text: # Hard Puzzles
  
  ### Sample 1
  
  At a party you overhear another guest asking the age of the hosts' 3 children. The host tells the guest that the product of his kids' ages is 72 and the sum of their ages is the same as the guest's house number. Now, the other guest obviously knows her own house number but, after thinking for a bit, she is forced to ask the host for additional information. The host then says that his oldest child likes strawberry ice cream and the other guest is able to say their ages. Even though you don't know the guest's house number, you have enough information.
  
  Assuming that all of the ages are whole numbers (1, 2, 3 not 2 1/2), How old is his oldest child?
  
  Solution:  8 years old
  
  This is much like Student Ages question although a bit easier. You first must factor 72 which breaks down into 1x2x2x2x3x3. Combining these numbers to make 3 ages reveals 11 combinations: (1 2 36), (1 3 24), (1 4 18), (1 6 12), (1 8 9), (2 2 18), (2 3 12), (2 4 9), (2 6 6), (3 3 8), and (3 4 6).
  
  The host says that their sum is equal to the house number of the guest. If the guest has to ask for more information then some of the combinations must add up to the same number. By adding up all of the combinations, we find that (2 6 6) and (3 3 8) both add up to 14 so the house number must be 14 and one of the combinations must be correct.
  
  But the guest is unable to get the answer. Then the host says that the oldest likes strawberry ice-cream which tells you that there is an oldest. Since (2 6 6) implies that two of his children have the same age, the answer must be (3 3 8).
  
  ### Sample 2
  
  How many ways are there to arrange the word “SYSTEMATIC” so that each S is followed by a vowel (including Y)?
  
  Solution:
  
  4 x 3 x P(8,6) = 241920
  
  (4 ways to assign a vowel to follow the first “S”, 3 ways to assign a vowel to follow the second “S”; note that because the S’s aren’t distinct, this is double-counting some pairs. Then if we consider these pairs as single characters, we have a total of 8 objects, 6 of which – the non-pairs – we need to place.  The double counting is canceled out by the fact that we will have two ways to insert the pairs into the permutation.)