How Many Ladies Had All Four Items at the Church Fete?

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Discussion Overview

The discussion revolves around a combinatorial problem involving a group of 100 ladies at a church fete, specifically focusing on how many of them must have had all four items: a white handbag, black shoes, an umbrella, and a ring. The scope includes mathematical reasoning and problem-solving strategies.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests that the problem is under-determined and seeks clarification on how to approach it.
  • Another participant proposes interpreting "must have had" as a lower bound, arguing that at least 10 ladies must have had all four items based on their reasoning involving overlaps among the groups.
  • A different participant recommends simplifying the problem by first solving a version that only considers two items to build understanding for the original problem.
  • One participant acknowledges a misunderstanding in their initial approach, realizing they were looking for a unique solution rather than a bound.
  • Another participant introduces a complementary approach, calculating the number of ladies without each item and concluding that 10 ladies must have all four items, while noting the maximization of those with three out of four items.

Areas of Agreement / Disagreement

Participants express differing views on how to interpret the problem and arrive at potential solutions. There is no consensus on a definitive answer, as multiple approaches and interpretations are presented.

Contextual Notes

Some assumptions about the distribution of items among the ladies may not be explicitly stated, and the problem's conditions lead to various interpretations regarding the minimum number of ladies with all four items.

ilvpat
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Q:
Out of 100 ladies attending the church fete, 85 had a white handbag; 75 had black shoes; 60 carried an umbrella; 90 wore a ring, How many ladies must have had all four items?


I feel like the problem is under-determined. But if anybody can explain it to me it would be great. Thank you very much.
 
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It certainly looks that way, but we can reinterpret "must have had" as a lower bound, that is, we are interested in finding out what is the least number of ladies that had to have all for items!

Now, consider the ladies with rings, and those with handbags.
At least 75 of them must have had both (why?)

Now, furthermore consider the R&H ladies (those with rings AND handbg), and those with black shoes (S). Now, of the S population, only 25 max could not be wearing both R and H, since the 100 ladies can be divided into those having both R and H (at least 75) and those not having both (at most 25).

Thus, of the S ladies, at least 50 of them are R&H&S ladies.

Finally, only max 50 of the umbrella ladies U could not be R&H&S ladies; hence, at least 10 U's are also R&H&S ladies.

Thus, at least 10 ladies had all four objects, up to a maximum of 60.
 
Welcome to physicsforums ilvpat,

my advice is to try a simplified version of the problem.
Out of 100 ladies, 85 have a white handbag, 75 have black shoes.
How many ladies must have both items?

I think if you can solve this simplified problem, you can also
solve the original one.
 
Last edited:
I see now. The way I interpreted the problem leads me to the wrong direction. I was looking for a unique solution rather than a bound. Thanks guys!
 
It might be simpler to look at complements:
15 are w/o a white handbag
25 are w/o black shoes
40 w/o umbrella
10 w/o ring

That's 90 without something, leaving 10 with everything.

(what happened? We maximized the number with 3 of the 4)
 

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