Farm auction distribution

At a farm auction, various pieces of farm equipment and livestock are bid on and sold.

Today, there are 7 distinct items up for auction, and there are 3 farmers bidding on them. One of the farmers declares that he will be bid on at least 2 of the items (not specifying which ones), and he won't be outbid.

If this farmer is telling the truth, then how many ways can the items be distributed among farmers?

Assume that all items are sold.


The answer is 1611.

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1 solution

Andy Hayes
May 10, 2016

Relevant wiki: Distinct Objects into Distinct Bins

In this problem, the farming items are modeled as "distinct objects", and the farmers are modeled as "bins."

Let U U be the set of all distributions of 7 7 distinct objects into 3 3 distinct bins. U = 3 7 = 2187 |U|=3^7=2187 .

Let A A be the set of all distributions in which the 1st bin receives exactly 0 0 items, and the 7 7 items are distributed among the remaining bins. This is equivalent to the set of all distributions of 7 7 distinct objects into 2 2 distinct bins. A = 2 7 = 128 |A|=2^7=128

Let B B be the set of all distributions in which the 1st bin receives exactly 1 1 item, and the remaining 6 6 items are distributed among the remaining bins. This is equivalent to the set of all distributions of 6 6 objects, chosen from 7 7 distinct objects, into 2 2 distinct bins. B = ( 7 6 ) 2 6 = 448 |B|=\binom{7}{6}2^6=448 .

U ( A B ) U\setminus(A\cup B) is the set of all distributions in which the 1st bin receives at least 2 items.

Note that A B = A\cap B=\emptyset , so A B = A + B = 128 + 448 = 576 |A\cup B|=|A|+|B|=128+448=576 .

Then U ( A B ) = 2187 576 = 1611 |U\setminus(A\cup B)|=2187-576=\boxed{1611} .

Since, one of the farmers is supposed to be telling the truth, he will get at least two items in 7C2 ways. The remaining 5 can be distributed among 3 farmers in 3^5 ways, So, total number of distributions = 7c2*3^5= 5103 .....What is wrong in this approach?

Nagbhushan Gangule - 2 years, 2 months ago

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further, if we don't fix up the first farmer as bin A..then the solution becomes 5103*3 = 15309

Nagbhushan Gangule - 2 years, 2 months ago

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Actually one particular farmer is fixed. So bin A is fixed. If it was “any farmer” has to have at least 2 items, then 3 factor will be multiplied but otherwise not. Or more precisely, here no because farmer/bin is fixed. Anyways, answer given by Mr Gangule is wrong. I have explained it in reply to him.

Akash Tyagi - 2 years, 1 month ago

This is not the correct way to approach these “at least” questions in DISTINCT in DISINCT cases. It causes overcounting. Here is the correct way to go about it. Since the particular farmer[Lets call him John!] (And yes we have to assume John is telling the truth)has promised to outbid others on at least 2 items: It means he will go home with either 2 or 3 or 4 or 5 or 6 or 7 items.

Case 1: When John wins only 2 items in C(7,2)=21 ways, 5 will be distributed to other 2 farmers in 2^5=32 ways. So total in this one case=21*32=672

Case 2: John wins only 3 items in C(7,3)=35 ways and remaining 4 items can be distributed to the other 2 in 2^4=16. So total ways in this case=35*16=560

Case 3: John wins only 4 items. So, C(7,4)=35 ways and remaining 3 items can be distributed to the other 2 in 2^3=8 ways. So total on this case=35*8=280 ways.

Case 4: John wins only 5 items. So, C(7,5)=21 ways and remaining 2 items can be distributed to the other 2 in 2^2=4 ways. So total on this case=21*4=84 ways.

Case 5: John wins only 6 items. So, C(7,6)=7 ways and remaining 1 item can be distributed to the other 2 in 2^1=2 ways. So total on this case=7*2=14 ways.

Case 6: John wins all 7 items. So that’s C(7,7)=1 way.

Hence total is= 672+560+280+84+14+1=1611.

Akash Tyagi - 2 years, 1 month ago

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this is the best explanation for me to understand👍

Justin Tang - 1 year, 10 months ago

The farmer auctioning for 2 items can have those in 7C2 ways (since he doesn't specify which). The remaining 5 items can then be distributed to the 2 farmers in 2^5 ways, thereby giving the number of possibilities as (7C2)*(2^5) = 672. What am I getting wrong?

Ali Ahmed - 2 months ago

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