.
Eg: {mashed potatoes, mashed potatoes, string beans}
======================================================================================================================
Method 2
We have n = 15 dishes and r = 3 selections.
We'll use the Combination with Repetition formula mentioned in this link here
https://math.libretexts.org/Courses/Monroe_Community_College/MTH_220_Discrete_Math/7%3A_Combinatorics/7.5%3A_Combinations_WITH_Repetitions
Scroll down to where it says "Theorem 7.5.1"
The portion n+r-1 takes the role of the original n.
Meaning that we really have a pool of n+r-1 = 15+3-1 = 17 items and r = 3 selections.
Let's calculate the value of 17C3
n C r = (n!)/(r!(n-r)!)
17 C 3 = (17!)/(3!*(17-3)!)
17 C 3 = (17!)/(3!*14!)
17 C 3 = (17*16*15*14!)/(3!*14!)
17 C 3 = (17*16*15)/(3!)
17 C 3 = (17*16*15)/(3*2*1)
17 C 3 = (4080)/(6)
17 C 3 = 680
Interpretation:
Let's say you labeled the side dishes 1 to 15.
You'll have 15 slips of paper with those numbers to be placed in a hat.
Then let's say you add slips of paper labeled 16 and 17 to represent situations when we want a repeat dish.
16 will mean we repeat the lower number dish.
17 will mean we repeat the higher number dish.
Getting both 16 and 17 at the same time will lead to 3 dishes of the same type (since we have 2 repeats).
An outcome like {1,5,16} will mean "select dish1 twice, and dish5 once"
So we have {1,5,16} translate to "dish1,dish1,dish5"
The 16 told us to repeat the lower dish (which is dish1)
If that 16 was a 17, then we'd repeat the higher dish (which is dish 5)
Meaning that {1,5,17} translates to "dish1,dish5,dish5".
In other words:
{1,5,16} ---> dish1,dish1,dish5
{1,5,17} ---> dish1,dish5,dish5
Other examples:
{1,2,3} ---> dish1,dish2,dish3
{4,6,12} ---> dish4,dish6,dish12
{8,16,17} ---> dish8,dish8,dish8
{2,3,16} ---> dish2,dish2,dish3
{4,7,17} ---> dish4,dish7,dish7
If 16 and 17 aren't present, then list the 3 unique dish numbers as mentioned. Otherwise you'll have a repeat of some sort.
Having both 16 and 17 at the same time leads to two repeats, aka all 3 dishes are the same for that scenario.
To summarize, we have 17 numbers placed into the hat and randomly pull out 3 of them without replacement. The order of selections doesn't matter. That is why we have 17C3 = 680 possible outcomes.
Further Reading:
https://en.wikipedia.org/wiki/Combination#Number_of_combinations_with_repetition
https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)
https://www.mathsisfun.com/combinatorics/combinations-permutations.html
For the MathIsFun link, scroll down to the "combinations with repetition" section.
|
|
|
