<p>A,B,C,D,E,F,G </p>
<p>A list consists of all possible three-letter arrangements formed by using the letters above such that the first letter is and one of the remaining letters is If no letter is used more than once in an arrangement in the list and one three-letter arrangement is randomly selected from the list, what is the probability that the arrangement selected will be </p>
<p>(A) 1/5
(B) 1/6
(C) 1/9
(D) 1/10
(E) 1/12</p>
<p>how do i do this problem without actually writing out the individual 3-letter arrangements?</p>
<p>Please fix the question..ur missing pieces..."one of the remaining letters is ?X?"</p>
<p>yea... what are the restictions on the first and last letters?</p>
<p>what is the probability that the arrangement selected will be <strong><em>DCA</em></strong>.</p>
<p>so sorry about that.</p>
<p>Sorry, but you still haven't answered the question in post#3. What is the restriction on the first letter? What is the restriction on 'one of the remaining letters'?</p>
<p>A list consists of all possible three-letter arrangements formed by using the letters above such that the first letter is <strong>D</strong> and one of the remaining letters is <strong>A</strong>. If no letter is used more than once in an arrangement in the list and one three-letter arrangement is randomly selected from the list, what is the probability that the arrangement selected will be <strong><em>DCA</em></strong></p>
<p>thanks</p>
<p>The #possible 3-letter arrangements that satisfy the conditions are:</p>
<h1>with A in pos<em>2 + #with A in pos</em>3</h1>
<p>= #(DA<em>) + #(D</em>A)
= (7-2) + (7-2) = 10</p>
<p>The chance of picking one specific arrangement DCA out of the 10 possible is 1/10 .</p>
<p>There were 7 original letters; two of them (D & A) are not legal candidates for this slot, since they are reserved for other slots.</p>
<p>optimizerdad, i thought only D isn't a legal candidate for spots 2 and 3. A can be in either spot 2 or 3.</p>
<p>by the way, you got the right answer.</p>
<p>Evanescent:
'A can be in either spot 2 or 3'..</p>
<p>Right - that's why we compute all possibilities where it's in spot 2 ( = # DA<em>) and all possibilities where it's in spot 3 ( = # D</em>A), and then add the number of such possibilities.</p>