Permutation Question.

forevergone

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.

DaleSpam

You are right on the right track. There are 8 possibilities for the first digit (0-7). 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

forevergone

I see, ok.

Thanks!