Question

A horse has one sock and one shoe in each of its four legs. The number of different orders in which the horse can put all its socks and shoes is (assuring that on each leg the sock must be put on before the shoe)
A. 2520
B. 630
C. 576
D. 1152

Answer 1

We are given with a condition that there are four socks and shoes that a horse has to put on its legs. Each leg can be put with the combination of sock and shoe only the restriction is sock should be put before the shoe. So the ratio of number of possible conditions to the product of possible cases for the four legs will be the correct answer.

Complete step by step solution:
Given that there are 4 legs. So the numbers of socks and shoes will be 4 each. So the possible number of combinations will be $8!$. Now each leg will have a pair such that one sock and one shoe. So for four legs the combination will be,
$2 \times 2 \times 2 \times 2$
Thus the ratio will be,
$\dfrac{{8!}}{{2 \times 2 \times 2 \times 2}}$
That is
$\dfrac{{8!}}{{{2^4}}}$
Now putting the values we get,
$\dfrac{{40320}}{{16}}$
On dividing we get,
$2520$

Hence,option A is correct.

Note: the condition if not noticed then the number of ways will be $8!$ directly.Also note that each leg will have a combination of sock and shoe so we divided the possible outcomes by ${2^4}$ for four legs.If the number of legs increase or decrease like a spider has 8 legs and same condition is applied then the answer will be, $\dfrac{{16!}}{{{2^8}}}$