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.

Let the act of putting on a sock be $1,2,3,4,5,6,7,8$, each corresponding to a leg. Let the act of putting on a shoe be $A,B,C,D,E,F,G,H$, each corresponding to a leg as well.

This problem is now bijective to counting the number of strings such that $A,B,C,D,E,F,G,H$ comes after $1,2,3,4,5,6,7, 8$ respectively.

We start with $1$ leg. Then there is only $1$ way: $1A$. Now we want to add the next leg. There are $3 \times 4$ ways to add $2$ and $B$ into the string, without any order. Why? Because there are $3$ ways to insert $2$ into $1A$, and $4$ ways to insert $B$ into each of the the resultant string.

But there are as many ways to insert $2$ and $B$ such that $2$ comes after $B$ as there are ways of insertion such that $B$ comes after $2$. So there are $\frac{1}{2} \times 3 \times 4$ ways such that $B$ comes after $2$, i.e. second shoe comes after second sock.

Generalize this to get a recursion: Let $a_n$ be the number of ways for $n$ legs. Then,

$$a_{n + 1} = \frac{1}{2}\times a_n \times (2n + 1)\times(2n + 2)$$

  • $\begingroup$ nice Yiyuan. nice solution. but is my process right? $\endgroup$ – krishan acton Apr 5 '14 at 14:27
  • $\begingroup$ @krishan I am unable to follow through your solution, it seems a bit too complicated to me... $\endgroup$ – Yiyuan Lee Apr 5 '14 at 14:42
  • $\begingroup$ Also see: artofproblemsolving.com/Wiki/index.php/2001_AMC_12_Problems/… $\endgroup$ – gar Apr 6 '14 at 14:45
  • $\begingroup$ am i right or not? answer me please.yes or no . $\endgroup$ – krishan acton Apr 9 '14 at 6:50

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