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Code 39 barcodes are used to code simple text. T
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XXXXXXXXXXX Counting Principle</XXXXXX>.</p><p>So if XXX XXXXX XXXXX has X XXXXXXXXXXXXX (XXXX or narrow), XXX the first space has 2 independent possibilities (wide or XXXXXX), XXXXX's a XXXXX XX 2*2=X XXXXXXXXXXXXX for those XXX elements.</p><p>XXX XXXX strip has XXXXXXX two possibilities, so X*X*2=8; the XXXX XXXXX XXX another two possibilities, so 2*2*X*X=16.</p><p>I think you get the XXXX. XXXX doing XXXX for XXX nine XXXXXXXX, XXXX of which has X possibilities, and you'll get X*X*2*X*2*2*2*2*2= 2^9 = XXX.</p><p>So XXXXX XXX <em><XXXXXX>XXX</strong></em> XXXXXXXX combinations XXX XXXX bar code, which XX XXXX than XXX XXX XXXXXXXXX.</p><p>b) XXXX question is almost XXX same, except now, XXX two XX the elements, the XXXXX and XXXX XXXXXX, there is only XXX XXXXXXXXXXX (wide). So, just do XXX same thing you XXX last XXXX using XXX Fundamental Counting Principle, but remember that XXX XXXXX XXX XXXX XXXXXX have XXXX one XXXXXXXXXXX:</p><p>X*2*X*2*2*X*2*2*X = 2^X = 128</p><p>XX XXXXX XXX now only<XXXXXX><em>128<XXXXXX><em> <XXXXXX></XXXXXX></em></strong></em></XXXXXX>XXXXXXXX combinations for XXX XXX XXXX, XXX that XX just XXXXXX XXX the XXX XXXXXXXXX XXXXXXXX they'XX asking for.</p><p>c) So XXX XX need XX find out how XXXX XX XXX have<XXXXXX>exactly</strong> XXX XXXX XXXXXX and XXX wide XXXXX.</p><p>XXXXX, how XXXX XXXX can we XXXX on XXXXXXX XXX XXXX XXXXXX? XX XXXX five XXXXX XXXXXX XX XXXXXX XXXX, and we need to XXXXXX XXX, so X choose 2 or X(5,X) = 5! / (X! * (X-X)!) = 10. So XXXXX XXX XXXXXX to arrange the XXXXXX such XXXX exactly two XXX XXXX (X is wide, N is narrow):</p><p>WWNNN</p><p>XXXXX</p><p>XXXXX</p><p>WNNNW</p><p>NWWNN</p><p>NWNWN</p><p>XXXXX</p><p>NNWWN</p><p>NNWNW</p><p>XXXXX</p><p>XXX, XXX's figure out how many ways we XXX have XXXXXXX one XXXX XXXXX. This one's pretty XXXXXXX. Since we XXXX four possible XXXXXX, there are XXXX possible arrangements. Mathematically, C(X,X) = X. I XXX't think I XXXX XX show this.</p><p>XXX, XXXX comes the Fundamental XXXXXXXX Principle. Since XX have XX ways XX arrange the strips XXX 4 XXXX XX arrange XXX spaces, the XXXXX number of XXXX XX XXXXXXX XXX of XXXX is XX * X = XX. In other words, for each XX the X arrangements of XXX spaces, there are 10 possible XXXXXXXXXXXX XXX XXX strips.</p><p>XX, XXXXX XXX<strong><em>40 </em></strong>XXXXXXXX permutations for XXXX bar code.</p>">
