Question 1191631
.


The word EAGLES has 6 letters; of them, one letter E is repeated and have a multiplicity 2.


When we analyze the number of different words of 4 letters formed from the given word (symbols),

we should distinct two different cases.



<pre>
<U>Case 1</U>.  All 4 letters in the final word are different.

         In this case, we have only 5 distinct letters to choose from, (E, A, G, L S);

         therefore, the number of possible words to form is  5*4*3*2 = 120  in this case  (the order of letters does matter !);



<U>Case 2</U>.  In the final word, we have 2 identical letters E and any 2 of the remaining 4 letters.

         In this case, we can select these two remaining letters by  {{{C[4]^2}}} = 6 different ways,

         and we can arrange then 4 letters with two repeating undistinguishable Es by  

             {{{4!/2!}}} = {{{24/2}}} = 12 different distinguishable ways.


         Combining everything altogether, we have 6*12 = 72 different words in Case 2.



Cases (1) and (2) are the disjoint sets of words; therefore, the answer to the problem's question is 120 + 72 = 192.


<U>ANSWER</U>.  192 different / (distinguished) words of the length 4 can be formed.
</pre>

Solved.