There are 5 types of toys, with 2 identical toys of each type. In how many…

When arranging n objects that include groups of identical items, the count is not simply n! — swapping two identical items within a group produces the exact same physical arrangement, so those swaps must not be counted as distinct. The general rule for such multiset arrangements is: total arrangements = n! divided by the product of the factorial of each identical group's size (n1! · n2! · ... · nk!).

1. There are 5 toy types with 2 identical toys in each type, so the total number of toys (positions on the shelf) is 5 × 2 = 10.

2. Each of the 5 types forms one identical group of size 2, so the raw count of 10! must be divided by 2! once for every one of the 5 groups: total arrangements = 10! / (2!)5.

3. Evaluate the divisor: 2! = 2, and 2 raised to the 5 power (one factor of 2! per group, 5 groups) equals 32.

4. Substituting back, the number of distinguishable arrangements is 10!/32.

Cross-check with a smaller analogous case: 2 toy types with 2 identical toys each (4 toys total) gives 4!/(2!)2 = 24/4 = 6 distinguishable arrangements, which matches direct enumeration of the 6 distinct orderings of two identical pairs. Scaling the same reasoning up to 5 types of 2 confirms 10!/(2!)5 = 10!/32 is the correct count for this problem.