# Thread: Another Set Counting Question

1. ## Another Set Counting Question

This question is taken from a book. I have posted it because I would
like to understand how to complete it to do a similar problem for my homework.

In a survey of 1000 households, washing machines, vacuum
cleaners and refrigerators were counted. Each house had at least
one of these appliances.

400 had no refrigerator
380 had no vacuum cleaner
542 no washing machine
294 had both a vacuum cleaner and washing machine
277 both a refrigerator and a vacuum cleaner
190 both a refrigerator and a washing machine.

How many households had all three appliances?

How many had only a vacuum cleaner?

Thanks for any guidance.

2. Originally Posted by barc0de
This question is taken from a book. I have posted it because I would
like to understand how to complete it to do a similar problem for my homework.

In a survey of 1000 households, washing machines, vacuum
cleaners and refrigerators were counted. Each house had at least
one of these appliances.

400 had no refrigerator
380 had no vacuum cleaner
542 no washing machine
294 had both a vacuum cleaner and washing machine
277 both a refrigerator and a vacuum cleaner
190 both a refrigerator and a washing machine.

How many households had all three appliances?

How many had only a vacuum cleaner?

Thanks for any guidance.
Draw a Venn diagram.

Put a pronumeral a, b, c, d, e, f, g in each part.

You can figure out where I did this to get:

a + b + c = 400 .... (1)

a + e + g = 380 .... (2)

c + f + g = 542 .... (3)

b + d = 294 .... (4)

d + f = 277 .... (5)

d + e = 190 .... (6)

a + b + c + d + e + f + g = 1000 .... (7)

Solve equations (1), (2), (3) ...... (7) simultaneously for a, b, c, .... g (not too tough).

d = 83.

There are probably easier ways but this is how I generally do it.

3. Is there an easier way?

4. Originally Posted by barc0de
Is there an easier way?
No. That way is very easy.

5. ## An easier way

There is a quite easy way to do it, this question belongs to the set category.
<br><br>Note the following identity, if A, B, C are sets, then
n(A∪B∪C)=n(A)+n(B)+n(C)-n(A∩B)-n(B∩C)-n(A∩C)+n(A∩B∩C)
<br><br>in this question, let people having washing machine be set A, vacuum B, refrigerator C. so<br>n(A)=1000-542=458
<br>n(B)=1000-380=620
<br>n(C)=1000-400=600
<br>n(A∩B)=294<br>n(B∩C)=277<br>n(A∩C)=190
<br>n(A∪B∪C)=1000&nbsp; <br><br>n(A∩B∩C)=1000-(458+620+600)+(294+277+190)=83

<br><br>So next time you can just sub in the values into the equation for same kind of problem and it wont take too much time.

6. How would I then determine how many have a vacuum only?

I would expect n(B) - n(A n B) - n(B n C) - n(A n B n C) to give the correct result.

I get 620 - 294 - 277 - 83 = -34

Where am I going wrong?

7. I believe 83 to be correct but I cannot understand why the rest doesn't tally up.

Can anyone offer any guidance please?

8. the best way to solve this problem is to use Ven diagram

9. I have done. But since the figures I have posted are the ones on the diagram it is no help

10. Originally Posted by barc0de
I have done. But since the figures I have posted are the ones on the diagram it is no help
a = 57, b = 211, c = 132, d = 83, e = 107, f = 194, g = 216.

11. Originally Posted by barc0de
How would I then determine how many have a vacuum only?

I would expect n(B) - n(A n B) - n(B n C) - n(A n B n C) to give the correct result.

I get 620 - 294 - 277 - 83 = -34

Where am I going wrong?
you got wrong in the last step, coz following the above identity, it should be
n(B) - n(A n B) - n(B n C)+ n(A n B n C)=620-294-277+83=132

you can interpret it this way: as you have eliminated people who have both a vacuum and a washing machine+people who have both a vacuum and a refrigerator, you have eliminated twice people who have all the three appliances, so you should add back instead of eliminating it.

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# in a recent survey of 1000 houses. Each house had at least one of these products. 400 had refrigerator

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