Question

A combination question. Suppose I have 16 different gifts and I want to equally distribute them to 4 kids, In how many ways I can do that?
Please only give a hint. Don't give the answer :)
Thanks

Answer 1

You misunderstood the question, in how many ways I can do.
I'll explain I have gift A, B, C, D, E, F, G and so on
1st kid can get ABCD or, ADEC or ACFG and so on
and all other kids will have different variants.
I want to know, how many ways I can do that?

Answer 2

its like, given alphabets of size 16, and asked to find how many 4 letter words we can form w/o repitition

Answer 3

We have 16 children, and we want to have an equal number of gifts given to each of them.
4 each...

From the sounds of it, I would say that this is a permutations problem.

\(\displaystyle \frac{n!}{(n-r)!}\times\frac{1}{r!}=\frac{n!}{r!(n-r)!}\)

I believe this is what we would use.

Answer 4

@ganeshie8 How would I factor in the 4 kids?
I can find no. of words without repetitions

Answer 5

16c4

Answer 6

16c4 is not the answer

Answer 7

16c4*12c4*8c4*4c4

Answer 8

@madrockz that would give a huge answer

Answer 9

that is correct...though you could also use the multinomial coefficient

Answer 10

Where \(n\) is the number of things to choose from, and you choose \(r\) of them. Forgot an important bit, my apologies.

Answer 11

\[\frac{16!}{4!\cdot 4!\cdot 4!\cdot 4!}\]

Answer 12

@suyash011 Any comments?

Answer 13

I may have made a few typos.... 16 gifts distributed to 4 children.

So, how many ways can you distribute 4 gifts out of 16?

I believe my formula should work for this. I am going to mosey on now. Don't want to confuse.

Answer 14

@Zarkon Care to explain?

Answer 15

@austinL Why it would be a permutation problem?

Answer 16

I made an error. I believe it is referred to as a combination without repetition.

Answer 17

http://en.wikipedia.org/wiki/Multinomial_theorem#Interpretations

Answer 18

I may, and probably am incorrect for this problem. But I would apply the Binomial Coefficient.

Answer 19

I liked the Zarkon's interpretation :

**** **** **** ****
| | | |


above, * are gifts, | are kids.

the *'s by themselves can be picked in 16! ways.
but the place of * at a particular kid doesn't matter. so divide those combinations : 4! for each kid

Answer 20

\(\displaystyle \frac{n!}{(n-r)!}\times\frac{1}{r!}=\frac{n!}{r!(n-r)!}\)

Where \(n\) is the number of things to choose from, and you choose \(r\) of them.

This works because order doesn't matter, and there is NO repetition.

Here is an example of a
1) Permutation without repetition,

For example, what order could 16 pool balls be in?
After choosing, say, number "14" you can't choose it again.

So, your first choice would have 16 possibilites, and your next choice would then have 15 possibilities, then 14, 13, etc. And the total permutations would be:

\(16 × 15 × 14 × 13 × ... = 20,922,789,888,000\)

But maybe you don't want to choose them all, just 3 of them, so that would be only:

\(16 × 15 × 14 = 3,360\)

In other words, there are 3,360 different ways that 3 pool balls could be selected out of 16 balls.



2) Combinations without Repetition

The easiest way to explain it is to:

-Assume that the order does matter (i.e. permutations)
-Then alter it so the order does not matter.

Going back to the pool ball example, let us say that you just want to know which 3 pool balls were chosen, not the order.

We already know that 3 out of 16 gave us 3,360 permutations.

But many of those will be the same to us now, because we don't care what order!

For example, let us say balls 1, 2 and 3 were chosen. These are the possibilites:
Order does matter:
1 2 3
1 3 2
2 1 3
2 3 1
3 1 2
3 2 1

Doesn't matter:
123

So, the permutations will have 6 times as many possibilites.

In fact there is an easy way to work out how many ways "1 2 3" could be placed in order, and we have already talked about it. The answer is:

\(3! = 3 × 2 × 1 = 6\)

So, all we need to do is adjust our permutations formula to reduce it by how many ways the objects could be in order (because we aren't interested in the order any more):

\(\displaystyle \frac{n!}{(n-r)!}\times\frac{1}{r!}=\frac{n!}{r!(n-r)!}\)

It is often found in calculators as \(_nC_r\)

I sincerely hope that I didn't type for nothing.......

Answer 21

@austinL Yes, it's good. But I wanted explanation pertaining to this question.
I'm still trying to take in @ganeshie8 & @Zarkon 's explanation

Answer 22

@ash2326
I wish you the best of luck, my brain hurts trying to remember high school like this.
Hopefully they respond soon :)

Answer 23

\[16!/(15!)+16!/(14!*2!)+16!/(13!*3!)+16!/(12!*4!)+16!/(11!*5!)+16!/(10!*5!)+16!/(9!*7!)+16!/(8!*8!)\]

Answer 24

@suyash011 Please explain how you got this?

Answer 25

You have 16 distinct gifts, so line them up in a row.
Next, assign a colour to each kid, say, red, yellow, blue, green, whatever.

Now, take four small markers of each colour and assign one to each gift.
The colour the gift is marked with corresponds to the kid who gets the present...

Answer 26

@myininaya please Help here if you have sometime :P

Answer 27

Well, the partitioning methods (a consequence of the multinomial) states that the number of ways of diving n distinct groups into r groups of soze x1,x2.....xr each is giving by: