The first boy enter the room and found 6 hats. He get one and then the second boy enters and also get one hat. The next day, they both find out that they got someone's else hat. In how many ways they could have gotten the wrong hats?
The first boy enter the room and found 6 hats. He get one and then the second boy enters and also get one hat. The next day, they both find out that they got someone's else hat. In how many ways they could have gotten the wrong hats?
The first boy could have taken any 5 wrong hats out of a total of 6.
Given he didn't take the hat of the next boy, the next boy could have taken any 4 wrong hats out of 5.
Given he did take the hat of the next boy, the latter would have got the wrong hat for sure, that is 5 ways.
Can you give the problem a try now?
so in
case 1:
1st boy take the other boy hat = 1C1 = 1
2nd boy take any other hat = 5C1 = 5
case 2:
1st boy didn't take the hat of the other boy = 4C1 = 4
2nd boy take any other hat but not his hat = 4C1 = 4
(5x1) + (4x4) = 21 possible ways
is the solution correct? thanks for the help