Code 39 barcodes are used to code simple text. The barcodes are consisting of wi

Question

Code 39 barcodes are used to code simple text. The barcodes are consisting of wide or narrow black strips separated by wide or narrow white spaces. Each Code 39 element contains 9 elements (5 bars and 4 spaces). a) How many different combinations can you have for this barcode scheme? Is it enough to encode the 255 character ASCII sequence? b) Because some barcode readers have low performance, it is suggested that the first and last strips are always wide. Now how many different outputs (codes) can you have for this scheme? Is it enough to encode the 128 character printable ASCII sequence? c) Some barcode readers still have problem with this scheme. So it is proposed to have exactly 2 wide black strips and 1 wide whitespace in each character. However, the places for the wide strips and the wide space can vary. Now, how many different permutations can you have in this code?

Explanation / Answer

Let S denotes the strip and E denotes the spaces. Then, the codes will be of the form of:

S E S E S E S E S

a) For each strips, we have 2 options and similarly for each spaces, we have 2 options. So,

Number of different combinations possible = 25 * 24 = 29 = 512

So it would be enough to encode the 255 character ASCII sequence.

b) If we keep first and last strip wide then for remaining 3 strip and 4 spaces, we will have 2 options. So,

Number of different codes possible = 23 * 24 = 128

So it would be enough to encode 128 character printable ASCII sequence.

c) We can choose two places out of 5 for wide strips and then one place out of 4 for wide whitespace. For rest of 3 strip places, there would be narrow strip and for rest of the 3 space places, there would be narrow space. Hence,

Number of different permutations = 5C2 * 4C1 = 10*4 = 40

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