Pizza party distributions

Probability Level 1

You are at a party with 4 of your friends (5 distinct people in total). You order a 12-cut pizza (12 identical slices).

How many distributions of pizza slices are there if each person gets at least one slice of pizza?

The answer is 330.

1 solution

Andy Hayes
May 12, 2016

Relevant wiki: Identical Objects into Distinct Bins

This problem can be modeled as $n=12$ identical objects distributed into $r=5$ distinct non-empty bins.

Using the formula for these kind of distributions : $\binom{n-1}{r-1}=\binom{11}{4}=330$

Thus, there are $\boxed{330}$ distributions of the pizza slices.

Brilliant has taught me yet another concept in math .this type of problem was on a contest of fairly difficult level called amatyc . Probably the same level of difficulty as the Aime,and Amc. It was worth the usual 2 points . Worded differently it asked the number of ways to put slashes between the letters AMATYCSML such that it divides into 4 * non empty * groups . Back then I had not learned the 🌟 &bar method but looking back this problem reeked of it lol and it makes it so much easier !! If I see a problem like this again it will be easily conquered . ( By the way the answer would be 8c3 )

Randin Divelbiss - 2 years, 5 months ago

But in your question, 9 letters, only 2 are identical, rest are distinct. How would that effect? How would we solve this mixed type. I know the formula for “Distinct objects in Distinct bins” [For distribution of n Distinct objects in r Distinct groups, it is- [r^n-C(r,1)(r-1)^n+C(r,2)(r-2)^n...+ (-1)^(n-1).C(r,r-1)] and also for “Identical objects in Distinct bins”..[ C(n-1,r-1) ]... so how to use these in mixed cases? May be @Andy Hayes can help?

Akash Tyagi - 2 years, 1 month ago

In questions like these, it is best to implement the idea of combinations instead of relying on a generalized formula. To answer the question given by Ranadin, no the solution is not a simple 8C3. Solution--> distribute the letters in the following fashion------> A--M--A--T--Y--C--S--M--L
The groups need to be non-empty, hence the number of ways of selecting spaces for 3 bars will be 8C3. Now we have to arrange the remaining letters in 9 spots. The number of ways to arrange them would be 9!/2!2!, as 'A' and 'M' repeat 2 times each. The final answer will be 8C3*9!/2!2!. Do let me know if this seems correct or not.

Nakshat Pandey - 7 months, 2 weeks ago

Pizza party distributions

The answer is 330.

1 solution

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