Mr. X is a famous magician. He has 1 to 100 cards at his disposal. He puts them in 3 different boxes-red,green,blue. Now, he requests the audience to blindfold him and select 1 card each from any 2 different boxes and tell him the sum of the numbers obtained in those 2 cards. Say one card has 5 in it and the other 83, then the audience will say 88. Just by listening to the sum Mr. X points out the box that has not been selected.
In how many ways, can Mr. X distribute the cards in the 3 boxes such that each box has at least one card?
The answer given is 12, but I am getting 6.
The only way the magician can know before hand if he uses the remainders of three. Say he groups in such a way that 1st box contains say multiples of three, 2nd box numbers which give remainder 1 when divided by 3, 3rd box remainder 2. Now when the audience adds the two numbers he/she will get only 3 combinations: 1+0=1; 1+2=0; 2+0=2. So, the magician can immediately identify the boxes. This he can do 3!=6 ways in 3 different boxes (permutation of 012). Hence the answer is 6.