Question

Number of arrangements of the set \(\{a_1,a_2, \ldots, a_{10}\}\), so that \(a_1\) is always ranked above \(a_2\).

Answer 1

@satellite73 @eliassaab

Answer 2

@experimentX :D

Answer 3

dunno i hate counting
but if \(a_1\) is first then there are 9 choices for \(a_2\)
if \(a_1\) is second there are 8,
if \(a_1\) is third there are 7, so i guess we can use that

Answer 4

without thinking too hard i get an answer that could easily be wrong, namely
\[9+8+7+6+5+4+3+2\]

Answer 5

oops +1 at the end because \(a_1\) could be 9th leaving one possibility for \(a_2\)

Answer 6

oh crap that is all wrong

Answer 7

but idea is probably right.
if \(a_1\) is first there are 9! arrangements

Answer 8

I think it could be 1(9+8+7+6+5+4+3+2+1)8!

Answer 9

yeah

Answer 10

if \(a_1\) is second there are 8 times 8! arrangements