How many distinct words, consisting of 2 letters, can be formed from the word ALGEBRA? I have tried to reason generally by observing that the way of arranging n letters such that no letter is used more than once to make a word n-length was n!. As we have $m$ letters to choose $n$ from, and we are interested in distinct groups. This can be done in ${m \choose n}$ ways.

Thus, I claim

$$n! \cdot {m \choose n}= \frac{m!}{(m-n)!}$$ As I employ this on my special case I also remember that there are two A:s. So I multiply $\frac{m!}{(m-n)!}$ by $2!$, which gives 84. But it is plain wrong! Would anyone be so kind as to help this poor fellow.


Hint: Distinguish words composed of two different letters and words composed of twice the same letter.

Now what is the total number of two-letter words?

Hint 2 (different approach): What are the options for the first letter? Given a first letter, how many words of two letters can still be made (keep in mind that not all letters may be used twice)?

  • $\begingroup$ I'm afraid I don't follow when you say: "What is the total number of possibilities?" $\endgroup$ – user70800 Apr 4 '13 at 14:37
  • $\begingroup$ Well, should that not be $2^{7}=128$? $\endgroup$ – user70800 Apr 4 '13 at 14:46
  • $\begingroup$ There are 7 possibilities for the first letter. As for the second letter we have 6 possibilities, so in total we have $7\times6=42$ possibilities. But, since there are two A:s we have to divide (or so I suspect) by $2$. $\endgroup$ – user70800 Apr 4 '13 at 15:17
  • $\begingroup$ @user7080: following the first approach, there is one two letter word with both the same, AA. Then there are six distinct letters, you can choose one for the first position, and another for the second. $\endgroup$ – Ross Millikan Apr 4 '13 at 15:39
  • $\begingroup$ So I should subtract by 11 then? $\endgroup$ – user70800 Apr 4 '13 at 15:55


Ways of choosing $2$ letters to make a word $7 \choose 2$

You can permute them in $2!$ ways. Now that you have $2$ A's, delete the possible words formed by letter A.

  • $\begingroup$ Yes. That's why I'm deleting the extra words created by one more $A$. $\endgroup$ – Inceptio Apr 4 '13 at 15:42
  • $\begingroup$ Are you saying $2!\times{7 \choose 2}-10$? As I can make out, we have that there are 5 (2 letter) words that start with A and 5 (2 letter words) which end with A $\endgroup$ – user70800 Apr 4 '13 at 15:44

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