A one-to-many binary relationship allows an occurrence of the entity type on the "one side" of the relationship to be associated with . a a single occurrence of the entity type on the "many side" of the relationship b multiple occurrences of the entity type on the "one side" of the relationship c multiple occurrences of the same entity type d multiple occurrences of the entity type on the "many side" of the relationship

This question involves a class named textfile that represents a text file. public class textfile { private string filename; private string filename; private arraylist words; // constructors not shown // postcondition: returns the number of bytes in this file public int filesize() { } // precondition: 0 < = index < words.size() // postcondition: removes numwords words from the words arraylist beginning at // index. public void deletewords(int index, int numwords) { } // precondition: 0 < = index < = words.size() // postcondition: adds elements from newwords array to words arraylist beginning // at index. pub lic voidaddwords(int index, string[] newwords) { } // other methods not shown } complete the filesize() method. the filesize() is computed in bytes. in a text file, each character in each word counts as one byte. in addition, there is a space in between each word in the words arraylist, and each of those spaces also counts as one byte. for example, suppose the words arraylist stores the following words: { mary had a little lamb; its fleece was white as snow. } the filesize() method would compute 4 + 3 + 1 + 6 + 5 + 4 + 6 + 3 + 5 + 2 + 5 as the sum of the lengths of each string in the arraylist. the value returned would be this sum plus 10, because there would also be 10 spaces in between the 11 words. complete the filesize() method below: // postcondition: returns the number of bytes in this file public int filesize() { }