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Code 39 barcodes are used to code simple text. T
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XXXXXXXXXXX XXXXXXXX Principle</strong>.</p><p>So XX XXX XXXXX strip XXX 2 XXXXXXXXXXXXX (XXXX or narrow), XXX the XXXXX space XXX 2 independent possibilities (XXXX or narrow), there's a XXXXX XX X*X=4 XXXXXXXXXXXXX for those XXX XXXXXXXX.</p><p>The XXXX strip XXX XXXXXXX two possibilities, so X*2*X=X; XXX XXXX space XXX another two possibilities, so 2*X*X*2=XX.</p><p>I think you get the XXXX. XXXX XXXXX this for XXX XXXX XXXXXXXX, each XX XXXXX XXX X XXXXXXXXXXXXX, XXX you'll get X*2*X*X*2*2*X*2*2= X^9 = XXX.</p><p>So XXXXX XXX <em><XXXXXX>XXX</strong></em> possible XXXXXXXXXXXX XXX XXXX bar XXXX, which XX XXXX than the 255 necessary.</p><p>b) This question XX almost the same, XXXXXX now, XXX two of the XXXXXXXX, XXX XXXXX XXX XXXX strips, XXXXX XX XXXX XXX possibility (wide). So, XXXX XX the XXXX thing you did XXXX time XXXXX XXX XXXXXXXXXXX Counting Principle, but remember that the first XXX last XXXXXX have XXXX one possibility:</p><p>1*2*2*X*2*X*X*2*1 = X^X = XXX</p><p>XX XXXXX XXX now XXXX<XXXXXX><em>128<strong><em> <strong></strong></em></strong></em></XXXXXX>possible XXXXXXXXXXXX XXX the bar XXXX, but that is just XXXXXX XXX the XXX character XXXXXXXX XXXX'XX asking XXX.</p><p>c) XX now XX XXXX XX find out how XXXX XX XXX XXXX<strong>XXXXXXX</strong> XXX wide strips XXX XXX wide XXXXX.</p><p>First, how many XXXX can we have on XXXXXXX XXX wide strips? XX have five total XXXXXX XX XXXXXX XXXX, XXX XX need XX XXXXXX XXX, so 5 choose X or C(5,X) = X! / (2! * (X-X)!) = XX. So XXXXX XXX XXXXXX to XXXXXXX the XXXXXX XXXX XXXX XXXXXXX XXX are XXXX (X XX wide, N XX narrow):</p><p>WWNNN</p><p>XXXXX</p><p>WNNWN</p><p>WNNNW</p><p>XXXXX</p><p>XXXXX</p><p>NWNNW</p><p>XXXXX</p><p>XXXXX</p><p>NNNWW</p><p>Now, XXX's XXXXXX out how many ways XX can XXXX XXXXXXX one wide XXXXX. This one's pretty XXXXXXX. XXXXX we have XXXX XXXXXXXX spaces, XXXXX are XXXX XXXXXXXX XXXXXXXXXXXX. XXXXXXXXXXXXXX, C(X,X) = X. I don't XXXXX I need XX show this.</p><p>XXX, here XXXXX the XXXXXXXXXXX Counting XXXXXXXXX. Since XX have 10 ways XX arrange the XXXXXX XXX X ways to arrange XXX spaces, XXX total number of XXXX to arrange XXX of XXXX is 10 * 4 = XX. In XXXXX XXXXX, XXX each XX the X XXXXXXXXXXXX of XXX spaces, XXXXX are 10 possible arrangements for XXX strips.</p><p>XX, there are<strong><em>40 </em></XXXXXX>XXXXXXXX permutations XXX XXXX bar XXXX.</p>">
