a spider has one sock and one shoe for each of its 8 legs.in how many different orders can the spider put on its shocks and shoes; assuming that on each leg ;the shock must be put on before the shoe? i have tried the problem in following processes-
- we consider in k th step it has completed covering exactly its k legs with socks.now we find the total possibilities in (k+1) th step.
- in (k+1) th step either one of the remaining (8-k) legs will be covered by shocks or one of the (k-r) legs will be covered by shoe. where r denote the number of legs among the k legs which were already covered by shoes. so 0$\leq$r$\leq$k and 1$\leq$k$\leq$8
- so in the 1st case there are (8-k) possibilities and in the 2nd case ..........if 1 leg is covered with shoe then (k-1) possibilities; if 2 legs are covered with shoes then (k-2) possibilities; if no leg then k possibilities. so total possibilities is k.(k-1).(k-2)(k-3).........1=k!
- in (k+1)th step the total possibilities is (8-k).k!.[in each step it puts on either a shock or a shoe]
- so the solution is $\Sigma$(8-k).k! ;k=1 to 16
- if i am wrong in any step then show me please. if i am not then give an alternate solution of this.