| [ QuizWit ] in KIDS 글 쓴 이(By): guest (legacy) 날 짜 (Date): 1995년08월29일(화) 10시29분19초 KDT 제 목(Title): [R]Arbeit <- Application of above. Choose one car and construct non-empty Ga's from its plate numbers. From the ristrictions, |Ga| = 2 and 2<= |{Ga}| <= 4. 1. |{Ga}| = 2 if you choose a from Ga1 and b from Ga2, they should have only one common number a3 but by the requirement there should be another car c which also has a3 in its plate among Ga1 or Ga2. It violates the disting- tiveness of Ga's. 2. |{Ga}| = 4 The possible combinations between elements from different Ga's are 24 ways. However one common number cover only three cars. In other words, we need 8 more numbers. But since we already consumed 4 numbers, that is impossible. (Assume the plate number are chosen among from 0 to 9). 3. |{Ga}| = 3 If you try you may find that it is possible. e.g. 1123 1145 1167 2246 2257 3347 3356. Anyway, thus the number of cars in the parking lot is 1+2*3 = 7. |