Calculating Passengers without Caps or Eye Glasses: 36 Bus Riders

  • Thread starter Thread starter blunted
  • Start date Start date
  • Tags Tags
    Bus
AI Thread Summary
To determine how many passengers on the bus have neither caps nor eyeglasses, start with the total of 36 passengers. There are 8 passengers with caps, and 2 who have both caps and eyeglasses. Since 26 passengers do not have eyeglasses, this implies that 10 passengers do have eyeglasses (36 total minus 26 without). Using the formula for union of sets, calculate the total with either caps or eyeglasses, leading to the conclusion that 18 passengers have neither. The final answer is that 18 passengers do not have caps or eyeglasses.
blunted
Messages
23
Reaction score
0

Homework Statement


Out of 36 passengers in a bus. 8 have caps, 2 have both cap and eye glasses , 26 don't have eye glasses, how many passengers have neither caps nor eye glasses


Homework Equations





The Attempt at a Solution



A draw 2 intersected diagrams C and E. In C I write 6 dots, in the intersected point i write 2 dots, but I don't know how many should I write in Diagram E(Eyeglasses)
The formula: C U E = C + E - CinteresectE
 
Physics news on Phys.org
You could draw venn diagrams etc.. but also consider the algebraic approach using the relation you gave at the bottom of your post. Have you come across DeMorgan's Laws?
 
blunted said:

Homework Statement


Out of 36 passengers in a bus. 8 have caps, 2 have both cap and eye glasses , 26 don't have eye glasses, how many passengers have neither caps nor eye glasses


Homework Equations





The Attempt at a Solution



A draw 2 intersected diagrams C and E. In C I write 6 dots, in the intersected point i write 2 dots, but I don't know how many should I write in Diagram E(Eyeglasses)
The formula: C U E = C + E - CinteresectE

Perhaps you should think how many to put outside area E. You already have 6 outside that area.
 
Can't make a move..
 
36 passengers in a bus, 26 don't have eye glasses, so how many do?
 
Okay found it, thanks.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top