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John and Carlos play a game of Scrabble.

a. John has the tiles A, B, E, E, P, T and Z on his tile rack. How many words of seven tiles can he make if he can also make nonsense words like BEEAPTZ ?

b. Carlos has the tiles E, F, O, O, R and R on his tile rack. How many words of six tiles can he make if he can also make nonsense words?

c. How many (nonsense) words can Carlos make that begin with a vowel?

d. In the bag there are ten more tiles. They are all different. John's tile rack is empty so he draws five more tiles from the bag. He puts them on the tile rack in the order he drew them. How many different (nonsense) words can John get on his tile rack?

 Oct 14, 2019
 #1
avatar+1808 
+3

Solutions:

 

a) John has the tiles A, B, E, E, P, T and Z on his tile rack. How many words of seven tiles can he make if he can also make nonsense words like BEEAPTZ ?

 

Answer: He can make (7!/2!) = 2520 words.  The two (2) “E” letters are indistinguishable and the redundancies are factored out by dividing by the factorial(s) of the duplicates.

 

b) Carlos has the tiles E, F, O, O, R and R on his tile rack. How many words of six tiles can he make if he can also make nonsense words?

 

Answer: (6!/2!/2!) = 180 words. This is the same algorithm as described above.

 

 c) How many (nonsense) words can Carlos make that begin with a vowel?

 

Answer: This would depend on the number of tiles, the number of vowels, and the number of duplicates Carlos has on his rack.  With the sample space of E, F, O, O, R and R, then starting with E,  5!/2!/2! = 30. Starting with O, 5!/2! =60. The sum of these two sets is the number of words he can make.  

 

d) In the bag there are ten more tiles. They are all different. John's tile rack is empty so he draws five more tiles from the bag. He puts them on the tile rack in the order he drew them. How many different (nonsense) words can John get on his tile rack?

 

Answer: With all distinguishable tiles, then nPr(10, 5) = 30240 different (nonsense) words.

 

GA

 Oct 14, 2019
 #2
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+1

Thanks for the reply, I appreciate it a lot! It's very clear and straightforward. I don't understand one thing though. You say that the two E's are indistinguishable, so shouldn't it be (6!/2!) instead then? Sorry if this is a stupid question, I'm kind of confused.

 Oct 14, 2019
 #3
avatar+1808 
+3

The problem with reducing the count by dropping one of the letters is it reduces the sample space. Each letter switches place with every other letter multiple times (half of all counts) as the entropy decreases to zero. Each E will switch places with every other letter and all of these events are unique, except for when the E’s switch with each other. 

 

Dropping the count by one (1) and removing the E, means the one of the E’s will not switch with any other letter. By keeping the E and its counts with all letters, it remains only to factor out the redundancies caused by switches with its twin.

 

This brings up an interesting point: There are times when it should not be factored out.

I note the title to your post is “Scrabble probability questions” (Great descriptive title. It’s one of the reasons I answered your post.) There is no probability question here, but counting is a precursor to probability. When counting for probability the redundancies are always counted.

 

Consider these questions: John randomized his seven (7) tiles and creates a word.

What’s the probability the word starts with Z?   Answer: (1/7)

What’s the probability the word starts with E? Answer: (2/7)

It’s easy to see that it’s counted here, and it’s always counted in combination and permutation problems related to probability. 

 

 

GA

GingerAle  Oct 14, 2019
 #4
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+1

Ah thank you so much, I understand now!

Guest Oct 15, 2019

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