How many ways are there to distribute 15 objects into 5 boxes so that the boxes have 2,2,3,1,7 objects in them? (a) If both the objects and boxes are distinguishable (b) If objects are distinguishable but boxes are identical
Shouldn't the answer to (a) be 5! and to (b) be 5!/2? Since in case (a), the boxes are distinguishable, hence the two boxes containing 2 objects each will contain different objects and hence both the cases will be different. Whereas in case (b), the boxes are identical, hence both the cases will be the same.
(a)
If it is to be in the particular order $2,2,3,1,7,$ then
$$\frac{15!}{2!2!3!1!7!}$$
If it is to be in all possible orders, multiply the above by $$\frac{5!}{2!1!1!1!}$$
(b)
Any one order will be the same, say $2,2,3,1,7$
But since you further can't distinguish boxes with identical numbers of balls,, $$\frac{15!}{2!2!3!1!7!}\div 2!$$