hilarious hat
-
Five persons A,B,C,D and E are seated in a circular arrangement. If each of them is given a hat of one of the three colours red, blue and green, then the numbers of ways of distributing the hats such that the person seated in adjacent seats get different coloured hats is
-
Inspiration - JEE Advanced 2019
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
2 solutions
Nice solution +1
Great explanation, simple to remember, this is the takeaway.
Since all three colors are used and none used more than two times, the only combination is 2 + 2 + 1 . Here are the variations.
( 2 r + 2 b + 1 g ) or ( 2 r + 2 g + 1 b )
( 2 b + 2 r + 1 g ) or ( 2 b + 2 g + 1 r )
( 2 g + 2 b + 1 r ) or ( 2 g + 2 r + 1 b )
5 × 6 = 3 0
You have copied the solution from Resonance
If A can't be moved you can only to put B and C alternating them, so the position is unique.
A could choose 3 colors, 2 colors for B, 1 color for C so there are 3x2x1=6 ways.
But A can be in 5 positions so there are 6x5 = 30 ways.
Log in to reply
you should upload it as a solution
Log in to reply
I think he has not got the question right btw are you a python programmer
Log in to reply
@Srijan Singh – yes any doubt ?
Log in to reply
@Razing Thunder – No would you like to take part in this
@Srijan Singh – Yes, I was wrong to choose the answer. The last part of the solution came to my mind correct only after I see the solution.
Log in to reply
@Lu Ca – Would you like to take par in this
Log in to reply
@Srijan Singh – I'm not so much experienced in programming.
1) No hat color can be used more than twice. If any color is used three times, two people with the same color hat will be next to each other.
2) With five seats to be filled, there will be one color used once and two used twice each.
3) Let Person A be the unique person alone in her hat color. There are five seats where Person A can be seated.
4) Let Person B be to Person A's right. There are two choices for Person B's hat color.
5) The choice for Person B forces the choices for the other three people.
Counting principle: 3 choices of solitary color × 5 choices of Person A's set × 2 choices of Person B's hat color = 30 possibilities.