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    <p>First you need a method to check if one word is an anagram of a second word. There are many possible solutions (search for "Objective-C anagram"). This is essentially the method from <a href="https://stackoverflow.com/a/13465672/1187415">https://stackoverflow.com/a/13465672/1187415</a>, written slightly differently:</p> <pre><code>- (BOOL)does:(NSString*)longWord contain:(NSString *)shortWord { NSMutableString *longer = [longWord mutableCopy]; __block BOOL retVal = YES; // Loop over all characters (letters) in shortWord: [shortWord enumerateSubstringsInRange:NSMakeRange(0, [shortWord length]) options:NSStringEnumerationByComposedCharacterSequences usingBlock:^(NSString *substring, NSRange substringRange, NSRange enclosingRange, BOOL *stop) { // Check if letter occurs in longer word: NSRange letterRange = [longer rangeOfString:substring]; if (letterRange.location != NSNotFound) { // Yes. Remove from longer word and continue. [longer deleteCharactersInRange:letterRange]; } else { // No. Set return value to NO and quit the loop. retVal = NO; *stop = YES; } }]; return retVal; } </code></pre> <p>Examples:</p> <ul> <li><code>[self does:@"abandoned" contain:@"bond"] = YES</code></li> <li><code>[self does:@"abandoned" contain:@"sea"] = NO</code>, because there is no "s" in the first word.</li> <li><code>[self does:@"abandoned" contain:@"noon"] = NO</code>, because "noon" has 2 letters "o", but the first word has only one "o".</li> </ul> <p>Then you can proceed as follows:</p> <pre><code>NSArray *englishWords = ...; // Your array of english words NSString *inputWord = @"abandoned"; // The input string NSString *middleLetter = [inputWord substringWithRange:NSMakeRange([inputWord length]/2, 1)]; NSPredicate *predicate = [NSPredicate predicateWithBlock:^BOOL(NSString *word, NSDictionary *bindings) { // Word must have at least 4 letters: if ([word length] &lt; 4) return NO; // Word must contain the middle letter: if ([word rangeOfString:middleLetter].location == NSNotFound) return NO; // Word must contain only letters of the input word: if (![self does:inputWord contain:word]) return NO; return YES; }]; NSArray *matchingWords = [englishWords filteredArrayUsingPredicate:predicate]; NSLog(@"%@", matchingWords); </code></pre>
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    1. PORearrange Letters from Array and check if arrangement is in array
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    1. USMartin R
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    1. COBut will this allow me to go through all the possibilities of words in the bigger word... I don't quite understand the first part of your coding... Does that essentially go through every possibility of words in the greater 9 letter word?? Thanks very much for your help though!!
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    2. CO@falky: I have added some comments and examples to the code. If I understand your question correctly, this should do exactly what you need. Why don't you try it and check if it works for you? - But of course you are welcome to ask more questions if required!
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    3. COOh ok!!! Now I've got it. That makes much more sense. That's a really good idea actually! So your are very basically checking if a word from the dictionary is in the 9 letter word. Very smart! Thanks very much for your help with this and hte other question. I can upvote now! I'll test it out later on tonight!
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