Distributing gifts among children, any number of gifts per child possible.

[YusifNurizade]

Hello all,

My name is Yusif Nurizade; long time lurker, first time poster.

I am taking a summer class which is going at a rather quick pace and I could really use some help with the following problem:

A total of 11 different gifts are to be distributed among 15 children. How many different ways are there to distribute the gifts if (a) A child can receive any number of gifts. (b) each child can receive at most one gift.

I've confirmed the answer for part b to be 15C11 * 11!

The answer I came up with for part a is 15C11 * 12^11. I initially suspected a to have a greater number of choices than b because it encompasses the latter. My reasoning is that either child can have any number of gifts, including none, ergo 12 possible gifts scenarios per 11 children. I am, however, unsure of my answer and my mind keeps wandering asking whether I've considered some of the children getting less because one of them got a certain number.

I would really appreciate any help you could provide, particularly how I could arrive at the answer rather than the answer itself because it is really much more important that I understand the method.

Looking forward to your replies on what I am sure is the first of many posts.
Yusif Nurizade

 

[Dick]

Welcome to the forums, Yusif! Part (a) is probably the easier of the two. Take the gifts one by one. That gift can go to any child. How many ways can that happen?

 

[chiro]

Hey YusifNurizade and welcome to the forums.

For the first part of the question, are you familiar with the multinomial distribution?

http://en.wikipedia.org/wiki/Multinomial_distribution

 

[YusifNurizade]

Thank you for the hospitality and the advice fellas.

I've reread your suggestions several times over and I'm afraid I am still in the dark.

Dick: One by one the gifts can go to any of 11 children. But then you have to pick which child gets a gift. Does this mean that I was right or only partially so?

Chiro: I'm afraid I don't see the connection. Will try rereading after I've cleared my head.

 

[HallsofIvy]

(a)
Choose any child to give the first gift to. There are 15 ways to do that.
Choose any child to give the second gift to. There are still 15 ways to do that.
Choose any child to give the third gift to. There are still 15 ways to do that.
.
.
.
Choose any child to give the eleventh gift to. There are still 15 ways to do that.
"Fundamental principle of counting".

(b)
Choose any child to give the first gift to. There are 15 ways to do that.
Choose a child to give the second gift to. There are 14 ways to do that.
Choose a child to give the third gift to. There are 13 ways to do that.
.
.
.
Choose a child to give the eleventh gift to. There are 15- 10= 5 ways to do that.
"Fundamental theorem of counting".

 

[YusifNurizade]

Thank you, that cleared it up!

We went over it in class today and it was basically the same explanation.

Glad to have a place I can turn to, this class is pretty hard to begin with and going really fast because it's summer.

Thanks again for the help!
Yusif Nurizade

 

[HallsofIvy]

Notice, by the way, that the answer to the last problem is $(15)(14)(13)\cdot\cdot\cdot(6)(5)$ which can be written as
$$\frac{(15)(14)(13)\cdot\cdot\cdot(6)(5)(4)(3)(2)(1)}{(4)(3)(2)(1)}= \frac{15!}{4!}$$
$$_{15}C_{11}= \begin{pmatrix}15 \\ 1\end{pmatrix}= \frac{15!}{11!4!}$$
so that
$$_{15}C_{11}(11!)= \frac{15!}{4!}$$
as before.

 

[LCKurtz]

Another way to look at (b): There are $C(15,11)$ ways to pick the 11 children to get the gifts. Then there are $11!$ ways to permute the gifts among them, giving the same answer $C(15,11)\cdot 11!$.

 

[YusifNurizade]

Thank you all for the ongoing support, it has made me feel very welcome on the forum!

Yusif Nurizade