# Thread: problem on inclusion exclusion

1. ## problem on inclusion exclusion

Please describe the way to find the answer for the following question.

This problem is based on inclusion exclusion principle

50 cars were assembled in a factory. The options available were radio, AC,
power steering. It is known that 20 of the cars have radios, 12 of them have
AC and 10 of them have power steerings. 5 of them have all three options.
Determine at least how many cars do not have any options at all.

2. Originally Posted by sudeepmansh
Please describe the way to find the answer for the following question.

This problem is based on inclusion exclusion principle

50 cars were assembled in a factory. The options available were radio, AC,
power steering. It is known that 20 of the cars have radios, 12 of them have
AC and 10 of them have power steerings. 5 of them have all three options.
Determine at least how many cars do not have any options at all.
B - AC
C - power steering

Let n(cars that do not have any options at all)=x

n(A U B U C) + x = 50
so, x = 50 - n(A U B U C)
min(x) = 50 - max(n(A U B U C))

So what is max(n(A U B U C))?
n(A U B U C) = n(A) + n(B) + n(c) - n(AB) - n(BC) - n(CA) + n(ABC)

So all we need is min of each of n(AB), n(BC), n(CA) - So what is it?

3. Originally Posted by sudeepmansh
Please describe the way to find the answer for the following question.

This problem is based on inclusion exclusion principle

50 cars were assembled in a factory. The options available were radio, AC,
power steering. It is known that 20 of the cars have radios, 12 of them have
AC and 10 of them have power steerings. 5 of them have all three options.
Determine at least how many cars do not have any options at all.
It doesn't seem that complicated.

5 cars have all three options.
If all other cars have only one option apiece until no options are left to install, this gives 20-5=15 with radios only, 12-5=7 with AC only, and 10-5=5 with power steering only.

Add 5 with all three options, 15 with radio only, 7 with AC only, and 5 with power steering only to get 32 maximum cars with options. 50-32=18, so at least 18 will have no options. There could be more with no options if some of the options are doubled up, but that is beyond the scope of this question.

|AUBUC| = |A|+|B|+|C|-|AnB|-|BnC|-|AnC|+|AnBnC|

The question is little bit tricky.

6. Originally Posted by sudeepmansh

|AUBUC| = |A|+|B|+|C|-|AnB|-|BnC|-|AnC|+|AnBnC|

The question is little bit tricky.
Can you answer the question I asked in my post? After that it's just a matter of putting values in the above formula.

7. No more information has been given in the question. Its a University question paper problem.

But one thing, the problem makes use of "among them". Is there any hidden information in that?

8. Originally Posted by sudeepmansh
No more information has been given in the question. Its a University question paper problem.

But one thing, the problem makes use of "among them". Is there any hidden information in that?
I was referring to my first post plz. Have a look and see if you follow that.

9. yes i have seen. But whats the value for AnB, BnC and AnC. No more information about that has been given in the problem.

10. Originally Posted by sudeepmansh
yes i have seen. But whats the value for AnB, BnC and AnC. No more information about that has been given in the problem.
Plz see oldguynewstudent post

Hint: AnBnC is a subset of AnB. Correct? So what can be the min value of n(AnB)? What about others?

This question is all about the argument above. And oldguynewstudent has given you the ans (he might not have put it formally though)

11. Ok I got it. Thanks