# Permutation? how many ways can 3 different meals be served to 6 people?

6 people, 3 different types of meals. How do you find the total number of orders possible? Meals can be repeated

Update:

each dish can be repeated so its not exactly a permutation/combination problem but it still is a factorial type of problem.

my bad for making the problem kinda confusing.

I did find the answer from a friend the answer is something like (n+r-1)!/r!(n-1)!, in which r is the 6 (number of people), n is 3 (dishes to choose from), so 28

Thanks for the help though

Relevance

18, multiply them together

There are 3 possibilities for each person and there are 6 people, so 3x3x3x3x3x3 possibilities. But, the kitchen doesn't care what order they serve the meals, so you need to divide by the permutations. I'm a little rusty on how to do that right now :-)

• sv
Lv 7

Total ways are 3*3*3*3*3*3 = 3^6 =729

as any one of the three type can be served to any one of the people.

Since orders can be repeated, you will get 3^6 different permutations. That is 729.

P= n! / (n-r)!

P= n people !/ (people - dishes)!

P= 6!/ (6-3)!

P = 6*5*4*3*2*1/ 3!

P=6*5*4*3*2*1/ 3*2*1

3, 2 , 1 on both numerator and denominator will be cancelled...

so

P= 6*5*4

P=120

P = n! / (n - r)!

P = 6! / (6 - 3)!

P = 6! / 3! = 6 * 5 * 4 * 3 * 2 * 1 / 3 * 2 * 1

P = 6 * 5 * 4

P = 120 permutations.