
#1
Dec1505, 06:11 PM

P: 49

The question is:
A sequence of length 6 is formed from the digits {0,1,2...9}. If no repetition is allowed, how many of these sequences can be formed if: f) the sum of the first two terms is 7? So i set up my place holders: _ _ _ _ _ _ if the first 2 place holders have a sum of 7, are there 8 possibilities for each place holder because 0+1, 1+6, 2+5, 3+4, 4+3, 5+2, 6+1, 7+0? Or would there be 4 for each consisting only of 0+1, 1+6, 2+5, 3+4? So really teh answer would be 16 x 8P4 or 64 x 8P4? The answer was 13440, but I don't see why. 



#2
Dec1505, 07:26 PM

Mentor
P: 16,485

You are right on the right track. There are 8 possibilities for the first digit (07). Once that is picked the second digit is determined. That leaves 8 possible digits for the third, 7 possible for the fourth, 6 possible for the fifth, and 5 possible for the sixth.
8*1*8*7*6*5 8*8P4 13440 Dale 



#3
Dec1705, 04:37 PM

P: 49

I see, ok.
Thanks! 


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