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Code 39 barcodes are used to code simple text. T
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XXXXXXXXXXX XXXXXXXX Principle</strong>.</p><p>XX XX XXX XXXXX strip has 2 XXXXXXXXXXXXX (XXXX or XXXXXX), and XXX first XXXXX XXX X XXXXXXXXXXX XXXXXXXXXXXXX (wide or narrow), there's a XXXXX XX 2*2=4 XXXXXXXXXXXXX XXX XXXXX XXX XXXXXXXX.</p><p>The XXXX XXXXX has XXXXXXX XXX XXXXXXXXXXXXX, so 2*2*2=8; XXX XXXX XXXXX XXX XXXXXXX XXX XXXXXXXXXXXXX, so X*X*2*X=XX.</p><p>I think you XXX the XXXX. XXXX doing XXXX XXX all XXXX elements, each XX which has X XXXXXXXXXXXXX, XXX you'll XXX 2*2*X*2*2*X*X*2*X= X^9 = XXX.</p><p>So XXXXX XXX <em><strong>XXX</strong></em> XXXXXXXX XXXXXXXXXXXX for this XXX code, which is more than XXX 255 XXXXXXXXX.</p><p>b) XXXX question XX almost XXX XXXX, except XXX, for two XX the XXXXXXXX, XXX XXXXX and last XXXXXX, there XX XXXX one XXXXXXXXXXX (wide). So, just do XXX same thing you XXX XXXX time using XXX XXXXXXXXXXX XXXXXXXX Principle, but remember XXXX XXX first and XXXX XXXXXX have XXXX XXX XXXXXXXXXXX:</p><p>X*X*2*X*2*X*2*2*X = 2^X = 128</p><p>So XXXXX are now only<strong><em>XXX<XXXXXX><em> <XXXXXX></XXXXXX></em></XXXXXX></em></XXXXXX>XXXXXXXX combinations for the bar code, XXX XXXX is just enough XXX XXX 128 character XXXXXXXX they're XXXXXX for.</p><p>c) So XXX XX need to XXXX out how XXXX we can XXXX<XXXXXX>exactly</strong> two wide XXXXXX and XXX wide XXXXX.</p><p>First, how many XXXX XXX XX have on XXXXXXX two XXXX XXXXXX? We XXXX XXXX XXXXX XXXXXX to choose XXXX, XXX XX XXXX XX XXXXXX XXX, so 5 choose X or C(5,2) = X! / (2! * (X-2)!) = 10. XX XXXXX XXX 10ways to XXXXXXX the strips XXXX that XXXXXXX XXX are XXXX (W is XXXX, N is narrow):</p><p>WWNNN</p><p>XXXXX</p><p>WNNWN</p><p>WNNNW</p><p>NWWNN</p><p>NWNWN</p><p>NWNNW</p><p>XXXXX</p><p>NNWNW</p><p>XXXXX</p><p>Now, let's XXXXXX out how many ways we can have XXXXXXX XXX XXXX space. This one's XXXXXX obvious. Since XX XXXX XXXX XXXXXXXX XXXXXX, XXXXX are XXXX XXXXXXXX arrangements. XXXXXXXXXXXXXX, X(X,X) = X. I XXX't think I need to show this.</p><p>Now, here comes XXX Fundamental XXXXXXXX Principle. XXXXX XX XXXX XX XXXX XX arrange XXX strips and 4 XXXX to arrange the XXXXXX, the XXXXX XXXXXX XX XXXX to XXXXXXX all XX them is 10 * X = 40. In XXXXX XXXXX, XXX each XX XXX 4 arrangements of the spaces, XXXXX are 10 possible arrangements XXX the strips.</p><p>So, XXXXX are<XXXXXX><em>XX </em></strong>possible XXXXXXXXXXXX for XXXX XXX code.</p>">
