XX XX the XXXXX strip has X XXXXXXXXXXXXX (wide or narrow), and the first space has X XXXXXXXXXXX XXXXXXXXXXXXX (wide or XXXXXX), there's a XXXXX XX 2*X=4 XXXXXXXXXXXXX XXX those two elements.
XXX next strip XXX another two possibilities, so 2*2*X=X; the XXXX XXXXX XXX XXXXXXX XXX possibilities, so 2*2*2*X=XX.
I think you XXX XXX XXXX. Keep doing this for all nine XXXXXXXX, XXXX of XXXXX XXX X XXXXXXXXXXXXX, and you'll XXX 2*X*2*X*X*2*X*X*X= 2^9 = XXX.
XX XXXXX are XXX XXXXXXXX combinations XXX XXXX bar XXXX, XXXXX XX XXXX than XXX 255 XXXXXXXXX.
b) This XXXXXXXX XX almost the same, XXXXXX XXX, for XXX XX XXX XXXXXXXX, the XXXXX XXX XXXX XXXXXX, there is only one possibility (wide). XX, just XX the XXXX thing you did last XXXX using XXX Fundamental XXXXXXXX XXXXXXXXX, XXX remember XXXX XXX XXXXX XXX XXXX XXXXXX XXXX XXXX XXX XXXXXXXXXXX:
X*X*2*2*2*2*X*X*1 = X^7 = XXX
XX there XXX now XXXX128 XXXXXXXX combinations for XXX XXX XXXX, but that is XXXX XXXXXX XXX XXX XXX XXXXXXXXX sequence they'XX asking for.
c) XX now we need to XXXX out how many XX XXX XXXXXXXXXXX two XXXX strips XXX one wide space.
First, how many ways can we have XX exactly two wide strips? We XXXX five XXXXX strips XX XXXXXX from, XXX XX need XX XXXXXX two, so 5 XXXXXX 2 or X(5,2) = 5! / (X! * (5-2)!) = 10. So XXXXX are XXXXXX to XXXXXXX the XXXXXX XXXX that XXXXXXX two XXX wide (X is XXXX, N XX narrow):
XXXXX
WNNNW
NWWNN
NWNWN
NWNNW
NNWWN
NNWNW
XXX, XXX's XXXXXX out how XXXX XXXX we XXX XXXX exactly one wide XXXXX. XXXX one's XXXXXX XXXXXXX. XXXXX XX XXXX four XXXXXXXX spaces, there are four XXXXXXXX arrangements. XXXXXXXXXXXXXX, X(4,1) = X. I XXX't XXXXX I XXXX to show this.
Now, XXXX comes the XXXXXXXXXXX Counting XXXXXXXXX. Since we have 10 ways XX arrange the XXXXXX XXX X XXXX to XXXXXXX XXX XXXXXX, XXX total number of ways XX XXXXXXX all of them is XX * X = 40. XX XXXXX words, for each of XXX 4 arrangements XX the spaces, XXXXX XXX XX XXXXXXXX XXXXXXXXXXXX for the XXXXXX.
So, XXXXX are40 possible XXXXXXXXXXXX XXX this bar code.