<p>hey, I really need help with this problem. Plz tell me how to solve it
A spider has one sock and one shoe for each of its 8 legs. Assuming that the sock for each leg must be put on before the shoe, in how many different orders can the spider put on its socks and shoes?
(also, assume that only 1 shoe/sock goes with one leg--you can't mix and match shoes or socks)</p>
<p>haha nice, i got 8 factorial times 8 factorial. but then again, i am not sure at all.</p>
<p>8*8=64.......</p>
<p>I am not really sure</p>
<p>i'll go with mkevb1 on this one 8! x 8!</p>
<p>they are 2 independent events, so to determine the total combination you find the number of each event and multiply them</p>
<p>for socks: first leg you have 8 choices, second leg you'll have 7.. so on so its 8!
same applies to shoes</p>
<p>for each sock combination there are 8! shoe combination avilbable, so for 8! sock combination there are 8! x 8! combinations.</p>
<p>(unless I read the question wrong?)</p>