I am a potato ~_~

Probability Level 3

If you are given a dozen of identical potatoes and 4 boxes, in how many ways can you arrange the potatoes in boxes such that no box is empty ?

This problem is a part of the set Vegetable combinatorics

The answer is 165.

4 solutions

Asher Joy
Jul 27, 2014

Using Balls&Urns, you get a + b + c + d = 12. But balls&urns includes a, b, c, d being 0. So just add 1 to each of a, b,c,d to get a+1+b+1 + c + 1 + d +1 = 12, or that a+b+c+d = 8. Then use the regular balls&urns to get (8+3)C3 = 165.

get it by using,the formula:-(n+r-1)C(r-1)_where n=8;r=4

sachin mittal - 6 years, 10 months ago

Can you explain what is 'C' in (8+3)C3=165 and why 165?

jomari ison - 6 years, 10 months ago

nCr ('n' Choose "r") ${n \choose r}$ Refer to http://www.mathwords.com/c/combination_formula.htm

Sunil Bn - 6 years, 10 months ago

Cody Johnson
Jul 24, 2014

Generating functions.

Josh Speckman
Sep 6, 2014

We know that there must be at least $1$ potato in each box, so we put $1$ potato in each box. That leaves us with 8 more potatoes. Then, we use stars and bars to determine that the answer is $\dbinom{11}{8} = \dfrac{11!}{8! 3!} = \dfrac{11 \cdot 10 \cdot}{3 \cdot 2 \cdot 1} = 11 \cdot 5 \cdot 3 = \boxed{165}$ .

Edmund Berry
Sep 30, 2014

This question is badly worded. The answer clearly depends on whether or not the boxes are distinguishable, and this information is not supplied by the problem's text.

The answer supplied (165) and the worked-out solutions provided below assume that the boxes ARE distinguishable.

If the boxes ARE NOT distinguishable, then the problem is moderately harder and requires use of partitions: http://en.wikipedia.org/wiki/Partition (number theory)

I am a potato ~_~

The answer is 165.

4 solutions

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