Combinatorics question: Identical/nonidentical

[fignewtons]

Homework Statement

and attempt at a solution[/B]
If 5 gifts are to be given among 8 children:

a) if the gifts are identical (indistinguishable) and no child can receive more than 1 gift, there are 8P5 ways

b) if the gifts are non-identical (distinguishable) there are 5!(8P5) ways

In a), the only order is that of the way the children receiving it, while in b), the distinguishability of the objects adds another order.

Can someone tell me if my reasoning and calculations are correct?

 

[Ray Vickson]

Homework Statement

and attempt at a solution[/B]
If 5 gifts are to be given among 8 children:

a) if the gifts are identical (indistinguishable) and no child can receive more than 1 gift, there are 8P5 ways

b) if the gifts are non-identical (distinguishable) there are 5!(8P5) ways

In a), the only order is that of the way the children receiving it, while in b), the distinguishability of the objects adds another order.

Can someone tell me if my reasoning and calculations are correct?

Impossible to say, since you did not present your reasoning processes. In particular, how do you take account of the difference between scenarios (a) and (b)? (Here I mean with words of explanation, not just formulas.)

 

[fignewtons]

Impossible to say, since you did not present your reasoning processes. In particular, how do you take account of the difference between scenarios (a) and (b)? (Here I mean with words of explanation, not just formulas.)

Ok for a) it is 8 x 7 x 6 x 5 x 4. The 5 gifts are like placeholders each with 8 possibilities of children to be awarded to.
For b) aside from a) another order is introduced. For each of the 5 kids who were given gifts, there are 5 ways to order the gifts since they are not identical. So 5! in addition to the 8 x 7 x 6 x 5 x 4.

Let me know if it makes sense or not?

 

[haruspex]

Ok for a) it is 8 x 7 x 6 x 5 x 4. The 5 gifts are like placeholders each with 8 possibilities of children to be awarded to.

A useful check with such problems is to consider the smallest non-trivial case. Try 2 gifts and 3 children. You should be able to list the possibilities to check your answer.
You have correctly stated that it is a matter of which 5 children get the gifts (at least, I think that's what you mean), but that does not lead to 8P5.